chap 8

Confidence Interval Estimation

USING STAilSTICS @ Saxon Home Improvement

8.1

8.2

CONFIDENCE INTERVAL ESTIMATIONFOR THE MEAN (o KNOWN)

CONFIDENCE INTERVAL ESTIMATIONFOR THE MEAN (o UNKNOWN)Student's r DistributionProperties of the I DistributionThe Concept ofDegrees ofFreedomThe Confidence Interval Statement

CONFIDENCE INTERVAL ESTIMATIONFOR THE PROPORTION

DETERMINING SAMPLE SIZESample Size Determination for the MeanSample Size Determination for the Proportion

APPLICATIONS OF CONFIDENCEINTERVAL ESTIMATION IN AUDITINGEstimating the Population Total AmountDifference EstimationOne-Sided Confidence Interval Estimation of the

Rate of Noncompliance with Internal Controls

CONFIDENCE INTERVAL ESTIMATIONAND ETHICAL ISSUES

8.7 G) (CD-ROM TOPiC) ESTTMATTONAND SAMPLE SIZE DETERMINATIONFOR FINITE POPULATIONS

EXCEL COMPANION TO CHAPTER 8E8.l Computing the Confidence Interval Estimate

for the Mean (o Known)88.2 Computing the Confidence Interval Estimate

for the Mean (o Unknown)E8.3 Computing the Confidence Interval Estimate

for the ProportionE8.4 Computing the Sample Size Needed

for Estimating the MeanE8.5 Computing the Sample Size Needed

for Estimating the ProportionE8.6 Computing the Confidence Interval Estimate

for the Population TotalE8.7 Computing the Confidence Interval Estimate

for the Total DifferenceE8.8 Computing Finite Population Correction

Factors

tt

,tr

I

8.3

8.4

8.5

8.6

In this chapter, you learn:r To construct and interpret confidence interval estimates for the mean and the proportionI How to determine the sample size necessary to develop a confidence interval for the mean or

proportionI How to use confidence interval estimates in auditins

284 CHAPTER EIGHT Confidence Interval Estimation

Using Statistics @ Saxon Home Irprovement

Saxon Home Improvement distributes home improvement supplies innortheastern United States. As a company accountant, you arefor the accuracy of the integrated inventory management and sales inmation system. You could review the contents of each and every recordcheck the accuracy of this system, but such a detailed review wouldtime-consuming and costly. A better approach would be to use statistiinference techniques to draw conclusions about the population ofrecords from a relativelv small samole collected durins an audit. Atend of each month, you could select a sample of the sales invoicesdetermine the followins:

r The mean dollar amount listed on the sales invoices for the month.r The total dollar amount listed on the sales invoices for the month.r Any differences between the dollar amounts on the sales invoices

the amounts entered into the sales information system.r The frequency of occurrence of errors that violate the internal control policy of the wa

Such errors include making a shipment when there is no authorized warehouse removal slip, faiure to include the correct account number, and shipping the incorrect home improvement item.

How accurate are the results from the samples and how do you use this information? Arethe sample sizes large enough to give you the information you need?

Q tatistical inference is the process ofusing sample results to draw conclusions about the charac.tJteristics of a population. Inferential statistics enables youto estimale unknown population char-acteristics such as a population mean or a population proportion. Two types of estimates are usedto estimate population parameters: point estimates and interval estimates. A point estimate is thevalue of a single sample statistic. A confidence interval estimate is a range of numbers, called aninterval, constructed around the point estimate. The confidence interval is constructed such that theprobability that the population parameter is located somewhere within the interval is known.

Suppose you would like to estimate the mean GPA of all the students at your university.The mean GPA for all the students is an unknown population mean, denoted by p. You select asample of students and find that the sample mean is 2.80. The sample mean, X = 2.80, is apoint estimate of the population mean, p. How accurate is 2.80? To answer this question, youmust construct a confidence interval estimate.

In this chapter, you will learn how to construct and interpret confidence interval estimates.Recall that the sample mean, X, is a point estimate of the population mean, p. However, thesample mean varies from sample to sample because it depends on the items selected in the sam-ple. By taking into account the known variability from sample to sample (see Section 7.4 onthesampling distribution of the mean), you can develop the interval estimate for the populationmean. The interval constructed should have a specified confidence of correctly estimating thevalue of the population parameter p. In other words, there is a specified confidence that p issomewhere in the range of numbers defined by the interval.

Suppose that after studying this chapter, you find that a 95oh confidence interval for themean GPA at your university is(2.75 < p < 2.85).You can interpret this interval estimate bystating that you are 95o/o confident that the mean GPA at your university is between 2.75 and2.85. There is a 5o/o chance that the mean GPA is below 2.75 or above 2.85.

After learning about the confidence interval for the mean, you will learn how to develop aninterval estimate for the population proportion. Then you will learn how large a sample toselect when constructing confidence intervals and how to perform several important estimationprocedures accountants use when performing audits.

