Chapter 9

371

CHAPTER 9 ANSWERS

Exercises 9.1

9.1 A hypothesis is a statement that something is true.

9.2 The decision criterion specifies whether or not the null hypothesis shouldbe rejected in favor of the alternative hypothesis.

9.3 (a) The population mean is equal to some fixed amount; i.e., = 0 .

(b) The population mean is greater than μ0; i.e., Ha: > 0 .

The population mean is less than 0 ; i.e., Ha: < 0.

The population mean is unequal to 0; i.e., Ha: 0 .

9.4 (a) Ha: 0 ; two-tailed

(b) Ha: < 0 ; left-tailed

(c) Ha: > 0 ; right-tailed

9.5 Let denote the mean cadmium level in Boletus pinicola mushrooms.

(a) H0: = 0.5 ppm (b) Ha: > 0.5 ppm (c) right-tailed test

9.6 Let denote the mean retail price of agriculture books.

(a) H0: = $66.52 (b) Ha: $66.52 (c) two-tailed

9.7 Let denote the mean daily intake of iron by adult females under age 51.

(a) H0: = 18 mg/day (b) Ha: < 18 mg/day (c) left-tailed

9.8 Let denote the mean age of early-onset dementia.

(a) H0: = 55 years old (b) Ha: < 55 years old

(c) left-tailed

9.9 Let denote the mean length of imprisonment for motor-vehicle theftoffenders in Australia.

(a) H0: = 16.7 months (b) Ha: 26.7 months

(c) two-tailed

9.10 Let denote the mean post-work heart rate of casting workers.

(a) H0: = 72 beats/min (b) Ha: > 72 beats/min

(c) right-tailed

9.11 Let denote the mean body temperature of healthy humans.

(a) H0: = 98.6ºF (b) Ha: 98.6ºF (c) two-tailed

9.12 Let denote the mean annual salary of classroom teachers in Hawaii.

(a) H0: = $45.9 thousand (b) Ha: < $45.9 thousand

(c) left-tailed

9.13 Let denote the mean local monthly bill for cell phone users in the U.S.

(a) H0: = $47.37 (b) Ha: > $47.37 (c) right-tailed

9.14 Let denote the mean annual energy consumed per U.S. household.

372 Chapter 9, Hypothesis Tests for One Population Mean

(a) H0: = 92.2 million BTU (mean western household energy consumption isthe same as all American households)Ha: 92.2 million BTU (mean western household energy consumptiondiffers from that of all American households)

(b) If the sample mean energy consumption x_differs by too much from 92.2

million BTU, then we should be inclined to reject H0 and conclude that

Ha is true. From the data, we compute x_

== 79.65 million BTU. Thequestion is whether the difference of 12.55 million BTU between thesample mean of 79.65 million BTU and the hypothesized population meanof 92.2 million BTU can be attributed to sampling error or whether thedifference is large enough to indicate that the population mean is not92.2 million BTU.

(c) The sampling distribution of x_

will be a normal distribution.

(d) It is quite unlikely that the sample mean x_

will be more than twostandard deviations away from the population mean . If x

_is more

than two standard deviations away from , then reject H0 and concludethat Ha is true. Otherwise, do not reject H0.

(e) We have = 15, n = 20, x_

= 79.65, and = 92.2 under H0 true. Thus,

(79.65 92.2)/(15/ 20) 3.74z . Since x_is more than two standard

deviations away from 92.2 million BTU, we reject H0 and conclude that Hais true.

9.15 (a) H0: = 5.6 radios per U.S. household

Ha: 5.6 radios per U.S. household

(b) If the sample mean number of radios per U.S. household x_

differs by toomuch from 5.6 radios, then we should be inclined to reject H0 and

conclude that Ha is true. From the data, we compute x_

== 5.89 radios.The question is whether the difference of 0.29 radios between thesample mean of 5.89 and the hypothesized population mean of 5.6 can beattributed to sampling error or whether the difference is large enoughto indicate that the population mean is not 5.6 radios.

(c) The sampling distribution of x_

will be approximately a normaldistribution.

(d) It is quite unlikely that the sample mean x_

will be more than twostandard deviations away from the population mean . If x

_is more

than two standard deviations away from , then reject H0 and concludethat Ha is true. Otherwise, do not reject H0.

(e) We have = 1.9, n = 45, x_= 5.89, and = 5.6 under H0 true. Thus,

(5.89 5.6)/(1.9/ 45) 1.02z . Since x_

is less than two standarddeviations away from 5.6 radios, we do not reject H0 and conclude thatH0 is reasonable.

9.16 (a) If the mean weight x_

of the 50 bags of pretzels sampled is more thanone standard deviation away from 454 grams, then reject the nullhypothesis that = 454 grams and conclude that the alternativehypothesis, which is μ 454 grams, is true. Otherwise, do not rejectthe null hypothesis.

Graphically, the decision criterion looks like:

Section 9.1, The Nature of Hypothesis Testing 373

Do notReject H0 Reject H0 Reject H0

454-1 454+1

-1 0 1 z

0.1587 0.6826 0.1587

454

Reject H0 Do not Reject H0Reject H0

454-1 454+1

-1 0 1 z

(b)

The lower figure shows that, using our decision criterion, theprobability is 0.3174 (= 1 - 0.6826 = 0.1587 + 0.1587) of rejecting thenull hypothesis if it is in fact true.

(c) We have = 7.9, n = 25, x_= 450, and = 454 if H0 is true. Thus,

56.2)25/8.7/()454450(z . The sample mean x_

is 2.56 standard

deviations below the null hypothesis mean of 454 grams. Since the mean

weight x_of 25 bags of pretzels sampled is more than one standard

deviation away from 454 grams, we reject the null hypothesis that μ =454 grams and conclude that the alternative hypothesis, which is μ454 grams, is true. In other words, the data provide sufficientevidence to conclude that the packaging machine is not workingproperly.

9.17 (a) If the mean weight x_of the 25 bags of pretzels sampled is more than

three standard deviations away from 454 grams, then reject the nullhypothesis that = 454 grams and conclude that the alternative

hypothesis, which is μ 454 grams, is true. Otherwise, do not rejectthe null hypothesis.

Graphically, the decision criterion looks like:


-4 -3 -2 -1 0 1 2 3 4

0.0013 0.9974 0.0013

Reject Do not Reject H0 Reject

H0 H0

454 – 3 454 454 + 3 Mean

454

Reject H0 Do not Reject H0Reject H0

454-3 454+3

-3 0 3 z(b)

The lower figure shows that, using our decision criterion, theprobability is 0.0026 (= 1 - 0.9974 = 0.0013 + 0.0013) of rejecting thenull hypothesis if it is in fact true.

(c) We have = 7.9, n = 25, x_= 450, and = 454 if H0 is true. Thus,

56.2)25/8.7/()454450(z . The sample mean x_

is 2.56 standarddeviations below the null hypothesis mean of 454 grams. Since the mean

weight x_of 25 bags of pretzels sampled is less than three standard

deviations away from 454 grams, we do not reject the null hypothesisthat μ = 454 grams and conclude that the null hypothesis, which is μ =454 grams, is reasonable. In other words, the data do not providesufficient evidence to conclude that the packaging machine is notworking properly.

9.18 If the null hypothesis is true, the chance of incorrectly rejecting it is0.0456 when using the 95.44% part of the 68.26-95.44-99.74 rule.

Exercises 9.2

9.19 (a) This statement is true: If it is important not to reject a true nullhypothesis, i.e., not to make a Type I error, then the hypothesis testshould be performed at a small significance level. This can beappreciated by considering the meaning of the significance level. Thesignificance level is equal to the probability of making a Type Ierror. The smaller the significance level, the smaller the probabilityof rejecting a true null hypothesis.

(b) This statement is true: Decreasing the significance level results inan increase in the probability of making a Type II error. This can beappreciated by considering the relation between Type I and Type IIerror probabilities. For a fixed sample size, the smaller the Type Ierror probability (which is equal to the significance level), the

Section 9.2, Terms, Errors, and Hypotheses 375

larger the Type II error probability.

9.20 (a) A test statistic is a statistic used to decide whether to reject thenull hypothesis.

(b) The rejection region is a set of values of the test statistic that leadto rejection of the null hypothesis.

(c) The nonrejection region is a set of values of the test statistic thatdo not lead to rejection of the null hypothesis.

(d) Critical values are values of the test statistic that separate therejection region from the nonrejection region. They are considered tobe part of the rejection region.

(e) The significance level is the probability of rejecting a true nullhypothesis.

9.21 A Type I error is made when a true null hypothesis is rejected. Theprobability of making this error is denoted by . A Type II error is madewhen a false null hypothesis is not rejected. We denote the probability ofa Type II error by .

9.22 (a) If the null hypothesis is rejected, we conclude that the alternativehypothesis is true.

(b) If the null hypothesis is not rejected, we conclude that the data didnot provide sufficient evidence to conclude that the alternativehypothesis is true. We cannot say that the null hypothesis is true,only that it is reasonable.

9.23 (a) Rejection region: z 1.645

(b) Nonrejection region: z < 1.645

(c) Critical value: z = 1.645

(d) Significance level: = 0.05

(e)

(f) Right-tailed test

9.24 (a) Rejection region: z -1.96 or z 1.96

(b) Nonrejection region: -1.96 < z < 1.96

(c) Critical values: z = ±1.96


Do notReject H0 Reject H0

0 1.645 z

0.9500 0.0500

Critical ValueNonrejection region |Rejec tion region



-1.96 0 1.96 zCritical value Critical value

Rejection | Nonrejectio n | RejectionRegion | Region | Region

0.0250 0.9500 0.0250

(e)

(f) Two-tailed test

9.25 (a) Rejection region: z -2.33

(b) Nonrejection region: z > -2.33

(c) Critical value: z = -2.33


(e)

(f) Left-tailed test

9.26 (a) Rejection region: z -1.645

(b) Nonrejection region: z > -1.645

(c) Critical value: z = -1.645



-2.33 0 zCritical value

Rejection | Nonrejecti onRegion | Region

0.01 0.99


Do notReject H 0 Reject H 0

-1.645 0 zCritical value

Rejection | Nonrejecti onRegion | Region

0.05 0.95

(e)

(f) Left-tailed test

9.27 (a) Rejection region: z -1.645 or z -1.645

(b) Nonrejection region: -1.645 < z < -1.645

(c) Critical values: z = -1.645 and z = 1.645


(e)

(f) Two-tailed test

9.28 (a) Rejection region: z 1.28






Rejection | Nonrejectio n | RejectionRegion | Region | Region

0.0500 0.9000 0.0500


(e)

(f) Right tailed test

9.29 (a) A Type I error would occur if, in fact, = 0.5 ppm, but the resultsof the sampling lead to the conclusion that > 0.5 ppm.

(b) A Type II error would occur if, in fact, > 0.5 ppm, but the resultsof the sampling fail to lead to that conclusion.

(c) A correct decision would occur if, in fact, = 0.5 ppm and theresults of the sampling do not lead to the rejection of that fact; orif, in fact, > 0.5 ppm and the results of the sampling lead to thatconclusion.

(d) If, in fact, the mean cadmium level in Boletus pinicola mushrooms isequal to 0.5 ppm, and we do not reject the null hypothesis that =0.5 ppm, we made a correct decision.

(e) If, in fact, the mean cadmium level in Boletus pinicola mushrooms isgreater than to 0.5 ppm, and we do not reject the null hypothesis that

= 0.5 ppm, we made a Type II error.

9.30 (a) A Type I error would occur if, in fact, = $66.52, but the results of

the sampling lead to the conclusion that $66.52.

(b) A Type II error would occur if, in fact, $66.52, but the resultsof the sampling fail to lead to that conclusion.

(c) A correct decision would occur if, in fact, = $66.52 and the resultsof the sampling do not lead to the rejection of that fact; or if, infact, $66.52 and the results of the sampling lead to thatconclusion.

(d) If, in fact, the mean retail price of agricultural books is equal to$66.52, and we reject the null hypothesis that = $66.52, we made aType I error.

(e) If, in fact, the mean retail price of agricultural books is not $66.52,and we reject the null hypothesis that = $66.52, we made a correctdecision.

9.31 (a) A Type I error would occur if, in fact, = 18 mg, but the results of

Do notReject H0 Reject H 0

0 1.28 z

0.9000 0.1000

Critical ValueNonrejection region | Rejection region


the sampling lead to the conclusion that < 18 mg.

(b) A Type II error would occur if, in fact, < 18 mg, but the results ofthe sampling fail to lead to that conclusion.

(c) A correct decision would occur if, in fact, = 18 mg and the resultsof the sampling do not lead to the rejection of that fact; or if, infact, < 18 mg and the results of the sampling lead to thatconclusion.

(d) If the mean iron intake equals the RDA of 18 mg, and we reject the nullhypothesis that = 18 mg, we made a Type I error.

(e) If, in fact, the mean iron intake is less than the RDA of 18 mg, and wereject the null hypothesis that = 18 mg, we made a correct decision.

9.32 (a) A Type I error would occur if, in fact, = 55 years old, but theresults of the sampling lead to the conclusion that < 55 years old.

(b) A Type II error would occur if, in fact, < 55 years old, but theresults of the sampling fail to lead to that conclusion.

(c) A correct decision would occur if, in fact, = 55 years old and theresults of the sampling do not lead to the rejection of that fact; orif, in fact, < 55 years old and the results of the sampling lead tothat conclusion.

(d) If the mean age of all people with early-onset dementia is 55 years,and we do not reject the null hypothesis that = 55 years old, wemade a correct decision.

(e) If, in fact, the mean age of all people with early-onset dementia isless than 55 years, and we do not reject the null hypothesis that =55 years old, we made a Type II error.

9.33 (a) A Type I error would occur if, in fact, = 16.7 months, but the

results of the sampling lead to the conclusion that μ 16.7 months.

(b) A Type II error would occur if, in fact, 16.7 months, but theresults of the sampling fail to lead to that conclusion.

(c) A correct decision would occur if, in fact, = 16.7 months and theresults of the sampling do not lead to the rejection of that fact; orif, in fact, 16.7 months and the results of the sampling lead tothat conclusion.

(d) If, in fact, the mean length of imprisonment equals 16.7 months, and wedo not reject the null hypothesis that = 16.7 months, we made acorrect decision.

(e) If, in fact, the mean length of imprisonment does not equal 16.7months, and we do not reject the null hypothesis that = 16.7 months,we made a Type II error.

9.34 (a) A Type I error would occur if, in fact, = 72 bpm, but the results ofthe sampling lead to the conclusion that > 72 bpm.

(b) A Type II error would occur if, in fact, > 72 bpm, but the resultsof the sampling fail to lead to that conclusion.

(c) A correct decision would occur if, in fact, = 72 bpm and the resultsof the sampling do not lead to the rejection of that fact; or if, infact, > 72 bpm and the results of the sampling lead to that


conclusion.

(d) If, in fact, the mean post-work heart rate of casting workers equaledthe normal resting heart rate of 72 bpm, and we rejected the nullhypothesis that = 72 bpm, we have made a Type I error.

(e) If, in fact, the mean post-work heart rate of casting workers didexceed the normal resting heart rate of 72 bpm, and we rejected thenull hypothesis that = 72 bpm, we have made a correct decision.

9.35 (a) A Type I error would occur if, in fact, = 98.6º F, but the resultsof the sampling lead to the conclusion that 98.6º F.

(b) A Type II error would occur if, in fact, 98.6º F, but the resultsof the sampling fail to lead to that conclusion.

(c) A correct decision would occur if, in fact, = 98.6º F and theresults of the sampling do not lead to the rejection of that fact; orif, in fact, 98.6º F and the results of the sampling lead to thatconclusion.

(d) If the mean temperature of all healthy humans equals 98.6º F, and wereject the null hypothesis that = 98.6º F, we made a Type I error.

(e) If, in fact, the temperature of all healthy humans is not equal to98.6º F, and we reject the null hypothesis that = 98.6º F, we made acorrect decision.

9.36 (a) A Type I error would occur if, in fact, = $45.9 thousand, but theresults of the sampling lead to the conclusion that < $45.9thousand.

(b) A Type II error would occur if, in fact, < $45.9 thousand, but theresults of the sampling fail to lead to that conclusion.

(c) A correct decision would occur if, in fact, = $45.9 thousand and theresults of the sampling do not lead to the rejection of that fact; orif, in fact, < $45.9 thousand and the results of the sampling leadto that conclusion.

(d) If the mean annual salary of classroom teachers in Hawaii equals thenational mean of $45.9 thousand, and we do not reject the nullhypothesis that = $45.9 thousand, we made a correct decision.

(e) If, in fact, the mean annual salary of classroom teachers in Hawaii isless than the national mean of $45.9 thousand, and we do not reject thenull hypothesis that = $45.9 thousand, we made a Type II error.

9.37 (a) A Type I error would occur if, in fact, = $47.37, but the results ofthe sampling lead to the conclusion that > $47.37.

(b) A Type II error would occur if, in fact, > $47.37, but the resultsof the sampling fail to lead to that conclusion.

(c) A correct decision would occur if, in fact, = $47.37 and the resultsof the sampling do not lead to the rejection of that fact; or if, infact, > $47.37 and the results of the sampling lead to thatconclusion.

(d) If the mean phone bill equals the 2001 mean of $47.37, and we do notreject the null hypothesis that = $47.37, we made a correctdecision.

(e) If, in fact, the mean cell phone bill is greater than the 2001 mean of


$47.37, and we do not reject the null hypothesis that = $47.37, wemade a Type II error.

9.38 (a) P(Type I error) = = 0.

(b) If = 0, the null hypothesis will never be rejected. If the nullhypothesis is not true, it will not be rejected. Therefore, P(Type II

error) = = 1.

9.39 (a) Exercise 9.31 is a situation in which it may be important to have asmall probability. Concluding that females under the age of 51 are,on the average, getting less than the RDA of 18 mg of iron could leadto remedial action by the nation’s health providers which would beexpensive and unnecessary if females under 51 are, in fact, getting theRDA of 18 mg of iron.

(b) Exercise 9.29 is a situation in which it may be important to have a

small probability. If the mean cadmium level in the mushrooms is,in fact, higher than the government recommended limit, eating themushrooms could have serious health consequences for those who eatthem. If a hypothesis test does not lead to the conclusion that thecadmium level is too high, the population would be led to believe thatthe mushrooms are safe to eat when, in fact, they are not. The

probability of this happening is and should be kept small.

(c) Exercise 9.40 provides a situation in which it is important to have

both small and probability. The null hypothesis is that thenuclear power plant is safe and the alternative is that it is not safe.

A discussion concerning the desirability of being small is found inthat exercise. Given the growing need for power and the power crisesin California in 2001 and in the eastern U.S. in 2003, it would also beimportant not to reject a power plant (of any kind) that was actuallysafe. Thus we would also want to be small in this case.

Another situation in which small and are desirable involvesdefective products. Suppose that a manufacturing company samples itemsfrom boxes to determine if the percentage of defective items is toohigh. The null hypothesis in each case is that the percentage ofdefects is, say, 3%, while the alternative hypothesis is that thepercentage is more than 3%. If the sample results in the conclusionthat the percentage of defects is too high when, in fact, it is not, aType I error has been committed and the manufacturer will unnecessarilyincur the expense of shutting down an assembly line to find anonexistent problem. If the sample results in the conclusion that thepercentage of defects is acceptable when, in fact, it is too high, aType II error has been committed and the manufacturer is likely toincur a loss of business and reputation at the hands of unhappycustomers.