FIGURE 8.1

Confidence intervalestimates for f ivedifferent samples ofn = 25 taken from apopulat ion wherep = 3 6 8 a n d o : 1 5

8.1: Confidence Interval Estimation for the Mean (o Known) 285

8.1 CONFIDENCEINTERVALESTIMATIONFOR THE MEAN (o KNOWN)In Section 7 .4, you used the Central Limit Theorem and knowledge of the population distribu-tion to determine the percentage of sample means that fall within certain distances of the pop-ulation mean. For instance, in the cereal-fill example used throughout Chapter 7 (see Example7 '6 on page 268), 95o/o of all sample means are between 362.12 and 373.88 grams. This state-ment is based on deductive reasoning. However, incluctive reasoning is what you need here.

You need inductive reasoning because, in statistical inference, you use the results ofa sin-gle sample to draw conclusions about the population, not vice versa. Suppose that in the cereal-fill example, you wish to estimate the unknown population mean, using the information fromonly a sample. Thus, rather than take p + (1.96l (o l1r f n) to f ind the upper and lower l imi tsaround p, as in Section 7 .4, you substitute the sample mean, X- , for the unknown I and useX t (1.96)(o l ( . ' ln) as an in terval to est imate the unknown p. Al though in pract ice youselect a single sample of size n and compute the mean, X, in order to understand the fullmeaning of the interval estimate, you need to examine a hypothetical set of all possible sam-ples of r va lues.

Suppose that a sample of n : 25 boxes has a mean of 362.3 grams. The interval developedto est imate p is 362.3 I ( l .96Xl 5) l (^125), or 362.3 + 5.g9. The est i rnate of p is

3 5 6 . 4 2 ( u ( 3 6 8 . 1 8

Because the population mean, p (equal to 368), is included within the interval, this sampleresults in a correct statement about p (see Figure g.l).

362.12 368 373.88I

Xt = 362.s

I = sos.s

x. = :oo

Xa= 362.12

Xu = eZa.eA356.24 362.12 398

356.42 362.3 368.18

36s.s + (1.s6)(t sy bl 2s )

or 369.5 + 5.88. The estimate is

3 6 3 . 6 2 S u S 3 7 5 . 3 8

Because the population mean, p (equal to 368), is also included within this interval. this state-ment about u is correct.

J O J . b Z 375.38

360

To continue this hypothetical example, suppose that for a different sample of n:25 boxes,the mean is 369.5. The interval developed from this samole is


Now, before you begin to think that correct statements about p are always made by devel-oping a confidence interval estimate, suppose a third hypothetical sample of n :25 boxesis selected and the_ffrmple mean is equal to 360 grams. The interval developed here is360 t (1.96)(15)l(a|25), or 360 + 5.88. In this case, the estimate of p is

3 5 4 . 1 2 ( p ( 3 6 5 . 8 8

This estimate is not aconect statement because the population mean, p, is not included in the inter-val developed from this sample (see Figure 8.1). Thus, for some samples, the interval estimate of pis correct, but for others it is incorrect. In practice, only one sample is selected, and because thepopulation mean is unknown, you cannot determine whether the interval estimate is correct.

To resolve this dilemma of sometimes having an interval that provides a correct estimate andsometimes having an interval that provides an incorrect estimate, you need to determine the pro-portion of samples producing intervals that result in correct statements about the population mean,p. To do this, consider two other hypothetical samples: the case in which X = 362.12 grams rndthecaseinwhich X =373.88 grams.I f X =362.12,theinterval is362.121(1.96X15)l(425),or362.12 + 5.88. This leads to the followins interval:

356.24(uS368.00

Because the population mean of 368 is at the upper limit of the interval, the statement is a cor-rect one (see Figure 8.1).

When X = 373.88, the interval is 373.88 t (1.96X15)l(425), or 373.88 + 5.88. Theinterval for the sample mean is

368.00 ap!319.76

In this case, because the population mean of 368 is included at the lower limit of the interval,the statement is correct.

In Figure 8.1, you see that when the sample mean falls anywhere between 362.12 and373.88 grams, the population mean is included somauhere within the interval. In Example 7.6on page 268, you found that 95o/o of the sample means fall between 362.12 and 373.88 grams.Therefore, 95%o of al| samples of n : 25 boxes have sample means that include the populationmean within the interval developed.

Because, in practice, you select only one sample and p is unknown, you never know forsure whether your specific interval includes the population mean. However, if you take all pos-sible samples of n and compute their sample means, 95o/o of the intervals will include the pop-ulation mean, and only 5o/o of them will not. In other words, you have 95% confidence that thepopulation mean is somewhere in your interval.

Consider once again, the first sample discussed in this section. A sample of n : 25 boxeshad a sample mean of 362.3 grams. The interval constructed to estimate p is:

362.3 t (1.e6Xls)/(J25)

362.3 + 5.88

356.42 S u( 368.18

The interval from356.42 to 368.18 is referred to as a 95% confidence interval.

"I am9ilo/oconfident that the mean amount of cereal in the population of boxes is some-where between356.42 and 368.18 grams."

8.1: Confidencc Interval E,stirnation for the Mean (o Known) 287

In some situations, you might want a higher degree of confidence (such as99{t/o) of includ-ing the population mean within the interval. In other cases, you rnight accept less confidence(such as 90%) of correctly estirnating the population mean. In general, the level of confidenceis symbolized by ( I - o) x 100%o, where u is the proportion in the tails of the distribution thatis outside the confidence interval. The proportion in the upper tail of the distribution is ul2, and,the proportion in the lower tail of the distribution is ul2.You use Equation (8.1) to construct a( I - ct) x 100% confidence interval estimate of the mean with o known.

CONFIDENCE INTERVAL FOR THE MEAN (o KNOWN)

x tz

(8.1)

where Z: the value corresponding to a cumulative area of 1 - al2 from the standardizednormal distribution (that is, an upper-tail probability of ul2).

The value ofZ needed for constructing a confidence interval is called the crit ical value forthe distribution.95o/o confidence corresponds to an o, value of 0.05. The crit ical Zvalue corre-sponding to a cumulative area of 0.9750 is L96 because there is 0.025 in the upper tail of thedistribution and the cumulative area less than Z: 1.96 is 0.975.

There is a different crit ical value for each level ofconfidence. I - cr. A level ofconfidenceof 95oh leads to a Z value of | .96 (see Figure 8.2). 99% confidence corresponds to an cr valueof 0.01 .The Z value is approximately 2.58 because the upper-tail area is 0.005 and the cumula-t ive area less than Z:2.58 is 0.995 (see Ficure 8.3) .

o--r=\l n

F-zf .u <n+2ft

FIGURE 8.2

Normal curve fordetermin ing theZvalue neededfor 95% confidence

FIGURE 8.3

Normal curve fordeterminrng theZvalue neededfor 99% confidence

'p, X-1 .96 0 +1 .96 Z

x+2.58 Z

288 CHAPTER EIGHT Confidence lnterval Estimation

Now that various levels of confidence have been considered whv not make the confilevel as close to I 00% as possible? Before doing so, you need to realize that any increase inlevel of confidence is achieved only by widening (and making less precise) the confiinterval. There is no "free lunch" here. You would have more confidence that themean is within a broader range of values; however, this might make the interpretation ofconfidence interval less useful. The trade-offbetween the width of the confidence intervalthe level ofconfidence is discussed in greater depth in the context ofdetermining thesize in Section 8.4. Example 8.1 illustrates the application of the confidence interval

To see the effect of using a99o/o confidence interval, examine Example 8.2.

EXAMPLE 8.1 ESTIMATING THE MEAN PAPER LENGTH WITH 95% CONFIDENCE

A paper manufacturer has a production process that operates continuously throughout anproduction shift. The paper is expected to have a mean length of l1 inches, and thedeviation of the length is 0.02 inch. At periodic intervals, a sample is selected towhether the mean paper length is still equal to 1l inches or whether something has gonein the production process to change the length ofthe paper produced. You select a randomple of 100 sheets, and the mean paper length is 10.998 inches. Construct a95%ointerval estimate for the population mean paper length.

SOLUTION Using Equation (8.1) on page 287 , with Z: |.96 for 95o/o confidence,

Xxz*=ro.essr(1.e6)g4n {100

= 10.998 + 0.00392

10.99408 Sp ( 11.00192

Thus, with 95% confidence, you conclude that the population mean is between 10.99408 and11.00192 inches. Because the interval includes 11, the value indicating that the productionprocess is working properly, you have no reason to believe that anything is wrong with theduction process.

EXAMPLE 8.2 ESTIMATING THE MEAN PAPER LENGTH WITH 99% CONFIDENCE

Construct a99%o confidence interval estimate for the population mean paper length.

SOLUTION Using Equation (8. I ) on page 287, with Z : 2.58 for 99o/o confidence,

Xtz*= lo .eest (2 .58)g4 n {100

= 10 .998 10 .00516

10.99284 (p ( 11.00316

Once again, because I I is included within this wider interval, you have no reason to belithat anything is wrong with the production process.

8.1 : Confidence Interval Estimation for the Mean (o Known) 289

As discussed in section 7.4,the sampling distribution of t is normally distributed if thepopulation ofXis a normal distribution. And" if the population ofXis not a normal distribution,the Central Limit Theorem ensures that X is normally distributed when r is large. However,when dealing with a small sample size and a population of X that is not a normal distribution,the sampling distribution of X is not normally distributed and therefore the confidence inter-val discussed in this section is inappropriate. In practice, however, as long as the sample size islarge enough and the population is not very skewed, you can use the confidence intervaldefined in Equation 8.1 to estimate the population mean when o is known. To assess theassumption of normality, you can evaluate the shape of the sample data by using a histogram,stem-and-leaf display, box-and-whisker plot, or normal probability plot.

M ?:,r, x = t25, o : 24. and n :ru,.:""::T.:,: ffi i;1"111.1H",'J.Til:11,[Tll"jJl[:?T;*:lAsslsTl 99% confidence interval estimate of the popula--- ---- r-r"'- | z seul shipment of lieht bulbs. The standard deviation isuon mean' p'

EU 100 hours. A-random sample of 64 light bulbs

Learning the Basics

8.1 I f X = 85, o : 8, and n:64, construct a95% confidence interval estimate of the popula-tion mean, p.

8.3 A market researcher states that she has 95oh confi-dence that the mean monthly sales of a product arebetween $170,000 and $200,000. Explain the meaning ofthis statement.

8.4 Why is it not possible in Example 8.1 on page 288 tohave 100% confidence? Explain.

8.5 From the results of Example 8.1 on page 288 regard-ing paper production, is it true that 95%o of the samplemeans will fall between 10.99408 and 11.00192 inches?Explain.