9.40 In this exercise, we are told that failing to reject the null hypothesiscorresponds to approving the nuclear reactor for use. This action —approving the nuclear reactor suggests that the null hypothesis must besomething like: "The nuclear reactor is safe." This further suggests thatthe alternative hypothesis is something like: "The nuclear reactor isunsafe." Putting things together, the Type II error in this situation is:"Approving the nuclear reactor for use when, in fact, it is unsafe." Thistype of error has consequences that are catastrophic. Thus, the propertythat we want the Type II error probability to exhibit is that it be small.

9.41 (a) A Type I error would occur if, in fact, the defendant is innocent, but


the jury concludes that the defendant is guilty.

(b) A Type II error would occur if, in fact, the defendant is guilty, butthe jury fails to conclude that the defendant is guilty.

(c) If I were a defendant, I would want to be small. Given that I aminnocent, I certainly want there to be a small probability of the juryrejecting my innocence (i.e., finding me guilty).

(d) If I were a prosecutor, I would want to be small. Given that thedefendant is guilty, I want there to be a small probability that thejury would declare the defendant not guilty.

(e) If = 0, then an innocent person would never be declared guilty. If = 0, then a guilty person would always be found guilty.

9.42 (a) The probability of a Type I error is the same as the significancelevel, = 0.0456.

(b) If the mean net weight being packaged is 447 g, then the distribution

of x_

is a normal distribution with mean 447 g and standard deviation

.56.125/8.7/ gn(c) is the probability of not rejecting the null hypothesis when it is

actually false. In this case, is the probability that x_falls

between 450.88 g and 457.12 g when = 447 grams and = 1.56 g.

Thus

P x P z( . . ) (..

..

)450 88 457 1245088 447

156

457 12 447

156

= P(2.49 < z < 6.49) = 1.0000 - 0.9936 = 0.0064(d) The probability of a Type II error is an area between the two critical

values of x_

(subscripts indicate left and right) above (i.e., between x_l

= 450.88 g and x_r = 457.12 g) assuming that the true mean is any one of

the thirteen values of presented in this part of the exercise. As a

probability statement, this is written P(450.88< x_

< 457.12).

Since the sample size in this exercise is large enough (i.e., n = 25),

the random variable x_is approximately normally distributed with mean

x and standard deviation /x n . Thus, in order to

calculate P(450.88 < <457.12), we implement the z-score formulas

zx

n

x

nl a r a

/ /and z ,

insert the necessary elements into the right-hand side of each formulaitself, and proceed with using Table II to find the appropriate areas.

Notice that x_l = 450.88 and x

_r = 457.12 and that the standard deviation

to be inserted into each formula has already been presented; i.e.,

.56.125/8.7/ gn Most importantly, the value of the populationmean to be inserted into each formula is not the value of assuming

that the null hypothesis is true; i.e., it is not 0 = 454. It is,

instead, an alternative value of , as indicated by the symbol a in


each of the formulas.

For this part of the exercise, we are given thirteen alternative "truemean" values for . This translates into 26 z-scores that need to becomputed (i.e., two for each value of the "true mean"). In turn, wecalculate the area associated with each pair of z-scores and then use

this information to compute , defined as the probability of a Type IIerror.

The appropriate calculations are:

True mean z-score P(Type II error)

computation

448

85.525/8.7

44812.457

85.125/8.7

44888.450

z

z1.0000 - 0.9678 = 0.0322

449

21.525/8.7

44912.457

21.125/8.7

44988.450

z

z1.0000 - 0.8869 = 0.1131

450

56.425/8.7

45012.457

56.025/8.7

45088.450

z

z1.0000 - 0.7123 = 0.2877

451

92.325/8.7

45112.457

08.025/8.7

45188.450

z

z1.0000- 0.4681 = 0.5319

452

28.325/8.7

45212.457

72.025/8.7

45288.450

z

z0.9995 - 0.2358 = 0.7637

453

64.225/8.7

45312.457

36.125/8.7

45388.450

z

z0.9959 - 0.0869 = 0.9090


True mean z-score P(Type II error)

computation

455

36.125/8.7

45512.457

64.225/8.7

45588.450

z

z0.9131 - 0.0041 = 0.9090

456

72.025/8.7

45612.457

28.325/8.7

45688.450

z

z0.7642 - 0.0005 = 0.7637

457

08.025/8.7

45712.457

92.325/8.7

45788.450

z

z0.5319 - 0.0000 = 0.5319

458

56.025/8.7

45812.457

56.425/8.7

45888.450

z

z0.2877 - 0.0000 = 0.2877

459

21.125/8.7

45912.457

21.525/8.7

45988.450

z

z0.1131 - 0.0000 = 0.1131

460

85.125/8.7

46012.457

85.525/8.7

46088.450

z

z0.0322 – 0.0000 = 0.0322

461

49.225/8.7

46112.457

49.625/8.7

46188.450

z

z0.0064 - 0.0000 = 0.0064

To summarize this part of the exercise, notice that the answer for eachvalue is presented in the third column of the previous table.


(e) Consider columns 1 and 3 of the table in part (e). Also consider agraph whose vertical axis is labeled and whose horizontal axis islabeled . Plot the points of in column 3 of the table versus therespective values of in column 1 and then connect the points with asmooth curve. This curve is presented below.

Recall that the value of , assuming that the null hypothesis is true,

is 0 = 454 g. The previous graph tells us that the farther the true

value of is from the null hypothesis value of 454 g, the smaller is

the probability of making a Type II error; i.e., the smaller is .

All of this is reasonable. It is more likely for a false null

hypothesis to be detected--and hence to be small--when the truevalue of is far from the null hypothesis value than when it isclose.

9.43 (a) Answers will vary. This exercise can easily be done with Minitab, butwe will describe a procedure using Excel. With a blank spreadsheet,from the Menu bar, select Tools, Data Analysis, Random NumberGeneration. Enter 100 for the number of variables, and 25 for theNumber of Random Numbers. Select Normal for the Distribution and enter454 for the Mean and 7.8 for the Standard Deviation. Click on OutputRange and enter A1. Then click OK. This will generate 100 columns of25 random normal numbers each.

(b) At the bottom of column A in A27, enter =AVERAGE(A1:A25) and copy thisformula into B27 through DV27.

(c) Then in A28, enter the formula =(A27-454)/(7.8/sqrt(25)), and copy thisformula into B28 through DV28. The null hypothesis will be rejectedwhenever the number in row 28 is less than or equal -2 or greater thanor equal to +2.

(d) Since the significance level is 0.0456, we would expect about 4 or 5 ofthe 100 samples to result in the rejection of the null hypothesis.

(e) Answers will vary. Our simulation led to rejection of the nullhypothesis 7 times.

(f) If your answer to part (d) is not 4 or 5, it is most likely the resultof sampling variation. On the average, we would expect 4.56% of allsamples to lead to rejection of the null hypothesis, but in any singleset of 100 samples, the percentage may differ from that amount.

Probabilityof aType IIError

00.10.20.30.40.50.60.70.80.91

445 450 455 460 465

Mean

Beta




Rejection | Nonreject ion | RejectionRegion | Region | Region

0.0500 0.9000 0.0500


0 1.645 z

0.9500 0.0500

Critical ValueNonrejection region | Rejection region

Do notReject H0 Reject H 0

-1.645 0 z

0.0500 0.9500

Critical ValueRejection region | Nonrejection region

Exercises 9.3

9.44 Critical values: +z0.05 = +1.645 9.45 Critical value: z0.05 = 1.645

9.46 Critical value: -z0.01 = -2.33 9.47 Critical value: -z0.05 = -1.645

9.48 Critical value: z0.01 = 2.33 9.49 Critical values: +z0.025 = +1.96

9.50 There are two reasons forconcern. When there are outliers, the normality assumption may be


0 2.33 z

0.9900 0.0100

Critical ValueNonrejection region |R ejection region



Rejection | Nonrejection | RejectionRegion | Region | Region

0.0250 0.9500 0.0250


-2.33 0 z

0.0100 0.9900

Critical ValueRejection region |No nrejection region

Section 9.3, Hyp. Tests for One Pop. Mean When is Known 387

questioned, and even for large samples, the presence of one or more outlierscan affect the results of a z-test because the sample mean can be highlyaffected by outliers.

9.51 (a) The z-test in not an appropriate method for highly skewed data when thesample size is less than 30.

(b) The z-test is appropriate for large samples with no outliers even ifthe data are mildly skewed.

9.52 (a) The z-test can be used for small samples that are close to beingnormally distributed, so it is appropriate in this case.

(b) The z-test is not appropriate for these data which have an outlier anda moderate sample size.

9.53 Reject H0 if z < -1.645; (20 22)/(4/ 32) 2.83z ; therefore, reject H0 andconclude that μ < 22.

9.54 Reject H0 if z < -1.645; (21 22)/(4/ 32) 1.41z ; therefore, do not rejectH0. The data do not provide sufficient evidence to support Ha: μ < 22.

9.55 Reject H0 if z > 1.645; (24 22)/(4/ 15) 1.94z ; therefore, reject H0 andconclude that μ > 22.

9.56 Reject H0 if z > 1.645; (23 22)/(4/ 15) 0.97z ; therefore, do not rejectH0. The data do not provide sufficient evidence to support Ha: μ > 22.

9.57 Reject H0 if z < -1.96 or z > 1.96; (23 22)/(4/ 24) 1.22z ; therefore, donot reject H0. The data do not provide sufficient evidence to supportHa: μ =/ 22.

9.58 Reject H0 if z < -1.96 or z > 1.96; (20 22)/(4/ 24) 2.45z ; therefore,reject H0 and conclude that μ =/ 22.

9.59 n = 12, = 0.37 ppm, x_

= 6.31/12 = 0.526 ppm

Step 1: H0: = 0.5 ppm, Ha: > 0.5 ppm

Step 2: = 0.05

Step 3: 24.0)12/37.0/()5.0526.0(zStep 4: Critical value = 1.645

Step 5: Since 0.24 < 1.645, do not reject H0.

Step 6: At the 5% significance level, the data do not provide sufficientevidence to conclude that the mean cadmium level of Boletuspinicola mushrooms is greater than the safety limit of 0.5 ppm.

9.60 n = 28, = $8.45, x_

= $1788.62/28 = $63.88

Step 1: H0: = $66.52, Ha: =/ $66.52

Step 2: = 0.10

Step 3: (63.88 66.52)/(8.45/ 28) 1.653zStep 4: Critical values = +1.645

Step 5: Since –1.653 < 1.645, reject H0.

Step 6: At the 10% significance level, the data do provide sufficientevidence to conclude that the mean retail price of agriculturalbooks is different from the 2000 mean price.


9.61 n = 45, x_

= 14.68, = 4.2

Step 1: H0: = 18 mg, Ha: < 18 mg

Step 2: = 0.01

Step 3: 30.5)45/2.4/()1868.14(zStep 4: Critical value = -2.33

Step 5: Since -5.30 < -2.33, reject H0.

Step 6: At the 1% significance level, the data provide sufficient evidenceto conclude that adult females under the age of 51 are, on theaverage, getting less than the RDA of 18 mg of iron. Consideringthat iron deficiency causes anemia and that iron is required fortransporting oxygen in the blood, this result could have practicalsignificance as well.

9.62 n = 21, x_

= 52.5 years, = 6.8 years

Step 1: H0: = 55 years, Ha: < 55 years

Step 2: = 0.01

Step 3: (52.5 55)/(6.8/ 21) 1.68zStep 4: Critical value = -2.33

Step 5: Since –1.68 > -2.33, do not reject H0.

Step 6: At the 1% significance level, the data do not provide sufficientevidence to conclude that the mean age of diagnosis of all peoplewith early-onset dementia is less than 55 years old.

9.63 n = 100, x_

= 17.8 months, = 6.0 months

Step 1: H0: = 16.7 months, Ha: =/ 16.7 months

Step 2: = 0.05

Step 3: 83.1)100/0.6/()7.168.17(zStep 4: Critical values = ±1.96

Step 5: Since –1.96 < 1.83 < 1.96, do not reject H0.

Step 6: At the 5% significance level, the data do not provide sufficientevidence to conclude that the mean length of imprisonment ofmotor-vehicle theft offenders in Sydney differs from the nationalmean in Australia.

9.64 n = 29, x_

= 78.3, = 11.2

Step 1: H0: = 72 bpm, Ha: > 72 bpm

Step 2: = 0.05

Step 3: (78.3 72)/(11.2/ 29) 3.03zStep 4: Critical value = 1.645

Step 5: Since 3.03 > 1.645, reject H0.

Step 6: At the 5% significance level, the data provide sufficient evidenceto conclude that the mean post-work heart rate for casting workingsexceeds the normal resting heart rate of 72 beats per minute.


GAIN

Frequency

1.00.50.0-0.5-1.0

5

4

3

2

1

0 _X

Ho

Histogram of GAIN(wi th Ho and 95% Z-confidence interval for the Mean, and S tDev = 0.42)

GAIN1.00.50.0-0.5-1.0

_X

Ho

Boxplotof GAIN(with Ho and 95% Z-confidence interval for the Mean, and StDev = 0.42)

GAIN

Percent

1.51.00.50.0-0.5-1.0

99

95

90

80

70

6050

40

30

20

10

5

1

Mean

0.137

0.295S tDev 0.5000N 20

AD 0.549P -V alue

Probability Plot of GAINNormal

9.65 (a) Using Minitab, with the data in a column named GAIN, we choose Stat Basic Statistics 1-Sample z..., click in the Samples in columns text

box and specify GAIN, click in the Standard deviation text box and type0.42, and click in the Test mean text box and type 0.2. Click theOptions… button, enter 95 in the Confidence level text box, click thearrow button at the right of the Alternative drop-down list box andselect greater than and click OK. Click on the Graphs button and checkthe boxes for Histogram of Data and Boxplot of data. Then click OKtwice. The result of the test is

Test of mu = 0.2 vs > 0.2The assumed standard deviation = 0.42

95%Lower

Variable N Mean StDev SE Mean Bound Z PGAIN 20 0.295000 0.499974 0.093915 0.140524 1.01 0.156

(b) The histogram and boxplot were produced by the procedure in part (a).

Now choose Stat Basic Statistics Normality test and enter GAIN in

the Variable text box. Click OK. Then choose Graph Stem-and-Leaf ,

enter GAIN in the Graph variables text box and click OK. The resultsare

Stem-and-leaf of GAIN N = 20Leaf Unit = 0.10

LO -11

2 -0 53 -0 24 -0 16 0 0110 0 223310 0 458 0 66674 0 8889

(c) Repeating the procedure of part (a), we obtain


CHARGE

Frequency

1201008060

5

4

3

2

1

0 _X

Ho

Histogram of CHARGE(with Ho and 95% Z-confidence interval for the Mean, and StDev = 22.4)

CHARGE130120110100908070605040

_X

Ho

Boxplotof CHARGE(with Ho and 95% Z-confidence interval for the Mean, and StDev= 22.4)


95%Lower


(d) The original sample size is only 20. The plots in part (b) indicatethat the value –1.1 is a potential outlier. The z-test should not beused with the original data. This is further confirmed by the factthat using all of the data leads to z = 1.01, whereas, deleting theoutlier leads to z = 1.75. This is enough of a change to alter ourconclusion from not rejecting the null hypothesis to rejecting it. Ifthere is no good reason for deleting the outlier, then the z-test isinappropriate for these data.


box and specify CHARGE, click in the Standard deviation text box andtype 22.4, and click in the Test mean text box and type 75. Click theOptions… button, enter 95 in the Confidence level text box, click thearrow button at the right of the Alternative drop-down list box andselect greater than and click OK. Click on the Graphs button and checkthe boxes for Histogram of Data and Boxplot of data. Then click OKtwice. The result of the test is

Test of mu = 75 vs < 75The assumed standard deviation = 22.4

95%Upper

Variable N Mean StDev SE Mean Bound Z PCHARGE 15 69.3560 24.3201 5.7837 78.8693 -0.98 0.165


Now choose Stat Basic Statistics Normality test and enter CHARGE

in the Variable text box. Click OK. Then choose Graph Stem-and-

Leaf, enter CHARGE in the Graph variables text box and click OK. Theresults are


CHARGE

Percent

140120100806040200

99

95

90

80

70

6050

40

30

20

10

5

1

Mean

0.009

69.36StDev 24.32N 15

AD 0.993P- Valu e

Probability Plot of CHARGENormal

TEMP

Frequency

99.599.098.598.097.597.0

20

15

10

5

0 _X

Ho

Histogram of TEMP(wi th Ho and 99% Z-confidence interval for the Mean, and S tDev = 0.63)

TEMP99.599.098.598.097.597.0

_X

Ho

Boxplotof TEMP(with Ho and 99% Z-confidence interval for the Mean, and StDev = 0.63)

Stem-and-leaf of CHARGE N = 15Leaf Unit = 1.0

2 4 77(6) 5 0136787 6 195 7 44 8 13 9 52 10 6

HI 130

(c) Repeating the procedure of part (a), we obtainTest of mu = 75 vs < 75The assumed standard deviation = 22.4

95%Upper


(d) The sample size is small, there is an outlier, and the data are skewedright with and without the outlier 130.17. Therefore, the use of thez-test is inappropriate for these data.

9.67 (a) Using Minitab, with the data in a column named TEMP, we choose Stat Basic Statistics 1-Sample z..., click in the Samples in columns text

box and specify TEMP, click in the Standard deviation text box andenter 0.63, and click in the Test mean text box and enter 98.6. Clickthe Options… button, enter 95 in the Confidence level text box, clickthe arrow button at the right of the Alternative drop-down list box andselect not equal and click OK. Click on the Graphs button and checkthe boxes for Histogram of Data and Boxplot of data. Then click OK

twice. Now choose Stat Basic Statistics Normality test and enter

CHARGE in the Variable text box. Click OK. Then choose Graph Stem-and-Leaf, enter CHARGE in the Graph variables text box and click OK.The graphs are


TEMP

Percent

10099989796

99.9

99

95

90

80

706050403020

10

5

1

0.1

Mean

0.124

98.12

StDev 0.6468N 93

AD 0.585P- Value

Probability Plotof TEMPNormal

Stem-and-leaf of TEMP N = 93Leaf Unit = 0.10

1 96 73 96 898 97 0000113 97 2223319 97 44444426 97 666677731 97 8888945 98 00000000000111(10) 98 222222223338 98 444444555528 98 6666666667717 98 888888810 99 000015 99 22331 99 4

(b) Yes. The sample size 93 is large and the distribution of the data isquite symmetric.

(c) Yes. The procedure in part (a) also produced the results of the testwhich areTest of mu = 98.6 vs not = 98.6The assumed standard deviation = 0.63

Variable N Mean StDev SE Mean 99% CI Z PTEMP 93 98.1237 0.6468 0.0653 (97.9554, 98.2919) -7.29 0.000

The critical values are –2.575 and 2.575. Since z = -7.29, we rejectthe null hypothesis and conclude that the mean body temperature ofhealthy humans is different from the generally accepted value of98.6ºF.

9.68 (a) Using Minitab, with the data in a column named SALARY, we choose Stat Basic Statistics 1-Sample z..., click in the Samples in columns text

box and specify SALARY, click in the Standard deviation text box andenter 9.2, and click in the Test mean text box and enter 45.9. Clickthe Options… button, enter 95 in the Confidence level text box, clickthe arrow button at the right of the Alternative drop-down list box andselect less than and click OK. Click on the Graphs button and checkthe boxes for Histogram of Data and Boxplot of data. Then click OK


SALARY in the Variable text box. Click OK. Then choose Graph Stem-and-Leaf, enter SALARY in the Graph variables text box and click OK.The graphs are


SALARY

Percent

8070605040302010

99.9

99

9590

80706050403020

10

5

1

0.1

Mean

0.394

44.50StDev 9.181N 90

AD 0.381P- Valu e

Probability Plot ofSALARYNormal

SALARY

Frequency

645648403224

16

12

8

4

0 _X

Ho

Histogram of SALARY(with Ho and 95% Z-confidence interval for the Mean, and StDev = 9.2)

SALARY706050403020

_X

Ho

Boxplot of SALARY(with Ho and 95% Z-confidence i nterval for the Mean, and StDev = 9.2)

Stem-and-leaf of SALARY N = 90Leaf Unit = 1.0

1 2 36 2 7888918 3 01222344444428 3 566666789944 4 0011111223344444(20) 4 5556666677778888999926 5 00000112222344411 5 666888893 6 011 6 6


(c) Yes. The procedure in part (a) also produced the results of the testwhich areTest of mu = 45.9 vs < 45.9The assumed standard deviation = 9.2

95%Upper

Variable N Mean StDev SE Mean Bound Z PSALARY 90 44.5033 9.1806 0.9698 46.0985 -1.44 0.075

The critical value is –1.645. Since z = -1.44, we do not reject thenull hypothesis. The data do not provide sufficient evidence that themean annual salary of classroom teachers in Hawaii is less than thenational mean.

9.69 (a) Using Minitab, with the data in a column named BILL, we choose Stat Basic Statistics 1-Sample z..., click in the Samples in columns text

box and specify BILL, click in the Standard deviation text box andenter 25, and click in the Test mean text box and enter 47.37. Clickthe Options… button, enter 95 in the Confidence level text box, clickthe arrow button at the right of the Alternative drop-down list box andselect greater than and click OK. Click on the Graphs button and checkthe boxes for Histogram of Data and Boxplot of data. Then click OK



BILL

Percent

140120100806040200-20-40

99.9

99

95

90

80

7060504030

20

10

5

1

0.1

Mean

<0.005

50.64

StDev 23.75N 75AD 1.978

P- Value

Probability Plot ofBILLNormal

BILL

Frequency

12010080604020

20

15

10

5

0 _X

Ho

Histogramof BILL(with Ho and 95% Z-confidence interval for the Mean, and StDev= 25)

BILL120100806040200

_X

Ho

Boxplot of BILL(with Ho and 95% Z-confidence interval for the Mean, and StDev = 25)

BILL in the Variable text box. Click OK. Then choose Graph Stem-and-Leaf, enter BILL in the Graph variables text box and click OK. Thegraphs are

Stem-and-leaf of BILL N = 75Leaf Unit = 1.0

1 1 34 1 6695 2 014 2 56777788920 3 01233431 3 55677889999(7) 4 112233337 4 5567889929 5 001133422 5 921 6 04418 6 68815 7 1313 713 8 01129 8 77995 9 24 9 78

HI 114, 119

(b) The results of the test carried out by the procedure in part (a) are

Test of mu = 47.37 vs > 47.37The assumed standard deviation = 25

95%Lower

Variable N Mean StDev SE Mean Bound Z PBILL 75 50.6405 23.7497 2.8868 45.8922 1.13 0.129

The critical value for the test is z = 1.645. Since z = 1.13, we donot reject the null hypothesis. The data do not provide sufficientevidence that the mean local monthly cell phone bill has increased fromthe 2001 mean of $47.37.

After deleting the two outliers 119.61 and 114.98, the graphs and testresults are


BILL

Percent

1209060300

99.9

99

9590

80706050403020

10

5

1

0.1

Mean

<0.005

48.81StDev 21.28N 73

AD 1.688P-V alu e


BILL100908070605040302010

_X

Ho

Boxplotof BILL(with Ho and 95% Z-confidence interval for the Mean, and StDev= 25)

BILL

Frequency

10080604020

20

15

10

5

0 _X

Ho

HistogramofBILL(with Ho and 95% Z-confidence interval for the Mean, and StDev = 25)


1 1 34 1 6695 2 014 2 56777788920 3 01233431 3 55677889999(7) 4 112233335 4 5567889927 5 001133420 5 919 6 04416 6 68813 7 1311 711 8 01127 8 77993 9 22 9 78


95%Lower


Although the value of z has been changed from 1.13 to 0.49 by deletingthe two outliers, the conclusion remains the same. We do not rejectthe null hypothesis. Intuitively, we should expect this result, sincedeleting two large outliers can only reduce the sample mean, making zsmaller.

9.70 (a) n = 28, = $8.45, x_

= $1788.62/28 = $63.88

The 90% confidence interval is 63.88 1.645(8.45)/ 28 (61.25,66.51) .

The hypothesized mean ($66.52) lies outside of the confidence interval,so we should reject the null hypothesis. Since the test statistic is

(63.88 66.52)/(8.45/ 28) 1.653z , which is less than the lowercritical value of –1.645, the hypothesis test also leads to the


conclusion that we should reject the null hypothesis.

(b) n = 100, x_

= 17.8 months, = 6.0 months

The 90% confidence interval is 17.8 1.96(6.0)/ 100 (16.6,19.0).

The hypothesized mean (16.7) lies inside of the confidence interval, sowe should not reject the null hypothesis. Since the test statistic is

83.1)100/0.6/()7.168.17(z , which is less than the upper criticalvalue of 1.96, the hypothesis test also leads to the conclusion that weshould not reject the null hypothesis.

9.71 (a) n = 45, x_

= 14.68, = 4.2

The 99% upper level confidence bound is 14.68 2.33(4.2)/ 45 16.14.

The hypothesized mean (18 mg) lies above the upper confidence bound, sowe should reject the null hypothesis. Since the test statistic is

30.5)45/2.4/()1868.14(z , which is less than the critical value of–2.33, the hypothesis test also leads to the conclusion that we shouldreject the null hypothesis.

(b) n = 21, x_

= 52.5 years, = 6.8 years

The 95% upper level confidence bound is 52.5 2.33(6.8)/ 21 55.96.

The hypothesized mean (55) lies below the upper confidence bound, so weshould not reject the null hypothesis. Since the test statistic is

(52.5 55)/(6.8/ 21) 1.68z , which is greater than the critical valueof –2.33, the hypothesis test also leads to the conclusion that weshould not reject the null hypothesis.

9.72 (a) n = 12, = 0.37 ppm, x_

= 6.31/12 = 0.526 ppm

The 95% lower level confidence bound is 0.526 1.645(0.37)/ 12 0.350.

The hypothesized mean (0.5) lies above the lower confidence bound, sowe should not reject the null hypothesis. Since the test statistic is

24.0)12/37.0/()5.0526.0(z , which is less than the critical value of1.645, the hypothesis test also leads to the conclusion that we shouldnot reject the null hypothesis.

(b) n = 30, x_

= 78.3, = 11.2

The 95% lower level confidence bound is 78.3 1.645(11.2)/ 30 74.94.

The hypothesized mean (72) lies below the lower confidence bound, so weshould reject the null hypothesis. Since the test statistic is

08.3)30/2.11/()723.78(z , which is greater than the critical valueof 1.645, the hypothesis test also leads to the conclusion that weshould reject the null hypothesis.

Exercises 9.4

9.73 Hypothesis tests have built-in margins of error. Errors will occur due tothe uncontrollable randomness in the data observed.

Section 9.4, Type II Error Probabilities; Power 397

9.74 (a) A Type I error occurs if the data leads to rejecting the nullhypothesis when it is, in fact, true.

(b) A Type II error occurs if the data leads to not rejecting the nullhypothesis when it is, in fact, false.

(c) The significance level is the probability associated with the testprocedure of rejecting the null hypothesis when it is actually true,i.e., it is the probability of making a Type I error.

9.75 (a) = significance level = P(Type I error) = P(rejecting a true nullhypothesis)

(b) = P(Type II error) = P(not rejecting a false null hypothesis)

(c) 1 – = Power of the test = P(rejecting a false null hypothesis)

9.76 The power of a hypothesis test is the probability of making the correctdecision of rejecting a false null hypothesis; that is, the probability ofnot committing a Type II error. Power = 1 – .

9.77 Since μ is unknown, the power curve enables one to evaluate theeffectiveness of a hypothesis test for a variety of values of .

9.78 The power of a hypothesis test increases if the sample size is increasedwithout changing the significance level. This makes sense because largersample sizes should make it possible to detect smaller differences from thenull hypothesis (or make it more likely to detect a difference of a givensize).

9.79 If the significance level is decreased without changing the sample size, therejection region is made smaller (in probability terms). This makes thenonrejection region larger, i.e., gets larger. This, in turn, makes the

power 1 - smaller.

9.80 Procedure B. Assuming that the assumptions underlying both procedures aresatisfied, we would choose the procedure with the greater power since itgives us a better chance of rejecting a false null hypothesis.

9.81 (a) Note:0

0/

/

xz x z n

n

Since this is a right-tailed test, we would reject H0 if z 1.645; or

equivalently if 6757.012/)37.0(645.15.0x

So we reject H0 if x_

0.676; otherwise do not reject H0.

(b) P(Type I error) = = 0.05


(c)

True mean z-score P(Type II error) Power

computation 1 –

0.55 18.112/37.0

55.0676.0z 0.8810 0.119 0

0.60 71.012/37.0

60.0676.0z 0.7611 0.2389

0.65 24.012/37.0

65.0676.0z 0.5948 0.4052

0.70 22.012/37.0

70.0676.0z 0.4129 0.5871

0.75 69.012/37.0

75.0676.0z 0.2451 0.7549

0.80 16.112/37.0

80.0676.0z 0.1230 0.8770

0.85 63.112/37.0

85.0676.0z 0.0516 0.9484

(d)

9.82 (a) Note: 00

//

xz x z nn

Since this is a two-tailed test, we would reject H0 if |z| 1.645; or

equivalently if 66.52 1.645(8.45)/ 28 69.15x or

66.52 1.645(8.45)/ 28 63.89x .

So reject H0 if x_

63.89 or x_



PowerCurve

0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

0.00 0.50 1.00

TrueMean


(c)


computation 1 –

62

63.89 621.18

8.45/ 28

69.15 624.48

8.45/ 28

z

z1.0000 – 0.8810 = 0.1190 0.8810

63

63.89 630.56

8.45/ 28

69.15 633.85

8.45/ 28

z

z0.9999 – 0.7123 = 0.2876 0.7124

64

63.89 640.07

8.45/ 28

69.15 643.22

8.45/ 28

z

z0.9994 – 0.4721 = 0.5273 0.4727

65

63.89 650.70

8.45/ 28

69.15 652.60

8.45/ 28

z

z0.9953 – 0.2420 = 0.7533 0.2467

66

63.89 661.32

8.45/ 28

69.15 661.97

8.45/ 28

z

z0.9756 – 0.0934 = 0.8822 0.1178

67

63.89 671.95

8.45/ 28

69.15 671.35

8.45/ 28

z

z0.9115 – 0.0256 = 0.8859 0.1141

68

63.89 682.57

8.45/ 28

69.15 680.72

8.45/ 28

z

z0.7642 – 0.0051 = 0.7591 0.2409

69

63.89 693.20

8.45/ 28

69.15 690.09

8.45/ 28

z

z0.5374 – 0.0007 = 0.5367 0.4633



computation 1 –

70

63.89 703.83

8.45/ 28

69.15 700.53

8.45/ 28

z

z0.2981 – 0.0001 = 0.2980 0.7020

71

63.89 714.45

8.45/ 28

69.15 711.16

8.45/ 28

z

z0.1230 – 0.0000 = 0.1230 0.8770

(d)

9.83 (a) Note:0

0/

/

xz x z nn

Since this is a left-tailed test, we would reject H0 if z -2.33; or

equivalently if 54.1645/)2.4(33.218x

So reject H0 if x_



(c) Answers may differ slightly from those in the text due to intermediaterounding.


computation 1 -

15.50 66.145/2.4

50.1554.16z 1.000 – 0.9515 = 0.0485 0.9515

15.75 26.145/2.4

75.1554.16z 1.000 – 0.8962 = 0.1038 0.8962

Power Curve

0.00

0.20

0.40

0.60

0.80

1.00

60 62 64 66 68 70 72

True Mean μ

Power



computation 1 -

16.00 86.045/2.4

00.1654.16z 1.000 – 0.8051 = 0.1949 0.8051

16.25 46.045/2.4

25.1654.16z 1.000 – 0.6772 = 0.3228 0.6772

16.50 06.045/2.4

50.1654.16z 1.000 – 0.5239 = 0.4761 0.5239

16.75 34.045/2.4

75.1654.16z 1.000 – 0.3669 = 0.6331 0.3669

17.00 73.045/2.4

00.1754.16z 1.000 – 0.2327 = 0.7673 0.2327

17.25 13.145/2.4

25.1754.16z 1.000 – 0.1292 = 0.8708 0.1292

17.50 53.145/2.4

50.1754.16z 1.000 – 0.0630 = 0.9370 0.0630

17.75 93.145/2.4

75.1754.16z 1.000 – 0.0268 = 0.9732 0.0268

(d)

9.84 (a) Note:0

0/

/

xz x z n

n


equivalently if 55 2.33(6.8)/ 21 51.54x

So reject H0 if x_



(c) Answers may differ slightly from those in the text due to intermediate

PowerCurve

0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

15 16 17 18

TrueMean


rounding.


computation 1 -

4751.54 47

3.066.8/ 21

z 1.000 – 0.9989 = 0.0011 0.9989

4851.54 48

2.396.8/ 21

z 1.000 – 0.9916 = 0.0084 0.9916

4951.54 49

1.716.8/ 21

z 1.000 – 0.9964 = 0.0036 0.9964

5051.54 50

1.046.8/ 21

z 1.000 – 0.8508 = 0.1492 0.8508

5151.54 51

0.366.8/ 21

z 1.000 – 0.6406 = 0.3594 0.6406

5251.54 52

0.316.8/ 21

z 1.000 – 0.3783 = 0.6217 0.3783

5351.54 53

0.986.8/ 21

z 1.000 – 0.1635 = 0.8365 0.1635

5451.54 54

1.666.8/ 21

z 1.000 – 0.0485 = 0.9515 0.0485

(d)

9.85 (a) Note: 0

0/

/

xz x z n

n


equivalently if 876.17100/)0.6(96.17.16x or

524.15100/)0.6(96.17.16x .

So reject H0 if x_

15.524 or x_



Power Curve

0.00

0.20

0.40

0.60

0.80

1.00

46 47 48 49 50 51 52 53 54 55

True Mean μ

Power


(c)


computation 1 –

14.0

46.6100/0.6

0.14876.17

54.2100/0.6

0.14524.15

z

z1.0000 – 0.9945 = 0.0055 0.9945

14.5

63.5100/0.6

5.14876.17

71.1100/0.6

5.14524.15

z

z1.0000 – 0.9564 = 0.0436 0.9564

15.0

79.4100/0.6

0.15876.17

87.0100/0.6

0.15524.15

z

z1.0000 – 0.8078 = 0.1922 0.8078

15.5

96.3100/0.6

5.15876.17

04.0100/0.6

5.15524.15

z

z1.0000 – 0.5160 = 0.4540 0.5160

16.0

13.3100/0.6

0.16876.17

79.0100/0.6

0.16524.15

z

z0.9991 – 0.2148 = 0.7843 0.2157

16.5

29.2100/0.6

5.16876.17

63.1100/0.6

5.16524.15

z

z0.9890 – 0.0516 = 0.9374 0.0626

17.0

46.1100/0.6

0.17876.17

46.2100/0.6

0.17524.15

z

z0.9279 – 0.0069 = 0.9210 0.0790

17.5

63.0100/0.6

5.17876.17

29.3100/0.6

5.17524.15

z

z0.7357 – 0.0005 = 0.7352 0.2648



computation 1 –

18.0

21.0100/0.6

0.18876.17

13.4100/0.6

0.18524.15

z

z0.4168 – 0.0000 = 0.4168 0.5832

18.5

04.1100/0.6

5.18876.17

96.4100/0.6

5.18524.15

z

z0.1492 – 0.0000 = 0.1492 0.8508

19.0

87.1100/0.6

0.19876.17

79.5100/0.6

0.19524.15

z

z0.0307 – 0.0000 = 0.0307 0.9693

(d)

9.86 (a) Note:0

0 //

xz x z n

n

Since this is a right-tailed test, we would reject H0 if z > 1.645; or


So reject H0 if x_

> 75.42; otherwise do not reject H0.


(c) Answers may differ slightly from those in the text due to intermediaterounding.

Power Curve

0.00

0.20

0.40

0.60

0.80

1.00

13 14 15 16 17 18 19 20

True Mean μ

Power


Power Curve

0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

1.2000

72 74 76 78 80 82

True Mean μ

Power


computation 1 –

73 16.129/2.11

)7342.75(z 0.8777 0.1223

74 68.029/2.11

)7442.75(z 0.7526 0.2474

75 20.029/2.11

)754.75(z 0.5800 0.4200

76 28.029/2.11

)764.75(z 0.3902 0.6098

77 76.029/2.11

)774.75(z 0.2237 0.7763

78 24.129/2.11

)784.75(z 0.1074 0.8926

79 72.129/2.11

)794.75(z 0.0426 0.9574

80 20.229/2.11

)804.75(z 0.0138 0.9862

(d)

9.87 (a) Note:0

0 //

xz x z n

n

Since this is a right-tailed test, we would reject H0 if z 1.645; or

equivalently if 636.020/)37.0(645.15.0x

So we reject H0 if x_



(c) Answers may differ slightly from those in the text due to intermediate


rounding.


computation 1 –

0.55 04.120/37.0

55.0636.0z 0.8508 0.1492

0.60 44.020/37.0

60.0636.0z 0.6700 0.3300

0.65 17.020/37.0

65.0636.0z 0.4325 0.5675

0.70 77.020/37.0

70.0636.0z 0.2206 0.7794

0.75 38.120/37.0

75.0636.0z 0.0838 0.9162

0.80 98.120/37.0

80.0636.0z 0.0239 0.9761

0.85 59.220/37.0

85.0636.0z 0.0048 0.9952

(d)

The power curve with n = 20 rises more quickly as the true mean μincreases, resulting in a higher power at any given value of thanfor n = 12. This illustrates the principle that a larger sample sizehas a higher probability of rejecting the null hypothesis when thenull hypothesis is false and the significance level remains the same.

9.88 (a) Note: 00

//

xz x z nn



PowerCurve

0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

1.2000

0.00 0.50 1.00

TrueMean


66.52 1.645(8.45)/ 50 64.55x .