8.6 Is it true in Example 8.1 on page 288 that you do notknow for sure whether the population mean is between10.99408 and 11.00192 inches? Explain.

Applying the Concepts

8.7 The manager of a paint supply store wantsto estimate the actual amount of paint containedin l-gallon cans purchased from a nationally

known manufacturer. The manufacturer's specificationsstate that the standard deviation of the amount of paint is

to 0.02 gallon. A random sample of 50 cans isselected, and the sample mean amount of paint per l-galloncan is 0.995 sallon.a. Construct a99o/o confidence interval estimate of the pop-

ulation mean amount of paint included in a l-gallon can.On the basis of these results, do you think the managerhas a right to complain to the manufacturer? Why?

c. Must you assume that the population amount of paintper can is normally distributed here? Explain.

d. Construct a 95oh confidence interval estimate. Howdoes this change your answer to (b)?

indicated a sample mean life of 350 hours.a. Construct a 95o/o confidence interval estimate of the

population mean life of light bulbs in this shipment.b. Do you think that the manufacturer has the right to state

that the light bulbs last an average of 400 hours?Explain.

c. Must you assume that the population of light bulb life isnormally distributed? Explain.

d. Suppose that the standard deviation changes to 80 hours.What are your answers in (a) and (b)?

8.9 The inspection division of the Lee CountyWeights and Measures Department wants to esti-mate the actual amount of soft drink in 2-Iiter

bottles at the local bottling plant of a large nationallyknown soft-drink company. The bottling plant hasinformed the inspection division that the population stan-dard deviation for 2-liter bottles is 0.05 liter. A randomsample of 100 2-liter bottles at this bottling plant indicatesa sample mean of 1.99 liters.a. Construct a 95oh confidence interval estimate of the

population mean amount of soft drink in each bottle.b. Must you assume that the population of soft-drink fill is

normally distributed? Explain.c. Explain why a value of 2.02liters for a single bottle is

not unusual, even though it is outside the confidenceinterval you calculated.

d. Suppose that the sample mean is 1.97 liters. What isyour answer to (a)?

-

290 CHAPTER EIGHT Confidence lnterval Estimarron

8.2 CONFIDENCE INTERVAL ESTIMATIONFOR THE MEAN (o UNKNOWN)Just as the mean of the population, p, is usually unknown, you rarely know the actual standarddeviation of the population, o. Therefore, you often need to construct a confidence interval

estimate of p, using only the sample statistics X and S.

Student's t DistributionAt the beginning of the twentieth century, Will iam S. Gosset, a statistician for GuinnessBreweries in Ireland (see reference 3) wanted to make inferences about the mean when o wasunknown. Because Guinness employees were not permitted to publish research work undertheir own names, Gosset adopted the pseudonym "Student." The distribution that he developedis known as Student's r distribution and is commonly referred to as the t distribution.

If the random variable X is normally distributed" then the following statistic has a t distri-bution with n - I degrees of freedom:

Y - t t

This expression has the same form as the Z statistic in Equatron (7 .4) on page 266, except that

S is used to estimate the unknown o. The concept of degrees o.f'.freedom is discussed further onpages 291-292.

Properties of the t Distribution

In appearance, the / distribution is very similar to the standardized normal distribution. Both

distribrrtion.s are bell shapecl. Floweve-r, the t tlistribLrtiorr hits l77orc area irt the tirils arrd less ln

rfie ccnfcr lh:rn clocs thc sfandardized norntal distribufion (see Figtrre 8.4). Because Jis used to

estirnate the unknown o. the values of 1 are more variable than those for Z.

s,FN

FIGURE 8.4

Standardizednormal d is t r ibut ionand t d is t r ibut ion for5 degrees of freedom

- S tandard ized normal d is t r ibu t ion

- f distr ibutionfor 5 degreesof freedom

The degrees of freedom, n - 1, are directly related to the sample size, n. As the sample sizeand degrees of freedom increase, S becomes a better estimate of o, and the r distribution grad-ually approaches the standardized normal distribution, unti l the two are virtually identical.With a sample size of about 120 or more, S estimates o precisely enough that there is little dif-ference between the t and Z distributions.

As stated earlier, the t distribution assumes that the random variable X is normally distrib-uted. In practice, howeveq as long as the sample size is large enough and the population is notvery skewed, you can use the I distribution to estimate the population mean when o isunknown. When dealing with a small sample size and a skewed population distribution, thevalidity of the confidence interval is a concern. To assess the assumption of normality, you can

8.2: Confidence Interval E,stimation for the Mean (o Unknown) 291

evaluate the shape of the sample data by using a histogram, stem-and-leaf display, box-and-whisker plot, or normal probabil ity plot.

You find the critical values of r for the appropriate degrees of freedom from the table of ther distribution (see Table E.3). The columns of the table represent the area in the upper tail of ther distribution. The rows of the table represent the degrees of freedom. The cells of the table rep-resent the particular / value for each specific degree of freedom. For example, with 99 degreesof freedom, if you want 95% confidence, you find the appropriate value of l, as shown in Table8. L The 95% confidence level means that2.5oh of the values (an area of 0.025) are in each tailof the distribution. Looking in the column for an upper-tail area of 0.025 and in the row corre-sponding to 99 degrees of freedom gives you a critical value for t of 1.9842. Because r is a sym-metrical distribution with a mean of 0, if the upper-tail value is +1.9842, the value for thelower-tail area (lower 0.025) is - l .9842. A I value of -1.9842 means that the probability that tis less than -1.9842 is 0.025, or2.5oh (see Figure 8.5). Note that for a 95% confidence interval,you will always use an upper-tail area of 0.025. Similarly, for a99o/o confidence interval, use0.005, for 98%o use 0.01, 90% use 0.05, and 80% use 0.10.