So reject H0 if x_

64.55 or x_



(c) The details of the z-score computation are the same as in Exercise9.82, with 64.55 replacing 63.89, 68.49 replacing 69.15, and 50replacing 28. The results of the computations are

P(Type II) Powerμ z-left z-right P(z<z-left) P(z<z-rt) 1-62 1.60 4.06 0.9452 1.0000 0.0548 0.945263 0.97 3.44 0.8340 0.9997 0.1657 0.834364 0.34 2.81 0.6331 0.9975 0.3645 0.635565 -0.28 2.19 0.3897 0.9857 0.5960 0.404066 -0.91 1.56 0.1814 0.9406 0.7592 0.240867 -1.53 0.93 0.0630 0.8238 0.7608 0.239268 -2.16 0.31 0.0154 0.6217 0.6063 0.393769 -2.79 -0.32 0.0026 0.3745 0.3718 0.628270 -3.41 -0.95 0.0003 0.1711 0.1707 0.829371 -4.04 -1.57 0.0000 0.0582 0.0582 0.9418

(d)

The power curve with n = 50 rises more quickly as the true mean μincreases or decreases from $66.52, resulting in a higher power at anygiven value of than for n = 28. This illustrates the principlethat a larger sample size has a higher probability of rejecting thenull hypothesis when the null hypothesis is false and the significancelevel remains the same.

9.89 (a) Note: 00

//

xz x z nn



16.7 1.96(6.0)/ 40 14.841x .

So reject H0 if x_

14.841 or x_



Power Curve

0.00

0.20

0.40

0.60

0.80

1.00

60 62 64 66 68 70 72

True Mean μ

Power


(c) The details of the z-score computation are the same as in Exercise9.85, with 14.841 replacing 15.524, 18.559 replacing 17.876, and 40replacing 100. The results of the computations are

Powerμ z-left z-right P(z<z-left)P(z<z-rt) 1-

14.0 0.89 4.81 0.8133 1.0000 0.1867 0.813314.5 0.36 4.28 0.6406 1.0000 0.3594 0.640615.0 -0.17 3.75 0.4325 0.9999 0.5674 0.432615.5 -0.69 3.22 0.2451 0.9994 0.7543 0.245716.0 -1.22 2.70 0.1112 0.9965 0.8853 0.114716.5 -1.75 2.17 0.0401 0.9850 0.9449 0.055117.0 -2.28 1.64 0.0113 0.9495 0.9382 0.061817.5 -2.80 1.12 0.0026 0.8686 0.8661 0.133918.0 -3.33 0.59 0.0004 0.7224 0.7220 0.278018.5 -3.86 0.06 0.0001 0.5239 0.5239 0.476119.0 -4.38 -0.46 0.0000 0.3228 0.3228 0.6772

The power curve with n = 40 rises less quickly as the true mean μincreases or decreases from 16.7, resulting in a lower power at anygiven value of than for n = 100. This illustrates the principlethat a larger sample size has a higher probability of rejecting thenull hypothesis when the null hypothesis is false and the significancelevel remains the same.

9.90 Note:0

0/

/

xz x z nn


equivalently if 55 2.33(6.8)/ 15 50.91x

So reject H0 if x_



(c) The details of the z-score computation are the same as in Exercise9.85, with 50.91 replacing 51.54 and 15 replacing 21. The results ofthe computations are

Power Curve

0.00

0.20

0.40

0.60

0.80

1.00

13 14 15 16 17 18 19 20

True Mean μ

Power


Power

1

μ0

Power

1

μ0

P(TypeII) Power

μ z P(Z<z) 1-47 2.63 0.9957 0.0043 0.995748 1.96 0.9750 0.0250 0.975049 1.29 0.9015 0.0985 0.901550 0.61 0.7291 0.2709 0.729151 -0.06 0.4761 0.5239 0.476152 -0.73 0.2327 0.7673 0.232753 -1.41 0.0793 0.9207 0.079354 -2.08 0.0188 0.9812 0.0188

(d)

The power curve with n = 15 rises less quickly as the true mean μ ordecreases from 55, resulting in a lower power at any given value ofthan for n = 21. This illustrates the principle that a larger samplesize has a higher probability of rejecting the null hypothesis whenthe null hypothesis is false and the significance level remains thesame.

9.91 (a)

(b) The curve in part (a) portrays that, ideally, one desires the value

for the power for any given a to be as close to 1 as possible.

9.92 (a)

Power Curve

0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

46 47 48 49 50 51 52 53 54 55

True Mean μ

Power


Power

1

μ0

(b) The curve in part (a) portrays that, ideally, one desires the value

for the power for any given a to be as close to 1 as possible.

9.93 (a)

(b) The curve in (a) portrays that, ideally, one desires the value for the

power for any given a to be as close to 1 as possible.

9.94 (a) Note:0

0/

/

xz x z nn

Since this is a left-tailed test, reject H0 if z < -1.645; or


So reject H0 if x_

25.6; otherwise do not reject H0. If is really

25.4, then P(Type II error) = P(x_

25.6) =

25.6 25.4( ) ( 0.78) 1.0000 0.7823 0.2177.

1.4/ 30P z P z

(b)-(c) Answers will vary. You can use the procedure outlined in thesolution to Exercise 9.43 to generate the samples using the Excelspreadsheet.

(d) Since the probability of a Type II error when = 25.4 (i.e., offailing to reject the null hypothesis when μ = 25.4) is 0.2177 frompart (a), we would expect non-rejection of the null hypothesis about 21or 22 times in the 100 samples.

(e)-(f) Answers will vary. Our simulation contained 20 samples in whichthe null hypothesis was not rejected when =25.4. This is very close towhat we expected.

Exercises 9.5

9.95 (1) It allows the reader to assess significance at any desired level, and(2) it permits the reader to evaluate the strength of the evidence againstthe null hypothesis.

9.96 The P-value of a test is, assuming the null hypothesis to be true, theprobability of observing a value of the test statistic as extreme or moreextreme than the one that was actually observed. When the P-value is small,it provides evidence against the null hypothesis.

9.97 In the critical value approach, we determine critical values based on thesignificance level. The critical values determine where the rejection andnonrejection regions lie for the test statistic. If the value of the teststatistic falls in the rejection region, the null hypothesis is rejected. Inthe P-value approach, the test statistic is computed and then theprobability of observing a value as extreme or more extreme than the valueobtained is determined. If the P-value is smaller than the significance

Section 9.5, P-Values 411

level, the null hypothesis is rejected. Reporting a P-value allows a readerto draw his/her own conclusion based on the strength of the evidence.

9.98 The P-value for a one-sample z-test is obtained as:

(a) P(z < observed z value) for a left-tailed test

(b) P(z > observed z value) for a right-tailed test

(c) 2P(z > absolute value of the observed z value) for a two-tailed test.

9.99 True

9.100 (a) Do not reject the null hypothesis.

(b) Reject the null hypothesis.

(c) Reject the null hypothesis.

9.101 (a) Do not reject the null hypothesis.

(b) Reject the null hypothesis.

(c) Do not reject the null hypothesis.

9.102 A P-value of 0.02 provides stronger evidence against the null hypothesisthan does a value of 0.03. It says that if the null hypothesis is true, thedata are less likely than they are when the P-value is 0.03.

9.103 (a) Strength of the evidence against the null hypothesis is moderate.

(b) There is weak or no evidence against the null hypothesis.

(c) Strength of the evidence against the null hypothesis is strong.

(d) Strength of the evidence against the null hypothesis is very strong.

9.104 (a) Strength of the evidence against the null hypothesis is weak or none.

(b) Strength of the evidence against the null hypothesis is moderate.

(c) Strength of the evidence against the null hypothesis is very strong.

(d) Strength of the evidence against the null hypothesis is strong.

9.105 (a) ( )/( / ) (20 22)/(4/ 32) 2.83z x n ; P-value = P(z < -2.83) = 0.0023

(b) The evidence against the null hypothesis is very strong.

9.106 (a) ( )/( / ) (21 22)/(4/ 32) 1.41z x n ; P-value = P(z < -1.41) = 0.0793

(b) The evidence against the null hypothesis is moderate.

9.107 (a) ( )/( / ) (24 22)/(4/ 15) 1.94z x n ; P-value = P(z > 1.94) = 0.0262

(b) The evidence against the null hypothesis is strong.

9.108 (a) ( )/( / ) (23 22)/(4/ 15) 0.97z x n ; P-value = P(z > 0.97) = 0.1660

(b) The evidence against the null hypothesis is weak or none.

9.109 (a) ( )/( / ) (23 22)/(4/ 24) 1.22z x n ; P-value = 2P(z > |1.22|) = 0.2224

(b) The evidence against the null hypothesis is weak or none.

9.110 (a) ( )/( / ) (20 22)/(4/ 24) 2.45z x n ; P-value = 2P(z > |-2.45|) = 0.0142

(b) The evidence against the null hypothesis is strong.

9.111 (a) z = 2.03, P-value = 1.0000 - 0.9788 = 0.0212

(b) z = -0.31, P-value = 1.0000 - 0.3783 = 0.6217

9.112 (a) z = -1.84, P-value = 0.0329


(b) z = 1.25, P-value = 0.8944

9.113 (a) z = -0.74, P-value = 0.8770

(b) z = 1.16, P-value = 0.0329

9.114 (a) z = 3.08, Right-tail probability = 1.0000 - 0.9990 = 0.0010

P-value = 0.001 x 2 = 0.0020

(b) z = -2.42, Left-tail probability = 0.0078

P-value = 0.0078 x 2 = 0.0156

9.115 (a) z = -1.66, Left-tail probability = 0.0485

P-value = 0.0485 x 2 = 0.0970

(b) z = 0.52, Right-tail probability = 1.0000 - 0.6985 = 0.3015

P-value = 0.3015 x 2 = 0.6030

9.116 (a) z = 1.24, P-value = 1.0000 - 0.8925 = 0.1075

(b) z = -0.69, P-value = 1.0000 - 0.2451 = 0.7549

9.117 (See Exercise 9.59 for classical approach results.)

Step 1: H0: = 0.5 ppm, Ha: > 0.5 ppm

Step 2: = 0.05

Step 3: z = 0.24

Step 4: P-value = P(z 0.24) = 1.0000 - 0.5948 = 0.4052

Step 5: Since 0.4052 > 0.05, do not reject H0.

Step 6: At the 5% significance level, the data do not provide sufficientevidence to conclude that the mean cadmium level of Boletuspinicola mushrooms is greater than the safety limit of 0.5 ppm.

Using Table 9.12, we classify the strength of evidence against the nullhypothesis as weak or none because P > 0.10.


Step 1: H0: = $66.52, Ha: $66.52

Step 2: = 0.10

Step 3: z = -1.65

Step 4: P-value = 2P(z 1.65) = 2(1.0000 - 0.9505) = 0.0990

Step 5: Since 0.0990 < 0.10, reject H0.

Step 6: At the 10% significance level, the data do provide sufficientevidence to conclude that the mean retail price of agriculturebooks has changed from the 2000 mean.

Using Table 9.12, we classify the strength of evidence against the nullhypothesis as moderate because 0.05 < P < 0.10.


Step 1: H0: = 18 mg, Ha: < 18 mg

Step 2: = 0.01

Step 3: z = -5.30

Step 4: P-value = P(z < -5.30) = 0.0000 (to four decimal places)


Step 6: At the 1% significance level, the data provide sufficientevidence to conclude that adult females under the age of 51 are,


on the average, getting less than the RDA of 18 mg of iron.

Using Table 9.12, we classify the strength of evidence against the nullhypothesis as very strong because P < 0.01.


Step 1: H0: = 55 years, Ha: < 55 years

Step 2: = 0.01

Step 3: z = -1.69

Step 4: P-value = P(z < -1.69) = 0.0455


Step 6: At the 1% significance level, the data do not provide sufficientevidence to conclude that the mean age at diagnosis of allpeople with early-onset dementia is less than 55 years old.

Using Table 9.12, we classify the strength of evidence against the nullhypothesis as strong because 0.01 1.83) = 2(0.0336) =0.0672


Step 6: At the 5% significance level, the data do not provide sufficientevidence to conclude that the mean length of imprisonment formotor-vehicle theft offenders in Sydney, Australia differs fromthe Australian national mean of 16.7 months.

Using Table 9.12, we classify the strength of evidence against the nullhypothesis as moderate since 0.05 72

Step 2: = 0.05

Step 3: z = 3.08

Step 4: P-value = P(z > 3.08) = 1.0000 - 0.9990 = 0.0010


Step 6: At the 5% significance level, the data provide sufficientevidence to conclude that the post-work heart rate for castingworkers is greater than the normal resting heart rate of 72 bpm.

Using Table 9.12, we classify the strength of evidence against the nullhypothesis as very strong because P < 0.01.


box and specify GAIN, click in the Standard deviation text box and type0.42, and click in the Test mean text box and type 0.2. Click theOptions… button, enter 95 in the Confidence level text box, click thearrow button at the right of the Alternative drop-down list box andselect greater than and click OK. Click on the Graphs button and check


GAIN

Frequency

1.00.50.0-0.5-1.0

5

4

3

2

1

0 _X

Ho

Histogram of GAIN(with Ho and 95% Z-confidence interval for the Mean, and StDev = 0.42)

GAIN1.00.50.0-0.5-1.0

_X

Ho

Boxplot of GAIN(with Ho and 95% Z-confidence interval for the Mean, and StDev= 0.42)

GAIN

Percent

1.51.00.50.0-0.5-1.0

99

95

90

80

70

60

5040

30

20

10

5

1

Mean

0.137

0.295StDev 0. 5000

N 20AD 0.549P-V alu e

Probability Plot of GAINNormal

the boxes for Histogram of Data and Boxplot of data. Then click OKtwice. The result of the test is


95%Lower


The P-value is reported in the last line of the output as 0.156. Since0.156 > 0.05, we do not reject the null hypothesis. The evidenceagainst the null hypothesis is weak or nonexistent.


Now choose Stat Basic Statistics Normality test and enter GAIN in

the Variable text box. Click OK. Then choose Graph Stem-and-Leaf ,

enter GAIN in the Graph variables text box and click OK. The resultsare

Stem-and-leaf of GAIN N = 20Leaf Unit = 0.10

LO -11

2 -0 53 -0 24 -0 16 0 0110 0 223310 0 458 0 66674 0 8889

(c) Repeating the procedure of part (a) after removing the outlier, weobtain


95%Lower


Now the P-value is 0.04, which is less than the significance level,leading to rejection of the null hypothesis. Now the data do provide


CHARGE

Frequency

1201008060

5

4

3

2

1

0 _X

Ho

Histogram of CHARGE(with Ho and 95% Z-confidence interval for the Mean, and StDev = 22.4)

CHARGE130120110100908070605040

_X

Ho

Boxplot of CHARGE(with Ho and 95% Z-confidence interval for the Mean, and StDev = 22.4)

sufficient evidence to conclude that, on average, the net percentagegain exceeds 0.2.

(d) The original sample size is only 20. The plots in part (b) indicatethat the value –1.1 is a potential outlier. The z-test should not beused with the original data. This is further confirmed by the factthat using all of the data leads to z = 1.01, whereas, deleting theoutlier leads to z = 1.75. This is enough of a change to alter ourconclusion from not rejecting the null hypothesis to rejecting it. Ifthere is no good reason for deleting the outlier, then the z-test isinappropriate for these data.


box and specify CHARGE, click in the Standard deviation text box andtype 22.4, and click in the Test mean text box and type 75. Click theOptions… button, enter 95 in the Confidence level text box, click thearrow button at the right of the Alternative drop-down list box andselect greater than and click OK. Click on the Graphs button and checkthe boxes for Histogram of Data and Boxplot of data. Then click OKtwice. The result of the test is

Test of mu = 75 vs < 75The assumed standard deviation = 22.4

95%Upper


The P-value is reported in the last line of the output as 0.165. Since0.165 > 0.05, we do not reject the null hypothesis. The evidenceagainst the null hypothesis is weak or nonexistent.


Now choose Stat Basic Statistics Normality test and enter CHARGE

in the Variable text box. Click OK. Then choose Graph Stem-and-

Leaf, enter CHARGE in the Graph variables text box and click OK. Theresults are


CHARGE

Percent

140120100806040200

99

95

90

80

70

60

5040

30

20

10

5

1

Mean

0.009

69.36StDev 24.32

N 15AD 0.993P- Value

Probability Plot of CHARGENormal

TEMP

Frequency

99.599.098.598.097.597.0

20

15

10

5

0 _X

Ho

Histogram of TEMP(with Ho and 95% Z-confidence interval for the Mean, and StDev = 0.63)

TEMP99.599.098.598.097.597.0

_X

Ho

Boxplot of TEMP(wi th Ho and 95%Z-confidence interval for the Mean, and StDev = 0.63)

Stem-and-leaf of CHARGE N = 15Leaf Unit = 1.0

2 4 77(6) 5 0136787 6 195 7 44 8 13 9 52 10 6

HI 130

(c) Repeating the procedure of part (a) after deleting the outlier, weobtainTest of mu = 75 vs < 75The assumed standard deviation = 22.4

95%Upper


The P-value is now 0.048, which is less than the significance level of0.05, leading to rejection of the null hypothesis.

(d) The sample size is small, there is an outlier, and the data are skewedright with and without the outlier 130.17. Therefore, the use of thez-test is inappropriate for these data.

9.125 (a) Using Minitab, with the data in a column named TEMP, we choose Stat Basic Statistics 1-Sample z..., click in the Samples in columns text

box and specify TEMP, click in the Standard deviation text box andenter 0.63, and click in the Test mean text box and enter 98.6. Clickthe Options… button, enter 95 in the Confidence level text box, clickthe arrow button at the right of the Alternative drop-down list box andselect not equal and click OK. Click on the Graphs button and checkthe boxes for Histogram of Data and Boxplot of data. Then click OK


TEMP in the Variable text box. Click OK. Then choose Graph Stem-and-Leaf, enter TEMP in the Graph variables text box and click OK. Thegraphs are


TEMP

Percent

10099989796

99.9

99

95

90

80

7060504030

20

10

5

1

0.1

Mean

0.124

98.12

StDev 0.6468N 93AD 0.585

P- Value

Probability Plotof TEMPNormal

Stem-and-leaf of TEMP N = 93Leaf Unit = 0.10

1 96 73 96 898 97 0000113 97 2223319 97 44444426 97 666677731 97 8888945 98 00000000000111(10) 98 222222223338 98 444444555528 98 6666666667717 98 888888810 99 000015 99 22331 99 4


(c) Yes. The procedure in part (a) also produced the results of the testwhich areTest of mu = 98.6 vs not = 98.6The assumed standard deviation = 0.63

Variable N Mean StDev SE Mean 99% CI Z PTEMP 93 98.1237 0.6468 0.0653 (97.9554, 98.2919) -7.29 0.000

The P-value is shown in the last line of the output as 0.000. This isless than the significance level of 0.01, leading to rejection of thenull hypothesis. The evidence against the null hypothesis iscategorized as very strong since P < 0.01.

9.126 (a) Using Minitab, with the data in a column named SALARY, we choose Stat Basic Statistics 1-Sample z..., click in the Samples in columns text

box and specify SALARY, click in the Standard deviation text box andenter 9.2, and click in the Test mean text box and enter 45.9. Clickthe Options… button, enter 95 in the Confidence level text box, clickthe arrow button at the right of the Alternative drop-down list box andselect less than and click OK. Click on the Graphs button and checkthe boxes for Histogram of Data and Boxplot of data. Then click OK


SALARY in the Variable text box. Click OK. Then choose Graph Stem-and-Leaf, enter SALARY in the Graph variables text box and click OK.The graphs are


SALARY

Percent

8070605040302010

99.9

99

9590

80706050403020

10

5

1

0.1

Mean

0.394

44.50StDev 9.181N 90

AD 0.381P- Value

Probability Plot ofSALARYNormal

SALARY

Frequency

645648403224

16

12

8

4

0 _X

Ho

Histogram of SALARY(with Ho and 95% Z-confidence interval for the Mean, and StDev = 9.2)

SALARY706050403020

_X

Ho

Boxplot of SALARY(with Ho and 95% Z-confidence interval for the Mean, and StDev = 9.2)

Stem-and-leaf of SALARY N = 90Leaf Unit = 1.0

1 2 36 2 7888918 3 01222344444428 3 566666789944 4 0011111223344444(20) 4 5556666677778888999926 5 00000112222344411 5 666888893 6 011 6 6


(c) Yes. The procedure in part (a) also produced the results of the testwhich areTest of mu = 45.9 vs < 45.9The assumed standard deviation = 9.2

95%Upper

Variable N Mean StDev SE Mean Bound Z PSALARY 90 44.5033 9.1806 0.9698 46.0985 -1.44 0.075

The P-value in the last line is 0.075 which is greater than thesignificance level of 0.05. Therefore, we do not reject the nullhypothesis. The data do not provide sufficient evidence to concludethat the mean teacher salary in Hawaii is less than the nationalaverage.

9.127 (a) Using Minitab, with the data in a column named BILL, we choose Stat Basic Statistics 1-Sample z..., click in the Samples in columns text

box and specify BILL, click in the Standard deviation text box andenter 25, and click in the Test mean text box and enter 47.37. Clickthe Options… button, enter 95 in the Confidence level text box, clickthe arrow button at the right of the Alternative drop-down list box andselect greater than and click OK. Click on the Graphs button and checkthe boxes for Histogram of Data and Boxplot of data. Then click OK


BILL in the Variable text box. Click OK. Then choose Graph Stem-

and-Leaf, enter BILL in the Graph variables text box and click OK. Thegraphs are


BILL

Percent

140120100806040200-20-40

99.9

99

9590

80706050403020

10

5

1

0.1

Mean

<0.005

50.64StDev 23.75N 75

AD 1.978P-V alu e


BILL120100806040200

_X

Ho

Boxplot ofBILL(with Ho and 95% Z-confidence interval for the Mean, and StDev = 25)

BILL

Frequency

12010080604020

20

15

10

5

0 _X

Ho

Histogramof BILL(with Ho and 95% Z-confidence interval for the Mean, and StDev= 25)


1 1 34 1 6695 2 014 2 56777788920 3 01233431 3 55677889999(7) 4 112233337 4 5567889929 5 001133422 5 921 6 04418 6 68815 7 1313 713 8 01129 8 77995 9 24 9 78

HI 114, 119

(b) The results of the test carried out by the procedure in part (a) are


95%Lower


The P-value is shown in the last line as 0.129, which is greater than thesignificance level of 0.05, leading to nonrejection of the null hypothesis.The data do not provide sufficient evidence to conclude that the mean localmonthly cell phone bill has increased from the 2001 mean of $47.37.The evidence against the null hypothesis is weak or nonexistent.

After deleting the two outliers 119.61 and 114.98, the graphs and testresults are


BILL

Percent

1209060300

99.9

99

95

90

80

706050403020

10

5

1

0.1

Mean

<0.005

48.81

StDev 21.28N 73

AD 1.688P- Value


BILL100908070605040302010

_X

Ho

Boxplot of BILL(wi th Ho and 95% Z-confi dence interval for the Mean, and StDev = 25)

BILL

Frequency

10080604020

20

15

10

5

0 _X

Ho

Histogram of BILL(with Ho and 95% Z-confidence interval for the Mean, and StDev = 25)


1 1 34 1 6695 2 014 2 56777788920 3 01233431 3 55677889999(7) 4 112233335 4 5567889927 5 001133420 5 919 6 04416 6 68813 7 1311 711 8 01127 8 77993 9 22 9 78


95%Lower


Although the P-value has been changed from 0.129 to 0.311 by deletingthe two outliers, the conclusion remains the same. We do not rejectthe null hypothesis. Intuitively, we should expect this result, sincedeleting two large outliers can only reduce the sample mean, making theP-value larger.

9.128 (a) The P-value is expressed as P(z < z0) if the hypothesis test is left-tailed.

(b) The P-value is expressed as P(z > z0) if the test is two-tailed.

9.129 (a) Left-tailed: P-value = P(z < z0) = (z0)

(b) Right-tailed: P-value = P(z > z0) = 1 - P(z < z0) = 1 - (z0)

(c) Two-tailed: P-value = P(z > z0)

= P(z > z0) + P(z < - z0)

Section 9.6, Hyp. Tests for One Pop. Mean When Is Unknown 421

By symmetry = P(z > z0) + P(z > z0)

= 2 P(z > z0) = 2 [1 - P(z < z0)]

= 2 [1 - (z0)]

9.130 Given that x can be transformed to z (and x0 to z0), we have:

1. P(x > x0) = P(z > z0), for a right-tailed test

2. P(x < x0) = P(z < z0), for a left-tailed test

3. 2 min {P(x < x0), P(x > x0)} = 2min {P(z < z0), P(z > z0)}

2 0

2 0

0 0

0 0

Pz z

Pz z

( )

( )

if z

if z

By symmetry2 0

2 0

0

0

Pz z

P z z

( )

( )

if z

if z0

0

= 2P(z > z0)

By symmetry = P(z < -z0) + P(z > z0) = P(z > z0)

9.131 The P-value approach provides the actual significance of the hypothesistest, that is, the smallest significance level at which the results of thetest are significant. It also allows the reader to judge the strength ofthe evidence against the null hypothesis for himself/herself. The onlydisadvantage of the P-value approach is that one must compute the P-value(one extra step) after computing the z-value.

Exercises 9.6

9.132 In the z-test, it is assumed that is known. In the t-test, isunknown.

9.133 (a) 0.01 0.025; Do not reject H0 for < 0.01; Undecided for0.01 < < 0.025.

9.134 (a) P > 0.10

(b) Do not reject H0 for < 0.10; Undecided for > 0.10.

9.135 (a) P < 0.005

(b) Reject H0 for > 0.005; Undecided for < 0.005.

9.136 (a) 0.05 0.01; Undecided for < 0.01.

9.139 (a) df = 31; t = -2.828

(b) P < 0.005; Reject H0; Evidence against H0 is very strong.

9.140 (a) df = 31; t = -1.414

(b) 0.05 < P < 0.10; Do not reject H0; Evidence against H0 is moderate.


9.141 (a) df = 14; t = 1.936

(b) 0.05 0.20; Do not reject H0; Evidence against H0 is weak or none.

9.144 (a) df = 23; t = -2.449

(b) 0.02 1.729, we reject H0. The data provide sufficient evidence toconclude that the amount of television watched per day last year by theaverage person differed from that in 2002.

9.146 H0: μ = 180 yards, Ha: μ > 180 yards, = 0.05;

( )/( / ) (182.7 180)/(2.7/ 6) 2.449t x n ; 0.025 < P < 0.05

Since P-value < 0.05, we reject H0. The data provide sufficient evidence toconclude that the club will hit the ball more than 180 years at a club headvelocity of 85 mph.

9.147 n = 10, df = 9, x_

= 2.5, s = 0.149

Step 1: H0: = 2.3, Ha: > 2.3

Step 2: = 0.01

Step 3: 251.410/149.0

3.25.2t

Step 4: Critical value = 2.821

Step 5: Since 4.251 > 2.821, reject H0. Note: For the P-value approach,P-value < 0.01. So, since the P-value < , reject H0.

Step 6: At the 1% significance level, the data do provide sufficientevidence to conclude that the mean available limestone in soiltreated with 100% MMBL effluent is greater than 2.30%. Thepractical significance of this result probably depends on what cropis to be grown in the soil.

9.148 n = 25, df = 24, x_

= $1935.76, s = $350.90

Step 1: H0: = $1749, Ha: $1749

Step 2: = 0.05

Step 3:1935.76 1749

2.661350.9/ 25

t

Step 4: Critical values = +2.064

Step 5: Since 2.661 > 2.064, reject H0. Note: For the P-value approach,0.005 < P-value < 0.01. So, since the P-value < , reject H0.

Step 6: At the 5% significance level, the data do provide sufficientevidence to conclude that the mean annual expenditure on appareland services for consumer units in the Northeast differed from thenational mean in 2002.

Aaron

Note

Ignore the Pvalue discussion here, we only calculate Pvalues for Ztests in our class. The critical value for this test is part a. 5% alpha = 2.015 part b. 1% alpha = 3.365

Aaron

Cross-Out

Aaron

Cross-Out

Aaron

Cross-Out

Aaron

Note

The correct test stat is .3416 which would fail to reject Ho

Aaron

Cross-Out


AGE

Frequency

10090807060

5

4

3

2

1

0

Histogram of AGE

9.149 n = 187, df = 186, x_= 0.64, s = 0.15

Step 1: H0: = 0.9, Ha: < 0.9

Step 2: = 0.05

Step 3: 703.23187/15.0

90.064.0t

Step 4: Critical value = -1.653

Step 5: Since –23.703 < -1.653, reject H0. Note: For the p-valueapproach, P < 0.005. So, since the p-value < , reject H0.

Step 6: At the 5% significance level, the data provide sufficientevidence to conclude that the mean ABI for women withperipheral arterial disease is less than the healthy ABI of 0.9.Thus, we conclude that such women do have an unhealthy ABI. Thepractical significance of this result is that the ABI may be agood tool for determining the possibility of peripheral arterialdisease in women. There could also be other causes of a lowABI, so this test by itself may not be able to determine theprecise ailment.

9.150 n = 200, df = 199, x_= $2480, s = $766

Step 1: H0: = $2528, Ha: < $2528Step 2: = 0.05

Step 3:2480 2528

0.886766/ 200

t

Step 4: Critical value = -1.660Step 5: Since –0.886 > -1.660, do not reject H0. Note: For the p-value

approach, P > 0.10. So, since the P-value < , reject H0.Step 6: At the 5% significance level, the data do not provide sufficient

evidence to conclude that the mean cost of having a baby by AMLis less than the average cost of having a baby in a U.S.hospital.

9.151 We used Minitab to produce the following histogram.

The sample size is only 20. Although there are no outliers, thedistribution is not very close to being normally distributed. It does notappear to be reasonable to use a t-test with these data.

9.152 We used Minitab to produce the following normal probability plot.


BURIALS

Percent

3000200010000-1000-2000

99

95

90

80

70

605040

30

20

10

5

1

Mean

<0.005

292.8

S tDev 586.3N 17AD 3.364

P -V alu e

Probability Plotof BURIALSNormal - 95% CI

SPENDING

Percent

20000150001000050000-5000-10000

99

95

90

80

70

6050

4030

20

10

5

1

Mean

<0.005

3462

S tDev 4182N 27AD 2.069

P -V alu e

Probability Plot of SPENDINGNormal - 95% CI

It is clear from the probability plot that the data are far from beingnormally distributed. A t-test is clearly not appropriate for these datawith a sample size of only 17.

9.153 We used Minitab to produce the following normal probability plot.

9.154 It is clear from the probability plot that the data are far from beingnormally distributed and the largest observation is a potential outlier. At-test is clearly not appropriate for these data with a sample size of only27.

9.155 (a) Using Minitab, with the data in a column named PRESSURE, we choose Stat

Basic Statistics 1-Sample t..., click in the Samples in columns

text box and specify PRESSURE, click in the Test mean text box andenter 80. Click the Options... button, enter 90 in the Confidence leveltext box, click the arrow button at the right of the Alternative drop-down list box and select greater than and click OK. Click on theGraphs button and check the boxes for Histogram of Data and Boxplot of

data. Then click OK twice. Now choose Stat Basic Statistics Normality test and enter PRESSURE in the Variable text box. Click OK.

Then choose Graph Stem-and-Leaf , enter PRESSURE in the Graph

variables text box and click OK. The results are


PRESSURE

Percent

1201101009080706050

99

95

90

80

70

60

5040

30

20

10

5

1

Mean

0.625

81.76StDev 10.32

N 41AD 0.281P-V alu e

Probability Plot ofPRESSURENormal - 95% CI

PRESSURE10090807060

_X

Ho

Boxplot ofPRESSURE(with Ho and 90% t-confi dence interval for the mean)

PRESSURE

Frequency

10090807060

8

6

4

2

0 _X

Ho

Histogram of PRESSURE(with Ho and 90% t-confidence interval for the mean)

Stem-and-leaf of PRESSURE N = 41Leaf Unit = 1.0

1 5 82 6 35 6 56910 7 0033417 7 5677999(8) 8 0011133416 8 589912 9 00013345 9 55591 10 0

(b) The large sample size, lack of potential outliers, and the nearlylinear probability plot all indicate that a t-test is reasonable forthese data.

(c) The first procedure in part (a) also yielded the following testresults:

Test of mu = 80 vs > 80

90%Lower

Variable N Mean StDev SE Mean Bound T PPRESSURE 41 81.7561 10.3242 1.6124 79.6551 1.09 0.141

We see that t = 1.09 and the P-value is 0.141. Since the P-value isgreater than the significance level 0.10, we do not reject the nullhypothesis and conclude that the data do not provide evidence that themean diastolic blood pressure of bus drivers in Stockholm exceeds thenormal pressure of 88 mm Hg.

9.156 (a) Using Minitab, with the data in a column named DISTANCE, we choose Stat


text box and specify DISTANCE, click in the Test mean text box and type80. Click the Options... button, enter 90 in the Confidence level textbox, click the arrow button at the right of the Alternative drop-downlist box and select greater than and click OK. Click on the Graphsbutton and check the boxes for Histogram of Data and Boxplot of data.

Then click OK twice. Now choose Stat Basic Statistics Normality


DISTANCE

Percent

403020100-10

99.9

99

95

90

80706050403020

10

5

1

0.1

Mean

<0.005

11.30St Dev 6.002N 500

AD 7.619P- Value

Probability Plot ofDISTANCENormal - 95% CI

DISTANCE

Frequency

36302418126

80

60

40

20

0 _X

Ho

Histogramof DISTANCE(with Ho and 95% t-confidence interval for the mean)

DISTANCE403020100

_X

Ho

Boxplot of DISTANCE(with Ho and 95% t-confidence interval for the mean)

test and enter DISTANCE in the Variable text box. Click OK. Then

choose Graph Stem-and-Leaf , enter DISTANCE in the Graph variablestext box and click OK. The results are

Stem-and-leaf of DISTANCE N = 500Leaf Unit = 1.0

4 0 111127 0 2222222233333333333333392 0 44444444444444444444444444444455555555555555555555555555555555555+169 0 66666666666666666666666666666666666777777777777777777777777777777+248 0 88888888888888888888888888888888888888889999999999999999999999999+(58) 1 0000000000000000000000000000000000111111111111111111111111194 1 22222222222222222222222222223333333333333333333333333333333135 1 4444444444444445555555555555555555101 1 666666666666666677777777777777777767 1 888888888899999999948 2 000000000001111111129 2 2222333321 2 4444445513 2 6

HI 26, 26, 27, 27, 27, 28, 28, 29, 29, 30, 35, 36

(b) The large sample size makes this set of data a reasonable candidate fora t-test. However, the large number of outliers (12) indicate thatsome caution is appropriate. If there is any doubt, remove the outliersand retest.

(c) The first procedure in part (a) also yielded the following test


results:

Test of mu = 11.9 vs not = 11.9

Variable N Mean StDev SE Mean 95% CI T PDISTANCE 500 11.3020 6.0023 0.2684 (10.7746, 11.8294) -2.23 0.026

The P-value of the test is 0.026, which is less than the significancelevel of 0.05. We reject the null hypothesis and conclude that thedata do provide sufficient evidence that the mean distance driven lastyear differs from that in 2000. Note: If the 12 large outliers areall removed, the t-value changes to –4.34 with a P-value of 0.000, soour conclusion does not change.

9.157 (a) Using Minitab, with the data in a column named DISTANCE, we choose Stat


text box and specify RENT, click in the Test mean text box and enter692. Click the Options... button, enter 95 in the Confidence leveltext box, click the arrow button at the right of the Alternative drop-down list box and select greater than and click OK.


95%Lower

Variable N Mean StDev SE Mean Bound T PRENT 100 703.960 89.846 8.985 689.042 1.33 0.093

The P-value for the test is 0.093, larger than the significance lever0.05, so we do not reject the null hypothesis. The data do not providesufficient evidence to conclude that the mean rent for a two-bedroomunit in Maine is greater than the FMR of $692.

(b) After removing the outlier 405 and following the procedure in part (a),the results are


95%Lower

Variable N Mean StDev SE Mean Bound T PRENT 99 706.980 85.049 8.548 692.786 1.75 0.041

(c) Now the P-value is 0.041, leading to rejection of the null hypothesis.The sample mean increased by $3.02 and the standard deviation decreasedby about 4.8.

(d) The sample size is large in both cases, yet the effect of the outlieris considerable. Caution should be used and perhaps the data should beanalyzed using a method that is not influenced by outlers.

9.158 (a) Neither the z-test nor the t-test is appropriate for small samplescontaining outliers or exhibiting extremely non-normal distributionalshapes.

(b) A non-parametric test that is neither affected by a few outliers nor bynon-normality might be appropriate.

9.159 (a) n = 20, df = 19, t/2 = 1.729, s = 2.291, x_

= 4.835


The 90% confidence interval is 4.835 1.729(2.291)/ 20 (3.949,5.721).

The hypothesized mean (4.66) lies within the confidence interval, so weshould not reject the null hypothesis. Since the test statistic is

(4.835 4.66)/(2.291/ 20) 0.342t , which is less than the critical valueof 1.729, the hypothesis test also leads to the conclusion that weshould not reject the null hypothesis.

(b) n = 25, df = 24, t/2 = 2.064, s = $350.90, x_= $1935.76

The 95% confidence interval is

1935.76 2.064(350.90)/ 25 (1790.91,2080.61).

The hypothesized mean ($1749) lies outside the confidence interval, sowe should reject the null hypothesis. Since the test statistic is

(1935.76 1749)/(350.90/ 25) 2.661t , which is greater than the criticalvalue of 2.064, the hypothesis test also leads to the conclusion thatwe should reject the null hypothesis.

9.160 (a) n = 187, df = 186, -t = -1.660, s = 0.15, x_

= 0.64

The 95% upper confidence bound is 0.64 1.660(0.15)/ 187 0.658.

The hypothesized mean (0.9) lies above the upper confidence bound, sowe should reject the null hypothesis. Since the test statistic is

(0.64 0.9)/(0.15/ 187) 23.703t , which is less than the critical valueof –1.660, the hypothesis test also leads to the conclusion that weshould reject the null hypothesis.

(b) n = 200, df = 199, -t = -1.660, s = $766, x_

= $2480

The 95% upper confidence bound is 2480 1.660(766)/ 200 2569.91.

The hypothesized mean ($2528) lies below the upper confidence bound, sowe not should reject the null hypothesis. Since the test statistic is

(2480 2528)/(766/ 200) 0.886t , which is greater than the criticalvalue of –1.660, the hypothesis test also leads to the conclusion thatwe should not reject the null hypothesis.

9.161 (a) n = 6, df = 5, t = 2.015, s = 2.7, x_

= 182.7

The 95% lower confidence bound is 182.7 2.015(2.7)/ 6 180.479 .