Upper-Tail Areas

Degrees of Freedom ? { .005.01.05.10

TABLE 8 .1

Determining the Cri t icalValue from the t Tablefor an Area of 0.025in Each Tai l with 99Degrees of Freedom

FIGURE 8.5

td is t r ibut ion wi th 99degrees of freedom

31.8207 63.65746.9646 9.92484.s407 5.84093.1469 4.604r3.3.649 4.0.322

2.3658 2.62802.3654 2.62752.36s0 2.62692.3646 2.62642.3642 2.6259

I2J

45

100

824164706

969798

1.00000 . 8 1 6 50.16490.74070.1267

0.67710.67700.6770

3.07771 .88s61.63171.s3321.41s9

1.2904t.2903t.2902

6 .31382.92002.3s342 .13 l 82 .01s0

12 .4 73 .2.2

1.98400.6770

rtS

en

Source; E.ytroctel ]inn Tuhle E.3

The Concept of Degrees of Freedom

In Chapter 3 you learned that the numerator of the sample variance, 52 [see Equation (3.9) onpage 107], requires the computation of

S - . )

L(x , -x r

+1.9842


In order to compute 52, you first need to know X . Therefore, only r - I of the sample values

are free to vary. This means that you have n - 1 degrees of freedom. For example, suppose a

sample of five values has a mean of 20. How many values do you need to know before you can

determine the remainder of the values? The fact that n: 5 and X = 20 also tells you that

il

Tr = too. L " t

x-tn-r#.u <X+,,.#

because

Thus, when you know four of the values, the fifth one is not free to vary because the sum mustadd to 100. For example, if four of the values are 18, 24, 19, and 16, the fifth value must be 23so that the sum equals 100.

The Confidence lnterval Statement

Equation (8.2) defines the (l - a) x 100% confidence interval estimate for the mean with ounknown.

CONFIDENCE INTERVAL FOR THE MEAN UNKNOWN)

V + t . . ,

(o

sG

n

\ vi = l ;

n

(8.2)

where /,_, is the critical value of the r distribution, with n - I degrees of freedom for anarea of al2 in the upper tail.

To illustrate the application of the confidence interval estimate for the mean when thestandard deviation, o, is unknown, recall the Saxon Home Improvement Company UsingStatistics scenario presented on page 284.You wanted to estimate the mean dollar amountlisted on the sales invoices for the month. You select a sample of 100 sales invoices from thepopulation of sales invoices during the month, and the sample mean of the 100 sales invoices rs$110.27, with a sample standard deviation of $28.95. For95o/o confidence, the crit ical valuefrom the / distribution (as shown in Table 8.1) is L9842. Using Equation (8.2),

i t . ' S^ l l n I T

1 n

= | t0 .27 + ( 1 .s842 ) 28.95

{100=110.27 + 5.74

$104 .53 ( u < $116 .01

A Microsoft Excel worksheet for these data is presented in Figure 8.6.

8.2: Confidence Interval Estimation for the Mean (o Unknown) 293

FIGURE 8.6

Microsoft Excelworksheet to computea conf idence intervalest imate for the meansa les invo ice amountfor the Saxon Homelmprovement Company

See Sectlon EB.2 to createthrs.

10 Standard Enor o f the Means of Freedr lm

12 | Va lue

Confidence Interval

:84/SaRT{86}= 8 5 - 1=Tri lv( l - 87,811)=812 " 810

:85 - 813:85 * 813

Thus, with 95% confidence, you conclude that the mean amount of all the sales invoices isbetween $ 104.53 and $ I 16.01 . The 95% confidence level indicates that if you selected all possi-ble samples of 100 (something that is never done in practice), 95o/o of the intervals developedwould include the population mean somewhere within the interval. The validity of this confi-dence interval estimate depends on the assumption of normality for the distribution of the amountof the sales invoices. With a sample of 100, the normality assumption is not overly restrictive, andthe use of the t distribution is likely appropriate. Example 8.3 further illustrates how you constructthe confidence interval for a mean when the population standard deviation is unklown.

f -A ' * - -qj 1 rEstimate for the Mean Sales Invoice Im-1 - - _j- Q Data; 4 f f i

5 f t6 tSamnle S ize7 Confidence LevelB9 ln te rmed ia te Ca lcu la t ions

57443

EXAMPLE 8 .3

TABLE 8 .2

Force ( in Pounds)Required to Breakthe Insu la to r

FIGURE 8.7

Microsoft Excelconf idence intervalest imate for the meanamount of forcerequired to breakelectr ic insulators

See Section EB.2 to createthis.