The hypothesized mean (180) lies below the lower confidence bound, sowe should reject the null hypothesis. Since the test statistic is

(182.7 180)/(2.7/ 6) 2.449t , which is greater than the critical valueof 2.015, the hypothesis test also leads to the conclusion that weshould reject the null hypothesis.

The 99% lower confidence bound is 182.7 3.365(2.7)/ 6 178.991.

The hypothesized mean (180) lies above the lower confidence bound, sowe should not reject the null hypothesis. Since the test statistic is

(182.7 180)/(2.7/ 6) 2.449t , which is less than the critical value of3.365, the hypothesis test also leads to the conclusion that we shouldnot reject the null hypothesis.

(b) n = 10, df = 9, t = 2.821, s = 0.149, x_= 2.5

The 99% lower confidence bound is 2.5 2.821(0.149)/ 10 2.367 .

Section 9.7, The Wilcoxon Signed-Rank Test 429

The hypothesized mean (2.30) lies below the lower confidence bound, sowe should reject the null hypothesis. Since the test statistic is

(2.5 2.3)/(0.149/ 10) 4.245t , which is greater than the critical valueof 2.821, the hypothesis test also leads to the conclusion that weshould reject the null hypothesis.

Exercises 9.7

9.162 Technically, nonparametric methods are inferential methods that are notconcerned with parameters. In practice, nonparametric methods are thosethat can be applied without assuming normality.

9.163 The advantages of nonparametric methods are that they do not requirenormality, they make use of fewer and simpler calculations than doparametric methods, and they are resistant to outliers. The disadvantage ofnonparametric methods is that they tend to give less accurate results thanparametric methods when the assumptions underlying the parametric methodsare actually met.

9.164 The population must be symmetric.

9.165 Because the D-value for such a data value equals 0, we cannot attach a signto the rank of |D|.

9.166 (a) Wilcoxon signed-rank test (b) t-test

(c) Neither

9.167 (a) Wilcoxon signed-rank test (b) Wilcoxon signed-rank test

(c) Neither

9.168 It is because the median and the mean are the same when considering asymmetric distribution.

9.169 (a) W0.05 = 30 (b) W0.95 = 8(8 + 1)/2 – 30 = 6

(c) W0.025 = 32 W0.975 = 8(8 + 1)/2 – 32 = 4

9.170 (a) W0.01 = 50 (b) W0.95 = 10(10 + 1)/2 – 50 = 5

(c) W0.005 = 52 W0.995 = 10(10 + 1)/2 – 52 = 39.171 (a) W0.10 = 128 (b) W0.90 = 19(19 + 1)/2 – 128 = 62

(c) W0.05 = 136 W0.995 = 19(19 + 1)/2 – 136 = 549.172 (a) W0.05 = 90 (b) W0.95 = 15(15 + 1)/2 – 90 = 30

(c) W0.025 = 95 W0.975 = 15(15 + 1)/2 – 95 = 25

9.173 H0: μ = 124.9 days, Ha: μ < 124.9 days; n = 8; = 0.05,W0.95 = 8(8 + 1)/2 – 30 = 6

Days D |D| Rank |D| R103 -21.9 21.9 4 -480 -44.9 44.9 5 -579 -45.9 45.9 7 -7

135 10.1 10.1 2 2134 9.1 9.1 1 177 -47.9 47.9 8 -880 -44.9 44.9 5 -5

111 -13.9 13.9 3 -3The sum of the positive ranks is W = 3. This is less than the criticalvalue 6, so we reject the null hypothesis and conclude that the data providesufficient evidence that the mean number of days of ice cover on Lake Wingrais less than it was in the late 1800s.

9.174 H0: μ = 46.7 years, Ha: μ =/46.7 years; n = 8; = 0.05, W0.025 = 32,


W0.975 = 8(8 + 1)/2 – 32 = 4Years D |D| Rank |D| R30.3 -16.4 16.4 7 -747.0 0.3 0.3 1 156.4 9.7 9.7 5 530.5 -16.2 16.2 6 -639.6 -7.1 7.1 4 -447.9 1.2 1.2 2 229.7 -17 17.0 8 -852.5 5.8 5.8 3 3

The sum of the positive ranks is W = 11. This is between the criticalvalues 4 and 32, so we do not reject the null hypothesis and conclude thatthe data do not provide sufficient evidence that the mean number of happylife years has changed from that in the 1900s.

9.1750= 35.7, n = 10, = 0.01

Step 1: H0: = 35.7, Ha: > 35.7Step 2: = 0.01Step 3:

x x- 0=D|D| Rank of |D| Signed Rank R

42 6.3 6.3 3 3

45 9.3 9.3 4 462 26.3 26.3 10 1049 13.3 13.3 6 614 -21.7 21.7 8 -839 3.3 3.3 2 257 21.3 21.3 7 711 -24.7 24.7 9 -936 0.3 0.3 1 126 -9.7 9.7 5 -5

Step 4: W = sum of the + ranks = 33

Step 5: Critical value = 50

Step 6: Since W < 50, do not reject H0.

Step 7: At the 1% significance level, the data do not provide sufficientevidence to conclude that the median age has increased over the2002 median age of 35.7 years. The P-value is 0.305.

9.176 0 = 254, n = 12, = 0.05

Step 1: H0: = 250, Ha: > 250

Step 2: = 0.05


Step 3:

x x- 0 =D|D| Rank of |D| Signed Rank R

395 141 141 12 12

274 20 20 1 1210 -44 44 8 -8307 53 53 10 10218 -36 36 4 -4293 39 39 6.5 6.5299 45 45 9 9283 29 29 3 3293 39 39 6.5 6.5228 -26 26 2 -2315 61 61 11 11292 38 38 5 5

Step 4: W = sum of the + ranks = 64


Step 6: Since W > 61, reject H0.

Step 7: At the 5% significance level, the data do provide sufficientevidence to conclude that last year's mean amount spent byconsumers on nonalcoholic beverages has increased over the 2002mean of $250. The P-value is 0.027.

9.177 0 = $12,850, n = 10, = 0.10

Step 1: H0: = $12,850, Ha: < $12,850

Step 2: = 0.10

Step 3:

Price D |D|Rank |D| R13480 630 630 8 812992 142 142 3 312988 138 138 2 212800 -50 50 1 -112599 -251 251 5 -512499 -351 351 7 -711500 -1350 1350 9 -910400 -2450 2450 10 -1012500 -350 350 6 -612600 -250 250 4 -4

Step 4: W = sum of the + ranks = 13Step 5: Critical value = 10(11)/2 - 41 = 14

Step 6: Since W < 14, reject H0.

Step 7: At the 10% significance level, the data do provide sufficientevidence to conclude that the mean asking price for a 2003 FordMustang is less than the 2006 Kelly Blue Book value. The P-valueis 0.077.

9.178 0 = 7.4, n = 13, = 0.05 (The value of 7.4 was deleted from the original

sample.)

Step 1: H0: = 7.4, Ha: 7.4


Step 2: = 0.05

Step 3:

x x- 0=DD Rank of D Signed Rank R

8.6 1.2 1.2 4 4

8.8 1.4 1.4 5.5 5.5

8.2 0.8 0.8 3 3

5.3 -2.1 2.1 11 -11

9.2 1.8 1.8 9 9

13.8 6.4 6.4 13 13

5.6 -1.8 1.8 9 -9

7.8 0.4 0.4 2 2

6.0 -1.4 1.4 5.5 -5.5

5.7 -1.7 1.7 7 -7

11.6 4.2 4.2 12 12

9.2 1.8 1.8 9 9

7.2 -0.2 0.2 1 -1

Step 4: W = sum of the + ranks = 57.5

Step 5: Critical values = 17, 74 [13(14)/2 - 74 = 17]

Step 6: Since 17 < W < 74, do not reject H0.

Step 7: At the 5% significance level, the data do not provide sufficientevidence to conclude that this year's median birth weight differsfrom the median birth weight of 7.4 in 2002. The P-value is 0.422.

9.179 0 = 2.30, n = 10, = 0.01

(a) Step 1: H0: = 2.30, Ha: > 2.30Step 2: = 0.01

Step 3:

x x-0=D |D| Rank of |D| Signed Rank R

2.41 0.11 0.11 3 -3

2.60 0.30 0.30 8 82.31 0.01 0.01 1 12.51 0.21 0.21 5.5 5.52.54 0.24 0.24 7 72.51 0.21 0.21 5.5 5.52.28 -0.02 0.02 2 -22.42 0.12 0.12 4 42.72 0.42 0.42 10 102.70 0.40 0.40 9 9





Step 7: At the 1% significance level, the data provide sufficientevidence to conclude that the mean available limestone insoil treated with 100% MMBL effluent exceeds 2.30%. TheP-value is 0.005.

(b) A Wilcoxon signed-rank test is permissible because a normallydistributed population is symmetric.

9.180 0 = 2, n = 13, = 0.05 (All values of 2 were deleted from the data.)

(a) Step 1: H0: = 2, Ha: > 2Step 2: = 0.05Step 3:

x x- 0=D D Rank of D Signed Rank R

3 1 1 6.5 6.5

3 1 1 6.5 6.53 1 1 6.5 6.51 -1 1 6.5 -6.53 1 1 6.5 6.53 1 1 6.5 6.53 1 1 6.5 6.54 2 2 13.0 13.03 1 1 6.5 6.51 -1 1 6.5 -6.53 1 1 6.5 6.53 1 1 6.5 6.51 -1 1 6.5 -6.5




Step 7: At the 5% significance level, the data provide sufficientevidence to conclude that Helsinkians do respond with anecological welfare food choice motive greater than 2. The P-value is 0.037.

(b) For the t test, we have n = 18, df = 17, x_

= 44/18 = 2.4444, and s =0.85559.

= 0.05 and the critical value is 1.740.

tx

s n0 2 4444 2

0 85559 182 204

/

.

. /.

Since 2.204 > 1.740, we reject the null hypothesis. There issufficient evidence to conclude that Helsinkians respond with anecological welfare food choice motive greater than 2.

(c) The two tests lead to the same conclusion. However, since the datawere limited to integer values between 1 and 4, it is not possible forthis data to have come from a normal distribution, or even anapproximately normal distribution. The Wilcoxon test is a betterchoice for a test with these data.

9.181 n = 16, x_

= 306, s = 8.671, = 0.05

(a) Step 1: H0: = 310, Ha: < 310


Step 2: = 0.05

Step 3: 845.116/671.8

310306t


Step 5: Since -1.845 < -1.753, reject H0. Note: For the P-valueapproach, 0.025 < p-value < 0.05. So, since P-value < ,reject H0.

Step 6: At the 5% significance level, the data do provide sufficientevidence to conclude that the mean content, μ, is less thanthe advertised content of 310 ml.

(b) Step 1: H0: = 310, Ha: < 310

Step 2: = 0.05

Step 3:

x x- 0 =D|D| Rank of |D| Signed Rank R

297 -13 13 14 -14

311 1 1 2 2

322 12 12 12.5 12.5

315 5 5 7 7

318 8 8 9 9

303 -7 7 8 -8

307 -3 3 5 -5

296 -14 14 15 -15

306 -4 4 6 -6

291 -19 19 16 -16

312 2 2 4 4

309 -1 1 2 -2

300 -10 10 10.5 -10.5

298 -12 12 12.5 -12.5

300 -10 10 10.5 -10.5

311 1 1 2 2



Step 6: Since W > 36, do not reject H0.

Step 7: At the 5% significance level, the data do not providesufficient evidence to conclude that the mean content, μ, isless than the advertised content of 310 ml. The P-value is0.054.

(c) Since the population is normally distributed, the t-test is morepowerful than the Wilcoxon signed-rank test; that is, the t-test ismore likely to detect a false null hypothesis.

9.182 n = 10, x_

= 8.9, s = 2.4698, = 0.10, 0 = 10.2

(a) Step 1: H0: = 10.2, Ha: 10.2

Step 2: = 0.10


Step 3: 664.110/4698.2

2.109.8t

Step 4: Critical values = ±1.833

Step 5: Since -1.833 < -1.664 < 1.833, do not reject H0. Note: Forthe P-value approach, 0.10 < p-value < 0.20. So, since P-value > , do not reject H0.

Step 7: At the 10% significance level, the data do not providesufficient evidence to conclude that this year's medianeducational attainment has changed from the median 25 yearsago.

(b) Step 1: H0: = 10.2, Ha: 10.2

Step 2: = 0.10

Step 3:


14 3.8 3.8 8 8

10 -0.2 0.2 2 -210 -0.2 0.2 2 -210 -0.2 0.2 2 -25 -5.2 5.2 10 -108 -2.2 2.2 6.5 -6.56 -4.2 4.2 9 -99 -1.2 1.2 4.5 -4.58 -2.2 2.2 6.5 -6.59 -1.2 1.2 4.5 -4.5

Step 4: W = sum of the + ranks = 8Step 5: Critical values = 11, 44Step 6: Since W < 11, reject H0.

Step 7: At the 10% significance level, the data do provide sufficientevidence to conclude that this year's median educationalattainment has changed from the median 25 years ago of 10.2years.

(c) Since the population is given to be symmetric and nonnormal, theWilcoxon signed-rank test is more powerful than the t-test and thus ismore likely to detect a false null hypothesis.

9.183 (a) Using Minitab, with the data in a column named SCORES, we choose Stat Nonparametrics 1-Sample Wilcoxon..., select SCORES in the Variablestext box, click in the Test median box and type 5, click the arrowbutton at the right of the Alternative drop down list box, and selectLess than, and click OK. The result is

Test of median = 5.000 versus median < 5.000

Nfor Wilcoxon Estimated

N Test Statistic P MedianSCORE 156 84 777.0 0.000 4.500

Since the P-value of 0.000 is less than the significance level of 0.01,we reject the null hypothesis and conclude that there is very strongevidence that professional golfers score better than par on the HoleO’Cross Out.


(b) To perform the t-test with the original data, we choose Stat Basic

statistics 1-Sample t..., select the Samples in columns box, enter

SCORES in the Samples in Columns text box, enter 5 in the Test Meantext box, click on the Options button, enter 99.0 in the Confidencelevel text box, select less than in the Alternative box, and click OKtwice. The results in the Session Window are

Test of mu = 5 vs < 599%Upper

Variable N Mean StDev SE Mean Bound T PSCORES 156 4.66667 0.76482 0.06123 4.81061 -5.44 0.000

Again, since the P-value of 0.000 is less than the significance levelof 0.01, we reject the null hypothesis and conclude that there is verystrong evidence that professional golfers score better than par on theHole O’Cross Out.

(c) The results are the same for the two tests. The Wilcoxon test is themore appropriate test, however, since the data are discrete and cannothave a normal distribution.

9.184 Using Minitab, with the data in a column named PRESSURE, we choose Stat Nonparametrics 1-Sample Wilcoxon..., select PRESSURE in the Variablestext box, click in the Test median box and type 80, click the arrow buttonat the right of the Alternative drop down list box, and select greater than,and click OK. The result is

Test of median = 80.00 versus median > 80.00


N Test Statistic P MedianPRESSURE 41 39 468.0 0.140 82.00

Since the P-value of 0.140 is greater than the significance level of 0.10,we do not reject the null hypothesis and conclude that there is weak or noevidence that the mean diastolic blood pressure of bus drivers in Stockholmexceeds the normal diastolic blood pressure of 80 mm Hg.

9.185 Using Minitab, with the data in a column named DISTANCE, we choose Stat Nonparametrics 1-Sample Wilcoxon..., select DISTANCE in the Variablestext box, click in the Test median box and type 11.9, click the arrow buttonat the right of the Alternative drop down list box, and select not equal,and click OK. The result is

Test of median = 11.90 versus median not = 11.90


N Test Statistic P MedianDISTANCE 500 497 49193.0 0.000 10.75

Since the P-value of 0.000 is less than the significance level 0.05, wereject the null hypothesis and conclude that the mean miles driven by carslast year differs from the mean distance driven in 2000.

9.186 Using Minitab, with the data in a column named RENT, we choose Stat Nonparametrics 1-Sample Wilcoxon..., select RENT in the Variables text


box, click in the Test median box and type 692, click the arrow button atthe right of the Alternative drop down list box, and select greater than,and click OK. The result is

Test of median = 692.0 versus median > 692.0


N Test Statistic P MedianRENT 100 99 2873.5 0.082 705.0

Since the P-value of 0.082 is greater than the significance level of 0.05,we do not reject the null hypothesis and conclude that the data do notprovide sufficient evidence that the mean monthly rent for two bedroom unitsin Maine is greater than the FMR of $692.

9.187 The Wilcoxon signed-rank test is likely to give better results. Thedistribution of marriage durations is unlikely to be normal, possibly noteven symmetric. Given that duration cannot be less than 0 years and thereare likely to be some fairly long marriages which might look like outliers,it would be better to use the Wilcoxon signed-rank test which is insensitiveto outliers than to use the t-test which assumes normality and is sensitiveto outliers.

9.188 The distribution of times to complete the U.S. Census Form may beapproximately normal, but it is also likely that there will be some peoplewho unfamiliar with the form, don’t know the answers to some questions, havepoor eyesight, or are very slow or poor readers, but one would not expectthat there would be many extremely long times recorded. An examination ofthe data might prove helpful in evaluating the symmetry of the population.Given the uncertainty of the normality assumption for the t-test and thepotential for outliers, it would be better to use the Wilcoxon signed-ranktest.

9.189 (a) If John is not unlucky, he should expect to wait 15 minutes for thetrain, on the average.

(b) If John is not unlucky, the distribution of the times he waits for thetrains should be a uniform distribution over the interval from 0 to 30minutes.

(c) Step 1: H0: = 15, Ha: > 15

Step 2: = 0.10

Step 4:

x x- 0=DD Rank of D Signed Rank

R

24 9 9 6.5 6.5

26 11 11 9.5 8.520 5 5 3.5 3.54 -11 11 9.5 -9.520 5 5 3.5 3.53 -12 12 11 -1119 4 4 2 45 -10 10 8 -828 13 13 12 1216 1 1 1 122 7 7 5 524 9 9 6.5 6.5




Step 6: Since W = 50.5 < 56, do not reject H0.

Step 7: At the 10% significance level, the data do not providesufficient evidence to conclude that John waits more than 15minutes for the train, on the average.

(d) Since the population is uniform (which is symmetric), the Wilcoxon testis appropriate.

(e) Since the population is symmetric and nonnormal, the Wilcoxon signed-rank test is more powerful than the t-test and more appropriate thanthe t-test, which assumes normality.

9.190 Step 1: State the null and alternative hypotheses.

Step 2: Decide on the significance level .

Step 3: Construct a worktable of the form:

DataValueX

DifferenceD = x - 0 D

Rankof D

SignedRankR

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Step 4: Compute the value of the test statistic:

( 1)/ 4

( 1)(2 1)/ 24

W n nzn n n

where W is the sum of the positive ranks.

Step 5: The critical value(s):

(a) for a two-tailed test are /2z .

(b) for a left-tailed test is - z(c) for a right-tailed test is z .

Use Table II to find the critical value(s).

Step 6: If the value of the test statistic falls in the rejection region,reject H0; otherwise, do not reject H0.

Step 7: State the conclusion in words.

9.191 (a) Step 1: H0: = 7.4 lb, Ha: 7.4 lb

Step 2: = 0.05

Step 3: See Step 3 in the solution to Exercise 9.178.

Step 4: From Step 4 in the solution to Exercise 9.178, W = 57.5. Now:

84.024/)1132)(113(13

4/)113(135.57

24/)12)(1(

4/)1(

nnnnnWz

Step 5: Critical values = ±1.96

Step 6: Since -1.96 < 0.84 < 1.96, do not reject H0.

Step 7: At the 5% significance level, the data do not provide sufficientevidence to conclude that this year's median birth weightdiffers from that in 2000.


6543210

0. 25

0. 20

0. 15

0. 10

0. 05

0. 00

W

FREQ

(b) Neither the Wilcoxon signed-rank test nor the normal approximation ledto rejection of the null hypothesis.

9.192 (a) Summing the ranks corresponding to the "+" signs in each row results ina value for W. All possible values for W are presented in the lastcolumn.

Rank

1 2 3 W

+ + + 6

+ + - 3

+ - + 4

+ - - 1

- + + 5

- + - 2

- - + 3

- - - 0

(b) Since there are eight equally likely outcomes, the probability that asample will match any particular row of the table is 1/8 = 0.125.

(c) The probability distribution (d) A histogram for the probability

of the random variable W distribution of W for n = 3 is

when n = 3 is:

(e) For left-tailed tests with n = 3 and = 0.125, the critical valueW = 0.

9.193 Summing the ranks corresponding to the "+" signs in each row results in avalue for W. All sixteen possible values for W are presented in the lastcolumn.

W P(W)

0 0.125

1 0.125

2 0.125

3 0.250

4 0.125

5 0.125

6 0.125


109876543210

0.10

0.05

0.00

W

FREQ

(a) (b) 1/16 = 0.0625

Rank (c)

1 2 3 4 W W P(W)

+ + + + 10 0 0.0625

+ + + - 6 1 0.0625

+ + - + 7 2 0.0625

+ + - - 3 3 0.1250

+ - + + 8 4 0.1250

+ - + - 4 5 0.1250

+ - - + 5 6 0.1250

+ - - - 1 7 0.1250

- + + + 9 8 0.0625

- + + - 5 9 0.0625

- + - + 6 10 0.0625

- + - - 2

- - + + 7

- - + - 3

- - - + 4

- - - - 0

(d)

(e) For a left-tailed test with n = 4 and = 0.125, the critical value Wequals 1 since W = 0 or 1 has probability 0.125.

9.194 (a) 0 is the hypothesized median. The median is that value which has halfof the population to its left and half to its right. Therefore, theprobability that a value exceeds the median is 0.5.

(b) Each observation either exceeds 0 or it doesn’t. The observations areindependent of each other and the probability p that an observationexceeds 0 is 0.5 for each observation. The number of observations(trials) is fixed at n. Thus, the number of observations exceeding 0satisfies all of the criteria for a binomial distribution.

9.195 (a) The Wilcoxon signed-rank test takes into account the sign and theabsolute size of each difference from the median, whereas, the sign


test only considers the sign. Since the Wilcoxon test uses moreinformation than the sign test, it is more likely to be able to detecta false null hypothesis.

(b) The sign test can be used with any distribution since the probabilitythat an observation exceeds the median is always 0.5, regardless of theshape of the distribution. The Wilcoxon signed-rank test is based onan assumption that the underlying distribution of the data issymmetric, a slightly more restrictive assumption than for the signtest.

9.196 If one or more observations equals 0, delete those observations andcontinue with the testing process, using a reduced sample size.

9.197 (a) For the sign test, x = 2 of the n = 8 observations are above 124.9.The P-value is P(x < 2) for a binomial distribution with n = 8 and p =0.5. From Table XII, this probability is 0.004 + 0.031 + 0.109 =0.144. Since this is greater than the 0.05 significance level, thedata do not provide sufficient evidence that the number of days thelake was frozen over is less now than in the late 1800s.

(b) The Wilcoxon signed rank test had W = 3.0 with a P-value of 0.021.Since this is less than the 0.05 significance level, the data doprovide sufficient evidence that the number of days the lake was frozenover is less now than in the late 1800s.

9.198 (a) For the sign test, x = 4 of the n = 8 observations are above 46.7. TheP-value is 2min{P(x < 4), P(x < 4)} for a binomial distribution with n= 8 and p = 0.5. From Table XII, both probabilities in the {} are0.636. Multiplying by 2, we get a probability greater than 1, which wejust adjust to 1.000. Since this is greater than the 0.05 significancelevel, the data do not provide sufficient evidence that the mediannumber of happy-life years has changed from that in the 1900s.

(b) The Wilcoxon signed rank test had W = 11.0 with a P-value of 0.363.Since this is greater than the 0.05 significance level, the data do notprovide sufficient evidence that the median number of happy-life yearshas changed from that in the 1900s. Same conclusion, smaller P-value.

9.199 (a) For the sign test, x = 7 of the n = 10 observations are above 35.7.The P-value is P(x > 7)} for a binomial distribution with n = 10 and p= 0.5. From Table XII, this probability is 0.117 + 0.044 + 0.010 +0.001 = 0.172. Since this is greater than the 0.05 significance level,the data do not provide sufficient evidence that the median age oftoday’s U.S. residents has increased from the 2002 median of 35.7years.

(b) The Wilcoxon signed rank test had W = 33.0 with a P-value of 0.363.Since this is greater than the 0.05 significance level, the data do notprovide sufficient evidence that the median age of today’s U.S.residents has increased from the 2002 median of 35.7 years. Sameconclusion, larger P-value.

9.200 (a) For the sign test, x = 9 of the n = 12 observations are above 254. TheP-value is P(x > 9)} for a binomial distribution with n = 12 and p =0.5. From Table XII, this probability is 0.054 + 0.016 + 0.003+ 0.000= 0.073. Since this is greater than the 0.05 significance level, thedata do not provide sufficient evidence that last year’s mean amountspent by consumers on nonalcoholic beverages has increased from the2002 mean of $254.

(b) The Wilcoxon signed rank test had W = 64.0 with a P-value of 0.027.Since this is less than the 0.05 significance level, the data do


provide sufficient evidence that last year’s mean amount spent byconsumers on nonalcoholic beverages has increased from the 2002 mean of$254. Different conclusion, smaller P-value.

9.201 (a) For the sign test, x = 3 of the n = 10 observations are above $12850.The P-value is P(x < 3)} for a binomial distribution with n = 10 and p= 0.5. From Table XII, this probability is 0.001 + 0.044 + 0.010 +0.001 = 0.172. Since this is greater than the 0.10 significance level,the data do not provide sufficient evidence that the mean asking pricefor 2003 Ford Mustang coupes in Phoenix is less than the 2006 KelleyBlue Book retail value of $12850.

(b) The Wilcoxon signed rank test had W = 13.0 with a P-value of 0.077.Since this is less than the 0.10 significance level, the data doprovide sufficient evidence that the mean asking price for 2003 FordMustang coupes in Phoenix is less than the 2006 Kelley Blue Book retailvalue of $12850. Different conclusion, smaller P-value.

9.202 (a) For the sign test, x = 8 of the n = 13 observations (the observation7.4 was deleted) are above 7.4 lbs. The P-value is 2min{P(x < 8), P(X> 8)}} for a binomial distribution with n = 13 and p = 0.5. From TableXII, P(X < 80) =.866 and P(X > 8) = 0.291. Therefore, P = 2(0.291) =0.582. Since this is greater than the 0.05 significance level, thedata do not provide sufficient evidence that this year’s median birthweight differs from the 2002 median birth weight.

(b) The Wilcoxon signed rank test had W = 57.5 with a P-value of 0.422.Since this is greater than the 0.10 significance level, the data do notprovide sufficient evidence that this year’s median birth weightdiffers from the 2002 median birth weight. Same conclusion, smaller P-value.

Exercises 9.8

9.203 (a) One-mean z-test, One-mean t-test, Wilcoxon signed-rank test

(b) The z-test assumes that is known, and that the population is normalor the sample is large. The t-test assumes that is unknown, andthat the population is normal or the sample is large. The signed-ranktest assumes only that the population is symmetric.

(c)0 0

( )/( / ) t ( )/( / )z x n x s n

W = the sum of the positive ranks

9.204 (a) Yes. The t-test can be used when the population is normal and thepopulation standard deviation is unknown.

(b) Yes. The normal distribution is symmetric and that is all that isrequired in order to use the Wilcoxon signed-rank test.

(c) The t-test is preferable in this situation since it is more powerful(more likely to detect a false null hypothesis) when the population isnormal.

9.205 (a) Yes. The t-test can be used when the sample size is large. It isalmost equivalent to the z-test in this situation.

(b) Yes. The Wilcoxon signed-rank test can be used when the populationdistribution is symmetric.

(c) The Wilcoxon signed-rank test is preferable in this situation since itis more powerful (more likely to detect a false null hypothesis) whenthe population is symmetric, but nonnormal.

9.206 (a) Yes. It is permissible to use the z-test when the sample size is large,

Chapter 9 Review Problems 443

even if the population is skewed.

(b) No. The Wilcoxon signed-rank test is valid when the populationdistribution is symmetric, but not when it is highly skewed.

9.207 Since we have normality and is known, use the z-test.

9.208 Since we have a large sample, use the z-test.

9.209 Since we have a large sample with no outliers and is unknown, use the t-test.

9.210 Since we have a symmetric non-normal distribution, use the Wilcoxon signed-rank test.

9.211 Since we have a symmetric non-normal distribution, use the Wilcoxon signed-rank test.

9.212 Since we have a symmetric distribution, use the Wilcoxon signed-rank test.

9.213 The distribution looks skewed and the sample size is not large. Consult astatistician.

9.214 The distribution looks skewed and the sample size is not large. Consult astatistician.

Review Problems for Chapter 9

1. (a) A null hypothesis always specifies a single value for the parameter ofa population which is of interest.

(b) The alternative hypothesis reflects the purpose of the hypothesis test,which can be to determine that the parameter of interest is greaterthan, less than, or different from the single value specified in thenull hypothesis.

(c) The test statistic is a quantity calculated from the sample, under theassumption that the null hypothesis is true, which is used as a basisfor deciding whether or not to reject the null hypothesis.

(d) The rejection region is a set of values of the test statistic that leadto rejection of the null hypothesis.

(e) The nonrejection region is a set of values of the test statistic thatlead to not rejecting the null hypothesis.

(f) The critical values are values of the test statistic that separate therejection region from the nonrejection region.

2. (a) The statement is expressing the fact that there is variability in thenet weights of the boxes’ content and some boxes may actually containless than the printed weight on the box. However, the net weights foreach day’s production will average a bit more than the printed weight.

(b) To test the truth of this statement, we would use a null hypothesisthat stated that the population mean net weight of the boxes was equalto the printed weight and an alternative hypothesis that stated thatthe population mean net weight of the boxes was greater than theprinted weight.

(c) Null hypothesis: Population mean net weight = 76 oz

Alternative hypothesis: Population mean net weight > 76 oz

or

H0: = 76 Ha: > 76

3. (a) Roughly speaking, there is a range of values of the test statisticwhich one could reasonably expect to occur if the null hypothesis weretrue. If the value of the test statistic is one that would not be


expected to occur when the null hypothesis is true, then we reject thenull hypothesis.

(b) To make this procedure objective and precise, we specify theprobability with which we are willing to reject the null hypothesiswhen it is actually true. This is called the significance level of thetest and is usually some small number like 0.05 or 0.01. Specifyingthe significance level allows us to determine the range of values ofthe test statistic that will lead to rejection of the null hypothesis.If the computed value of the test statistic falls in this “rejectionregion,” then the null hypothesis is rejected. If it does not fall inthe rejection region, then the null hypothesis is not rejected.

4. We would use the alternative hypothesis 0 if we wanted to determine

whether the population mean were different from the value μ0 specified in

the null hypothesis. We would use the alternative hypothesis > 0 if we

wanted to determine whether the population mean were greater than from thevalue μ0 specified in the null hypothesis. We would use the alternative

hypothesis < 0 if we wanted to determine whether the population mean

were less than the value μ0 specified in the null hypothesis.

5. (a) A Type I error is made whenever the null hypothesis is true, but thevalue of the test statistic leads us to reject the null hypothesis. AType II error is made whenever the null hypothesis is false, but thevalue of the test statistic leads us to not reject the null hypothesis.

(b) The probability of a Type I error is represented by and that of aType II error by .

(c) If the null hypothesis is true, the test statistic can lead us toeither reject or not reject the null hypothesis. The first is thecorrect decision, while the latter constitutes a Type I error. Thus aType I error is the only type of error possible when the nullhypothesis is true.

(d) If the null hypothesis is not rejected, a correct decision has beenmade if the null hypothesis is, in fact, true. But if the nullhypothesis is false, we have made a Type II error. Thus a Type IIerror is the only type of error possible when the null hypothesis isnot rejected.

6. Assuming that the null hypothesis is true, find the value of the teststatistic for which the probability of obtaining a value greater than thespecified value is 0.05.

7. (a) If the population standard deviation is unknown, and the population isnormal or the sample size is large, we can use the one-mean t-

statistic, )//()( 0 nsxt .

(b) If the population standard deviation is known, and the population isnormal or the sample size is large, we can use the one-mean z-

statistic, )//()( 0 nxz .

(c) If the population is symmetric, we can use the Wilcoxon signed-rankstatistic W = the sum of the positive ranks.

8. (a) A hypothesis test is exact if the actual significance level is the sameas the one that is stated.

(b) A hypothesis test is approximately correct if the actual significancelevel only approximately equals .


9. A statistically significant result occurs when the value of the teststatistic falls in the rejection region. A result has practicalsignificance when it is statistically significant and the result also isdifferent enough from results expected under the null hypothesis to beimportant to the consumer of the results. By taking large enough samplesizes, almost any result can be made statistically significant due to theincreased ability of the test to detect a false null hypothesis, but smalldifferences from the conditions expressed by the null hypothesis may not beimportant, that is, they may not have practical significance.

10. The probability of a Type II error is increased when the significance levelis decreased for a fixed sample size.

11. (a) The power of a hypothesis test is the probability of rejecting the nullhypothesis when the null hypothesis is false.

(b) The power of a test increases when the sample size is increased whilekeeping the significance level constant.

12. (a) The P-value of a hypothesis test is the probability, assuming that thenull hypothesis is true, of getting a value of the test statistic thatis as extreme or more extreme than the one actually obtained.

(b) True. If the null hypothesis were true, a value of the test statisticwith a P-value of 0.02 would be more extreme than one with a P-value of0.03.

(c) True. If the P-value is 0.74, this means that 74% of the time when thenull hypothesis is true, the value of the test statistic would be moreextreme than the one actually obtained.

(d) The P-value of a hypothesis test is also called the observedsignificance level since it represents the smallest possiblesignificance level at which the null hypothesis could have beenrejected.

13. In the critical-value approach, the null hypothesis is rejected if the valueof the test statistic falls in the rejection region that is determined bythe chosen significance level. In the P-value approach, the test statisticis computed and then the probability of obtaining a value as extreme or moreextreme than the one actually obtained is found. This is the P-value. Theadvantages of providing the P-value are that the observed significance levelof the of the test is given and the reader of the results can determine forhim/herself whether the results are strong enough evidence against the nullhypothesis to reject it.

14. Non-parametric methods have the advantages of involving fewer and simplercalculations than parametric methods and are more resistant to outliers andother extreme values. Parametric methods are preferred when the populationis normal or the sample size is large since they are more powerful than non-parametric methods and thus tend to give more accurate results than non-parametric methods under those conditions.

15. Let μ denote last year's mean cheese consumption by Americans.

(a) H0: = 30.0 lb

(b) Ha: > 30.0 lb

(c) This is a right-tailed test.

16. (a) Rejection region: z 1.28





(e)

(f) Right-tailed test

17. (a) A Type I error would occur if, in fact, = 30.0 lb, but the resultsof the sampling lead to the conclusion that > 30.0 lb.

(b) A Type II error would occur if, in fact, > 30.0 lb, but the resultsof the sampling fail to lead to that conclusion.

(c) A correct decision would occur if, in fact, = 30.0 lb and theresults of the sampling do not lead to the rejection of that fact; orif, in fact, > 30.0 lb and the results of the sampling lead to thatconclusion.

(d) If, in fact, last year's mean consumption of cheese for all Americanshas not increased over the 2001 mean of 30.0 lb, and we do not rejectthe null hypothesis that = 30.0 lb, we made a correct decision.

(e) If, in fact, last year's mean consumption of cheese for all Americanshas increased over the 2001 mean of 30.0 lb, and we fail to reject thenull hypothesis that = 30.0 lb, we made a Type II error.

18. (a) P(Type I error) = significance level = = 0.10

(b) The distribution of x_will be approximately normal with a mean of 30.0

and a standard deviation of 17.135/9.6 .

(c) Note: zx

nx z n0

0//


equivalently if 49.3135/)9.6(28.100.30x .

So reject H0 if x_


If = 30.5, then

P(Type II error) = P(x_

< 31.49)


0 1.28 z

0.9000 0.1000

Critical Value

Nonrejection region |Rej ectionreg ion


Power

0.0

0.2

0.4

0.6

0.8

1.0

1.2

30 31 32 33 34 35

Truem ean μ

Power

= P(z < (31.49 – 30.50)/(6.9/ 35 )

= P(z < 0.85) = 0.8023

(d-e) Assuming that the true mean μ is one of the values listed, the

distribution of x_

will be approximately normal with that mean and with

a standard deviation of 1.166. The computations of and the power 1

– are shown in the table below.


computation 1 –

30.5 85.035/9.6

50.3049.310.8023 0.1977

31.0 42.035/9.6

00.3149.310.6628 0.3372

31.5 01.035/9.6

50.3149.310.4960 0.5040

32.0 44.035/9.6

00.3249.310.3300 0.6700

32.5 87.035/9.6

50.3249.310.1922 0.8078

33.0 29.135/9.6

00.3349.310.0985 0.9015

33.5 72.135/9.6

50.3349.310.0427 0.9573

34.0 15.235/9.6

00.3449.310.0158 0.9842

(f)


(g) The distribution of x_will be approximately normal with a mean of 30.0

and a standard deviation of 6 9 60 0 891. / . .

(h) Note: zx

nx z n0

0//


equivalently if .14.3160/)9.6(28.100.30x

So reject H0 if x_


If = 30.5, then

P(Type II error) = P(x_

< 31.14)

= P(z < (31.14 - 30.5)/(6.9/ 60 )

= P(z < 0.72) = 0.7642

(i-j)Assuming that the true mean is one of the values listed, the

distribution of x_

will be approximately normal with that mean and witha standard deviation of 0.891. The computations of and the power 1 – are shown in the following table.


computation 1 –

30.5 72.060/9.6

50.3014.31z 0.7642 0.2358

31.0 16.060/9.6

00.3114.31z 0.5636 0.4364

31.5 40.060/9.6

50.3114.31z 0.3446 0.6554

32.0 97.060/9.6

00.3214.31z 0.1660 0.8340

32.5 53.160/9.6

50.3214.31z 0.0630 0.9370

33.0 09.260/9.6

00.3314.31z 0.0183 0.9817

33.5 65.260/9.6

50.3314.31z 0.0040 0.9960

34.0 21.360/9.6

00.3414.31z 0.0007 0.9993


Power

0.0

0.2

0.4

0.6

0.8

1.0

1.2

30 31 32 33 34 35

TrueMe anμ

Power

(k)

(l) The principle being illustrated is that increasing the sample size fora hypothesis test without changing the significance level increasesthe power.