ESTIMATING THE MEAN FORCE REOUIRED TO BREAK ELECTRIC INSULATORS

A manufacturing company produces electric insulators. If the insulators break when in use, ashort circuit is likely. To test the strength ofthe insulators, you carry out destructive testing todetermine how rnuch,lbrzc is required to break the insulators. You measure force by observinghow many pounds are applied to the insulator before it breaks. Table 8.2 l ists 30 values fromthis experiment, which are located in the file [![!fQ Construct a 95o/o confidence intervalestimate for the population mean force required to break the insulator.

1,870 1 ,128 1 ,656 I ,610 1 ,634 t ,784 1 ,5221,866 1 ,764 1 ,734 1 .662 1 ,734 1 ,774 1 ,5501,820 t ,744 1 ,788 1 ,688 1 ,810 1 ,752 1 ,680

1,696 1,592 t ,662| . t 56 t .7 62 r ,866I . 810 t . 652 t , 736

SOLUTION Figure 8.7 shows that the sample mean is F = 1,723.4 pounds and the samplestandard deviation is S: 89.55 pounds. Using Equation (8.2) on page 292 to construct the con-fidence interval, you need to determine the critical value from the / table for an area of 0.025 in

=BIISORT{85)- 8 6 - l-Tlilv(l - 87, 811)-812 " 810

-85 - 813-85 . 813

q_ ___-_,,___ LI i Intermediale Calculations

l lnlerval Half Widlh

294 CHAPTER EIGHT Confidence Intcrval Estin.ration

each tail, with 29 degrees of freedom. From Table 8.3, you see that t),\:2.0452. Thus, usiX = 1 ,123 .4 , S : 89 .55 , n - 30 , and t , o : 2 .0452 ,

x + t, t

= | - t )7 4 + t ) ors)r 89 '55- ' - " - " ' '60

= 1.123.4 + 33.44

1 , 6 8 9 . 9 6 { u S 1 . 7 5 6 . 8 4

You conclude with95% confidence that the mean breaking force required for the population rinsulators is between 1,689.96 and 1,756.84 pounds. The validity of this confidence intervalestmate depends on the assumption that the fbrce required is norrnally distributed. Remember, houeveq that you can slightly relax this assumption for large sample sizes. Thus, with a sample of 3(you can use the r distribution even if the amount of force required is only slightly left skewed. Frorthe notmal probability plot displayed in Figure 8.8 or the box-and-whisker plot displayed in Figur8.9, the amount of force required appears slightly left skewed. Thus, the t distribution is appropriat,for these data.

Force Required to Break Electrical Insulators

sG

FIGURE 8.8

Microsof t Excel normalprobabi l i ty p lot forthe amount of forceranr r i raA tn h ror lz

electr ic insulators

See Sectlon E6.2 to createthis.

FIGURE 8.9

Microsoft Excel box-and-whisker plot forthe amount of forcerequired to breakelectr ic insulators

See Section E3.4 to createthis.

2000 T

1800

1600

1400

1200

! rooo

800

600

400

200

-?.5

- - - #

2 2.5-1.5 .1 .0.5 0 0.5 1

Z Value

Force Required to Break Electrical Insulators

1 7501700 1800

8.2: Confidence Interval Estimation for the Mean (o Unknown) 295

learning the Basics

8.10 Determine the critical value of / in each ofthe following circumstances:

o " I - 0 : 0 . 9 5 . n : l 0b . I - o : 0 .99 . n : l0c . l - 0 : 0 . 9 5 , n : 3 2d. I - c r : 0 .95 , n : 65e . l - 0 = 0 . 9 0 . n : 1 6

8.11 I f X = 75, S :24, and n:36. and assum-ing that the population is normally distributed,construct a 95% confidence interval estimate of

the population mean, p.

8. ' ,2 l f X = 50,,S: 15, and n:16, and assum-ing that the population is normally distributed,construct a 99Yo confidence interval estimate of

population mean, p.

8.13 Construct a95%o confidence interval estimate for theiopulation mean, based on each of the followine sets of

assuming that the population is normally distributed:

S e t 1 l , l , l , 1 , 8 , 8 , 8 , 8

S e t 2 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8

lain why these data sets have different confidenceeven though they have the same mean and range.

14 Construct a 95o/o confidence interval for the popu-mean, based on the numbers l, 2, 3, 4, 5, 6, and20.

the number 20 to 7 and recalculate the confi-interval. Using these results, describe the effect of

outlier (that is, an extreme value) on the confidence

around time for stress tests. Turnaround time is defined asthe time from when the test is ordered to when the radiolo-gist signs off on the test results. Initially, the mean turn-around time for a strbss test was 68 hours. After incorporat-ing changes into the stress-test process, the qualityimprovement team collected a sample of 50 turnaroundtimes. In this sample, the mean turnaround time was 32hours, with a standard deviation of 9 hours (Extracted fromE. Godin, D. Raven, C. Sweetapple, and F, R. Del Guidice,"Faster Test Results," Quality Progress, January 2004,37(l), pp. 33-39).a. Construct a95o/o confidence interval for the population

mean turnaround time.b. Interpret the interval constructed in (a).c. Do you think the quality improvement project was a suc-

cess? Explain.