19. (a) n = 35, x_= 1078/35 = 30.8, = 6.9

Step 1: H0: = 30.0 lb, Ha: > 30.0 lb

Step 2: = 0.10

Step 3: z = (30.8 - 30.0)/(6.9/ 35 ) = 0.69

Step 4: Critical value = 1.28

Step 5: Since 0.69 < 1.28, do not reject H0.

Step 6: At the 10% significance level, the data do not providesufficient evidence to conclude that last year's mean cheeseconsumption for all Americans has increased over the 2001mean of 30.0 lb.

(b) Given the conclusion in part (a), if an error has been made, it must bea Type II error. This is because, given that the null hypothesis wasnot rejected, the only error that could be made is the error of notrejecting a false null hypothesis.

20. (a) Step 1: H0: = 30.0, Ha: > 30.0

Step 2: = 0.10

Step 3: z = 0.69

Step 4: P = 1 - 0.7549 = 0.2451


Step 6: At the 10% significance level, the data do not providesufficient evidence to conclude that last year's mean cheeseconsumption for all Americans has increased over the 2001mean of 30.0 lb.

(b) Using Table 9.12, we classify the strength of evidence against the nullhypothesis as weak or none because P > 0.10.

21. n = 12, x_= $284.10, s = $86.90


Step 1: H0: = $332, Ha: < $332

Step 2: = 0.05

Step 3: t = (284.1 – 332)/(86.90/ 12 ) = -1.909


Step 5: Since -1.909 < -1.782, reject H0.

Step 6: At the 5% significance level, the data do provide sufficientevidence to conclude that the mean value lost because of pursesnatching has decreased from the 2002 mean of $332.

22. (a) n=12

H0: = 332; Ha: < 332

= 0.05

Critical value = 12(13)/2 - 61 = 17


207 -125 125 10 -10237 -95 95 7 -7

422 90 90 6 6

226 -106 106 9 -9

272 -60 60 3 -3

205 -127 127 11 -11

362 30 30 2 2

348 16 16 1 1

165 -167 167 12 -12

266 -66 66 5 -5

269 -63 63 4 -4

430 98 98 8 8

W = sum of the positive signed ranks = 17

Since 17 is less than or equal to the critical value of 17, we rejectthe null hypothesis and conclude that there is evidence that lastyear’s mean value lost to purse snatching has decreased from the 2002mean.

(b) In performing the Wilcoxon signed-rank test, we are assuming that thedistribution of last year’s values lost to purse snatching issymmetric.

(c) If the distribution of values lost is, in fact, a normal distribution,it is permissible to use the Wilcoxon test since a normal distributionis also symmetric.

23. If the values lost last year do have a normal distribution, the t-test isthe preferred procedure for performing the hypothesis test since it isthe more powerful test when the distribution is normal, that is, it hasa greater chance of rejecting a false null hypothesis.

24. (a) If the odds-makers are estimating correctly, the mean point-spreaderror is zero.

(b) It seems reasonable to assume that the distribution of point spreaderrors is approximately normal. In any case, the sample size of 2109is very large, so the t-test of H0: = 0 vs. Ha: 0 is


appropriate. At the 5% significance level, the critical values are

±1.960. Since ( 0.2 0.0)/(10.9 / 2109) 0.843t , we do not reject H0.

(c) There is not sufficient evidence to conclude that the mean point-spreadis different from zero.

25. Since the distribution is symmetric, use the Wilcoxon signed-rank test. Thesample size is 50 and is known, so the z-test may be appropriate. Usecaution, however, since it appears that there may be outliers.

26. The distribution is far from normal, being left-skewed. However, since thesample size is 37 and is unknown, it is probably reasonable to use the t-test.

27. (a) The Wilcoxon signed-rank test is appropriate in Exercise 25 since thedistribution is symmetric. It is not appropriate in Exercise 26 sincethe distribution is highly skewed to the left.

(b) The sample size is large (n=50) and the distribution is symmetric inExercise 26, so either the z-test or the Wilcoxon signed-rank testcould be used. Since the distribution appears to be more peaked withlonger tails than a normal distribution would have, the Wilcoxon testis preferable.

28. Using Minitab, we choose Stat Basic Statistics 1-Sample t..., click in

the Samples in columns text box, enter COST in the Samples in columns textbox, click in the Test mean text box and enter 168 in the Test mean textbox, click on the Options button, type 90.0 in the Confidence level textbox, click on the arrow to the right of the Alternative box and select

greater than, and click OK. Then choose Stat Nonparametrics 1-Sample

Wilcoxon..., enter COST in the Variables text box, click on the Test medianbutton and type 168 in its text box, select greater than in the Alternativetext box, and click OK. Click on the Graphs button and check the boxes for

Histogram of data and Boxplot of data. Click OK twice. Then choose Stat Basic Statistics Normality test, enter COST in the Variable text box and

click OK. Finally, choose Graph Stem-and-Leaf and enter COST in the

Graph Variables text box and click OK. The results are

(a) Test of mu = 168 vs > 168

95%Lower

Variable N Mean StDev SE Mean Bound T PCOST 11 185.182 55.496 16.733 154.855 1.03 0.164

The p-value is greater than the significance level of 0.10, so we do not rejectthe null hypothesis. There is not sufficient evidence to claim that theaverage cost of a private room in a nursing home exceeded $168 per day.

(b) Test of median = 168.0 versus median > 168.0


N Test Statistic P MedianCOST 11 11 48.0 0.099 186.8

The p-value is less than the significance level of 0.10, so we reject the nullhypothesis. There is sufficient evidence to claim that the average cost of aprivate room in a nursing home exceeded $168 per day.


COST

Frequency

30025020015010050

6

5

4

3

2

1

0

-1

_X

Ho

HistogramofCOST(with Ho and 95% t-confidence interval for the mean)

COST300250200150100

_X

Ho

Boxplot of COST(with Ho and 95% t-confidence interval for the mean)

COST

Percent

35030025020015010050

99

95

90

80

70

6050

4030

20

10

5

1

Mean

0.398

185.2StDev 55.50N 11

AD 0.352P- Value

Probability Plot of COSTNormal

Stem-and-leaf of COST N = 11 Leaf Unit = 10

LO 7

2 1 23 1 53 1(5) 1 888993 2 02 22 2 5

HI 28

(d) Although the distribution may be symmetric, it is not bell-shaped andit contains two potential outliers (see boxplot). Since the t-testshould not be used when the sample size is small (11) and there areoutliers, the Wilcoxon test is the more appropriate one to use.

29. (a,b) Using Minitab, choose Stat Basic Statistics 1-Sample t, enter

CONSUMPTION in the Samples in columns text box and 64.5 in the Testmean text box. Click on the Graphs button and check the boxes forHistogram of data and Boxplot of data. Click OK. Click on theOptions button, enter 95 in the Confidence level text box and selectLess than from the Alternative drop down box. Click OK twice. Then

choose Stat Basic Statistics Normality test and enter CONSUMPTION

in the Variable text box and click OK. Finally, choose Graph Stem-

and-Leaf and enter CONSUMPTION in the Graph Variables text box andclick OK. The results are


CONSUMPTION

Frequency

806040200

15.0

12.5

10.0

7.5

5.0

2.5

0.0 _X

Ho

Histogramof CONSUMPTION(with Ho and 95% t-confidence interval for the mean)

CONSUMPTION9080706050403020100

_X

Ho

Boxplot of CONSUMPTION(with Ho and 95% t-confi dence interval for the mean)

CONSUMPTION

Percent

120100806040200

99

95

90

80

70

6050

4030

20

10

5

1

Mean

<0.005

58.4

StDev 20.42N 40

AD 2.212P-V alu e

Probability Plot of CONSUMPTIONNormal

Stem-and-leaf of CONSUMPTION N = 40Leaf Unit = 1.0

LO 0, 0, 8, 20

5 3 75 47 4 7910 5 01416 5 666667(7) 6 011222317 6 5778912 7 12348 7 55677891 81 8 9


Variable N Mean StDev SE Mean 95% CI T PCONSUMPTION 40 58.4000 20.4172 3.2282 (51.8703, 64.9297) -1.89 0.066

The P-value of the test is 0.066, which is greater than the 0.05significance level. We do not reject H0. The data do not providesufficient evidence to conclude that the mean beef consumption thisyear is less than the 2002 mean of 64.5 lbs.

(c) After removing the four outliers, the test results are


Variable N Mean StDev SE Mean 95% CI T PCONSUMPTION 36 64.1111 11.0163 1.8360 (60.3837, 67.8385) -0.21 0.833

(d) The outliers had a very large effect on the test results. Although thesample size was 40, four outliers is too many for the t-test to beappropriate. If the outliers are not recording errors, they representlegitimate observations from the population and should not be deleted.If they were deleted, the test results would not yield validconclusions about the population. With so many outliers, it is justpossible that the population itself is quite left skewed as is thesample.


(e) This is more appropriately done with a non-parametric test that isinsensitive to outliers.

30. (a) Using Minitab, choose Stat Nonparametrics 1-Wilcoxon, enter

CONSUMPTION in the Samples in columns text box and 64.5 in the Testmean text box. Click on the Options button, enter 95 in theConfidence level text box and select Less than from the Alternativedrop down box. Click OK twice. The results areTest of median = 64.50 versus median < 64.50


N Test Statistic P MedianCONSUMPTION 40 40 324.5 0.127 62.00

After removing the outliers and repeating part (a), the results areTest of median = 64.50 versus median < 64.50


N Test Statistic P MedianCONSUMPTION 36 36 324.5 0.450 64.50

(b) The results are similar to those in Problem 29 in that the P-valueincreases substantially. The results are different in that the P-value increases from 0.066 to 0.833 for the t-test, and only from 0.127to 0.450 for the Wilcoxon test when the four outliers are deleted. Inboth cases, we should expect the P-value to increase since the foursmallest observations were deleted and the alternative hypothesis wasμ < 64.5. At the 5% significance level, the conclusion does not changein either case.

(c) While the Wilcoxon test is less sensitive to outliers, it is based onan assumption that the data are symmetrically distributed. Thatassumption does not appear to be reasonable for these data, so cautionis advised in using the Wilcoxon test. The sign test would be a betterchoice.

31. (a) Using Minitab, choose Stat Basic Statistics 1-Sample z, enter BMI

in the Samples in columns text box and 25 in the Test mean text box.Click on the Graphs button and check the boxes for Histogram of dataand Boxplot of data. Click OK. Click on the Options button, enter 95in the Confidence level text box and select Greater than from the

Alternative drop down box. Click OK twice. Then choose Stat Basic

Statistics Normality test and enter BMI in the Variable text box and

click OK. Finally, choose Graph Stem-and-Leaf and enter BMI in the

Graph Variables text box and click OK. The graphic results are


BMI

Frequency

3530252015

16

12

8

4

0 _X

Ho

Histogramof BMI(with Ho and 95% Z-confidence i nterval for the Mean, and StDev = 5)

BMI403530252015

_X

Ho

Boxplot of BMI(with Ho and 95% Z-confidence interval for the Mean, and StDev = 5)

BMI

Percent

4540353025201510

99.9

99

9590

80706050403020

10

5

1

0.1

Mean

0.517

25. 99StDev 5.029N 75

AD 0.325P- Value

Probability Plot ofBMINormal

Stem-and-leaf of BMI N = 75Leaf Unit = 1.0

1 1 44 1 6776 1 8816 2 000011111129 2 2222222233333(9) 2 44444445537 2 66666677728 2 8888889999917 3 000011110 3 222335 3 45551 3 7

(b) Yes. There are no outliers, the sample size is large, and the data isreasonably normally distributed.

(c) The process in part (a) yielded the following test results:

Test of mu = 25 vs > 25The assumed standard deviation = 5

95%Lower

Variable N Mean StDev SE Mean Bound Z PBMI 75 25.9867 5.0293 0.5774 25.0370 1.71 0.044

Since the P-value of 0.044 is less than the significance level 0.05, wereject the null hypothesis. The data do provide sufficient evidence toconclude that the mean BMI of U.S. adults is greater than that for ahealthy weight.

32. (a) Using Minitab, we choose Graph Histogram, choose the Simple version,

enter BEER in the Graph variables text box, and click OK. The resultis


BEER

Frequency

1059075604530150

50

40

30

20

10

0

Histogramof BEER

(b) There may be an outlier at about 105.(c) Since the sample size is very large (300), we can use the z-test or the

t-test. Since the standard deviation is unknown, we use the t-test.Using Minitab, we obtained the following result using all of the data.Test of mu = 22 vs > 22

99%Lower

Variable N Mean StDev SE Mean Bound T PBEER 300 27.5200 19.4265 1.1216 24.8967 4.92 0.000

After eliminating the potential outlier (106), we repeated the processand obtained

Test of mu = 22 vs > 2299%

LowerVariable N Mean StDev SE Mean Bound T PBEER 299 27.2575 18.9187 1.0941 24.6985 4.81 0.000

Clearly, elimination of the value 106 had little effect on the value of t or onthe P-value. In either case, we reject the null hypothesis. The data doprovide sufficient evidence to claim that the mean annual consumption of beerin Washington, D.C. exceeds the national mean.

Using the Focus Database: Chapter 9

(a) Using Minitab, we choose Stat Basic Statistics 1-Sample t, enter COMP

in the Samples in columns text box, enter 20.8 in the Test mean text box,click on the Options button, enter 95 in the confidence level text box, andselect greater than from the Alternative drop down box. Click OK twice. Theresults are

Test of mu = 20.8 vs > 20.8

95%Lower

Variable N Mean StDev SE Mean Bound T PCOMP 200 23.7400 3.1371 0.2218 23.3734 13.25 0.000

Since the P-value is 0.000, we reject the null hypothesis. The data providesufficient to claim that the mean ACT score at UWEC is greater than thenational mean of 20.6.

Case Study: Sex and Sense of Direction 457

(b) To obtain the true mean from the FOCUS database, select Stat Basic

Statistics Display Descriptive Statistics, enter COMP in the Variables

text box and click OK. The results are

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3COMP 6738 0 23.621 0.0385 3.162 13.000 21.000 23.000 26.000

Variable MaximumCOMP 34.000

The mean is shown as 20.621.

(c) Yes. The population mean is, in fact, greater than 20.8. It would notnecessarily have to be correct. Another sample might yield a sample meansmall at random enough to not reject the null hypothesis.

(d-e) Following the same procedure as in part (a) with the variables ENGLISH andMATH, we get the results shown below

Test of mu = 20.3 vs > 20.3

95%Lower

Variable N Mean StDev SE Mean Bound T PENGLISH 200 23.1700 4.0364 0.2854 22.6983 10.06 0.000

Test of mu = 20.6 vs > 20.6

95%Lower

Variable N Mean StDev SE Mean Bound T PMATH 200 23.5650 3.9041 0.2761 23.1088 10.74 0.000

Since the P-values are both 0.000, we reject the null hypotheses. The dataprovide sufficient to claim that the mean MATH and ENGLISH scores at UWECare greater than the respective national means of 20.3 and 20.6. In bothcases, we see from the entire database that the conclusions are correct.

Case study: Sex and Sense of Direction

(a) If the women just randomly guessed at the direction of south, their meanpointing error would be zero with a range from -180 to +180 degrees. Theabsolute pointing errors would range from 0 to 180 with a mean of 90degrees.

(b) n = 15, df = 14, x 55 4. , s = 42.2Step 1: H0: = 90 Ha: < 90Step 2: = 0.01Step 3: Critical value = -2.624

Step 4: (55.4 90.0)/ (42.2/ 15) 3.175tStep 5: Since -3.175 < -2.624, we reject the null hypothesis.Step 6: We conclude that the women who claim to have a good sense of

direction really do better, on the average, than would be doneby just randomly guessing at the direction of south.

(c) The four plots are


ERROR

Frequency

120100806040200

5

4

3

2

1

0

Histogram of ERROR

ERROR

140

120

100

80

60

40

20

0

Boxplot ofERROR

ERROR

Percent

150100500-50

99

95

90

80

70

605040

30

20

10

5

1

Mean

0.071

55.4

St Dev 42.24N 15AD 0.652

P- Value

Probability Plot of ERRORNormalStem-and-leaf of ERROR N = 15

Leaf Unit = 10

4 0 0111(4) 0 22337 07 0 6674 0 93 1 02 1 22

Since this is a small sample and the data do not appear to come from anormal distribution, it is not reasonable to use a t-test.

(d) With the data in a column named ERROR, choose Stat Basic Statistics 1-Sample t..., click in the Samples in columns text box and enter ERROR, clickin the Test mean text box and type 90, click the Options button, enter 99 inthe Confidence level text box, select less than in the Alternative box and

click OK twice. To obtain the boxplot, choose Graph Boxplot..., select

Simple, click OK, enter ERROR in Graph variables text box, and click OK. To

obtain the histogram, choose Graph Histogram..., select Simple, click OK,enter ERROR in the Graph variables text box, and click OK. To obtain the

stem-and-leaf plot, we choose Graph Stem-and-leaf..., enter ERROR in the

Graph Variables text box, and click OK. The resulting graphs are shown inpart (c). The t-test result is

Test of mu = 90 vs < 9099%

UpperVariable N Mean StDev SE Mean Bound T PERROR 15 55.4000 42.2371 10.9056 84.0216 -3.17 0.003

(e) n=15

H0: = 90; Ha: < 90

Case Study: Sex and Sense of Direction 459

= 0.01

Critical value = 15(16)/2 - 100 = 20

x x-0=D D Rank of D Signed Rank R

14 -76 76 13 -13

91 1 1 1 127 -67 67 10 -10

122 32 32 6 68 -82 82 15 -15

68 -22 22 5 -5128 38 38 7 778 -12 12 2 -220 -70 70 11 -11

109 19 19 3 331 -59 59 9 -969 -21 21 4 -412 -78 78 14 -1436 -54 54 8 -818 -72 72 12 -12

W = sum of the positive signed ranks = 17

Since 17 is less than or equal to the critical value of 20, we reject thenull hypothesis and conclude that there is evidence that the women who claimto have a good sense of direction really do better, on the average, thanwould be done by just randomly guessing at the direction of south.

(f) If the women were merely guessing the direction of south at random, theirguesses would be uniformly distributed over the range from -180 to +180degrees from south and their absolute pointing errors would be uniformlydistributed over the range from 0 to 180 degrees. The normal probabilityplot shows a slight s-shaped curve that is also characteristic of a uniformdistribution although the curve might be a little more symmetric if the datawere exactly from a uniform distribution. The other plots show a slighttendency toward right-skewness, but this could easily be the result of thesmall sample size. Overall, the Wilcoxon signed-rank test seems to be areasonable choice given both the expectations under the null hypothesis andthe plots.

(g) Choose Stat Nonparametrics 1-Sample Wilcoxon..., select ERROR in the

Variables text box, click on the Test median button and enter 90 in its textbox, select less than in the Alternative text box, and click OK. The resultis

Test of median = 90.00 versus median < 90.00N for Wilcoxon Estimated

N Test Statistic P MedianERROR 15 15 17.0 0.008 52.50

This result is the same as that obtained in part (e).

Chapter 9

Documents

hypothesis

nonrejecti

power computation

gt

population

strengthoftheevidenceagainstthenullhypothesisisstrong

tailedtest9

00xz