8.17 The U.S. Department of Transportationrequires tire manufacturers to provide tire perfor-mance information on the sidewall of the tire to

better inform prospective customers when making purchas-ing decisions. One very important measure of tire perfor-mance is the tread wear index, which indicates the tire,sresistance to tread wear compared with a tire graded with abase of 100. This means that a tire with a grade of 200should last twice as long, on average, as a tire graded witha base of 100. A consumer organization wants to estimatethe actual tread wear index of a brand name of tires graded200 that are produced by a certain manufacturer. A randomsample of n : l8 indicates a sample mean tread wear indexof 195.3 and a sample standard deviation of 21.4.a. Assuming that the population of tread wear indices is

normally distributed, construct a95% confidence inter-val estimate of the population mean tread wear indexfor tires produced by this manufacturer under thisbrand name.

b. Do you think that the consumer organization shouldaccuse the manufacturer of producing tires that do notmeet the performance information provided on the side-wall of the tire? Explain.

c. Explain why an observed tread wear index of 2 l0 for aparticular tire is not unusual, even though it is outsidethe confidence interval developed in (a).

8.18 The following data (stored in the file !!@@represent the bounced check fee, in dollars, charged by asample of 23 banks for direct-deposit customers who main-tain a $100 balance:

26 28 20 20 21 22 2s 2s 18 25 15 2018 20 25 25 22 30 30 30 15 20 29

Source: Extractedfrom "The New Face of Banking," ConsumerReports, June 2000.

the Concepts

15 A stationery store wants to estimate the mean retailof greeting cards that it has in its inventory. A random

of 100 greeting cards indicates a mean value of55 and a standard deviation of $0.44.Assuming a normal distribution, construct a95o/o confi-dence interval estimate of the mean value of all greetingcards in the store's inventory.Suppose there were 2,500 greeting cards in the store'sinventory. How are the results in (a) useful in assistingthe store owner to estimate the total value of herinventory?

8.16 Southside Hospital in Bay Shore, NewYork, commonly conducts stress tests to study theheart muscle after a person has a heart attack.of the diagnostic imaging department conducted

lity improvement project to try to reduce the turn-


a. Construct a 95oh confidence interval for the populationmean bounced check fee.

b. Interpret the interval constructed in (a)

8.19 The data in the file @ft[! represent the total fat,in grams per serving, for a sample of 20 chicken sand-wiches from fast-food chains. The data are as follows:

7 8 4 5 1 6 2 0 2 0 2 4 1 9 3 023 30 25 t9 29 29 30 30 40 s6

Source: Extractedfrom "Fast Food: Adding Health to the Menu,"Consumer Reports, September 2004, pp. 28-31.

a. Construct a95o/o confidence interval for the populationmean total fat, in grams per serving.

b. Interpret the interval constructed in (a).

8.20 One of the major measures of the quality of serviceprovided by any organization is the speed with which itresponds to customer complaints. A large family-held depart-ment store selling furniture and flooring, including carpet,had undergone a major expansion in the past several years. Inparticular, the flooring department had expanded from 2installation crews to an installation supervisoq a measurer,and l5 installation crews. Last year, there were 50 complaintsconceming carpet installation. The following data, in the file

@EE, represent the number of days between the receiptof a complaint and the resolution of the complaint:

pilation stage in which the policy pages are generated andsent to the bank for delivery. The ability to deliver approvedpolicies to customers in a timely manner is critical to theprofitability of this service to the bank. During a period ofone month, a random sample of 27 approved policies wasselected, and the total processing time, in days, was asshown below and stored in the file@,

t3 19 t6 64 28 28 3 t 90 60 56 3 l s6 22 l8

45 48 t7 t7 l t 91 92 63 50 s l 69 16 17

a. Construct a 95oh confidence interval estimate of themean processing time.

b. What assumption must you make about the populationdistribution in (a)?

c. Do you think that the assumption made in (b) is seri-ously violated? Explain.

8.22 The data in the file [l$$@f!represent the bat-tery life (in shots) for three-pixel digital cameras:

300 180 85 170 380 460 260 35 380 t20 rr0 240Source: Extractedfrom "Cameras: More Features in the Mix,"Consumer Reports, July 2005, pp. 14 18.

a. Construct a 95o/o confidence interval for the populationmean battery life (in shots).

b. What assumption do you need to make about the popu-lation ofinterest to construct the interval in (a)?

c. Given the data presented, do you think the assumptionneeded in (a) is valid? Explain.

8.23 One operation of a mill is to cut pieces of steel intoparts that are used later in the frame for front seats in anautomobile. The steel is cut with a diamond saw and requiresthe resulting parts to be within +0.005 inch of the lengthspecified by the automobile company. The measurementreported from a sample of 100 steel parts (and stored in thefile @ is the difference, in inches, between the actuallength of the steel part, as measured by a laser measurementdevice, and the specified length ofthe steel part. For exam-ple, the first observation, -0.002, represents a steel part thatis 0.002 inch shorter than the specified length.a. Construct a 95% confidence interval estimate of the

mean difference between the actual length of the steelpart and the specified length ofthe steel part.



d. Compare the conclusions reached in (a) with those ofProblem 2.23 on page 53.

5 4 5 3 5 r 3 7 3 l

l l t 9 t 2 6 l l 0 1 1 0

t2 4 16s 32 29

1 3 1 0 5 2 7 4

33 68

27 152 2 t23 81 74 21

2 9 6 1 3 5 9 4 3 1 2 6 5

2 8 2 9 2 6 2 s 1 1 4 1 3

52 30 22 36 26 20 23

a. Construct a 95o/o confidence interval estimate of themean number of days between the receipt of a complaintand the resolution of the complaint.



d. What effect might your conclusion in (c) have on thevalidity of the results in (a)?

8.21 In New York State, savings banks are permitted tosell a form of life insurance called savings bank life insur-ance (SBLI). The approval process consists of underwrit-ing, which includes a review of the application, a medicalinformation bureau check, possible requests for additionalmedical information and medical exams, and a policy com-

8.3 CONFIDENCE INTERVAL ESTIMATION FOR THE PROPORTIONThis section extends the concept ofthe confidence interval to categorical data. Here you areconcerned with estimating the proportion of items in a population having a certain characteris-tic of interest. The unknown population proportion is represented by the Greek letter n. The

rtoanCS

ith)ntheral3ntm-hat

the,eel

eri-

: o f

FIGURE 8.10

Microsoft Excelworksheet to constructa conf idence intervalestimate for theproportion of salesinvoices that containerrors

8.3: Confidence Interval Estirration fbr the Pronortion 297

point estirnate for n is the sample proportion, p - Xln, where ir is the sarrple size and X is thenurnber of iter.t.rs in the sarnple having the characteristic of interest. E,quation (8.3)defines theconfidence interval estir.nate for the population proportion.

coNFTDENCE TNTERVAL ESTTMATE FOX TXSjROPORTTON

,+ 7 . lP ( t - P tIT

or

p - Z <n< p+z (8.3)

X Number of items havins the characteristicp: Sample proportion :

; =

Sr_pl. ,t^n : population proportion

Z : critical value from the standardized norrnal distribution

r = sample s ize

assuming that both X and n - X are greater than 5

You can use the confidence interval estimate of the proportion defined in Equation (8.3) toestimate the proportion of sales invoices that contain errors (see the Using Statistics scenarioon page 284). Suppose that in a sample of 100 sales invoices. l0 conta in errors. Thus. for theseda ta .p : X ln :10 /100 :0 .10 . Us ing Equa t i on (8 .3 ) andZ- 1 .96 fo r 95 '% con f i dcnce ,

n * 7

= 0 . 1 0 t ( 1 . 9 6 )

= 0 . 1 0 t ( 1 . 9 6 ) ( 0 . 0 3 )

: 0 . l 0 + 0 . 0 5 8 8

0 . 0 4 1 2 < n ( 0 . 1 5 8 8

Therefore, you havc 95% confidence that betwecn 4.12% and 15.88% of all the sales invoicescontain errors. Figure 8.10 shows a Microsoft Excel worksheet forthese data.

=85/84=NORMSTNV(I - 86)2)=sQRr(Be'(1 - Bey84l=ABS(810 '811)

*89 - 812= 8 9 ' 8 1 2

N, aresris-The

123,557II1011

I J

1 i1515

p ( t - p )

n

p ( I - p )

n

p ( t - p )

i l

(0 . r0x0 .e0)100

Slandard Enor of thelntewal Half Width


EXAMPLE 8 .4

Example 8.4 illustrates another application of a confidence interval estimate for the

proportion.

ESTIMATING THE PROPORTION OF NONCONFORMING NEWSPAPERS PRINTED

A large newspaper wants to estimate the proportion of newspapers printed that have a noncon-

forming attribute, such as excessive ruboff, improper page setup. missing pages, or duplicate

pages. A random sample of 200 newspapers is selected from all the newspapers printed during

a single day. For this sample of 200, 35 contain some type of nonconformance. Construct and

interfret a 90o/o confidence interval for the proportion of newspapers printed during the day

that have a nonconforming attribute.

SOLUTION Using Equation (8.3),

? 5D - ::- = 0.175. and with a90oh level of confidenceZ =' 200

8.iVC:

far(ErofJot.a.

b.

8.2smitheApusea . i

I

l

I

b . iI

I

I

c . .

8.2is :nol

wihei

phsaiagCIa.

p !z

. . . - . [o . r7s) (0.82s)= 0.175 t ( r .64s){ - ; :

= 0.175 r (1.645X0.0269)

t75 !0.0442

0 . 1 3 0 8 3 n 3 0 . 2 1 9 2

You conclude with 90% confidence that between 13.08% and2l.92oh of the newspapers

printed on that day have some type of nonconformance.

Equation (8.3) contains a Z statistic because you can use the normal distribution to approx'

imate the binomial distribution when the sample size is sufficiently large. In Example 8.4, the

confidence interval using Z provides an excellent approximation for the population proportion

because both X and n - X are greater than 5. However, if you do not have a sufficiently large

sample size, you should use the binomial distribution rather than Equation (8.3) (see references

1.2, and 6). The exact confidence intervals for various sample sizes and proportions of suc'

cesses have been tabulated by Fisher andYates (reference 2).

p(t - p)

n

@ElAssrsT I

ffi


8.26 The telephone company wants to estlmatethe proportion of households that would purchasean additional telephone line if it were made avail-able at a substantially reduced installation cost.A

random sample of 500 households is selected. The resultsindicate that 135 of the households would purchase the

additional telephone line at a reduced installation cost.

b.

8.M

Learning the Basics

8.24 lf n : 200 and X: 50, constrwt a 95o/o@I ASSTST I

ffi

confidenceproportion.

8.25 l f nconfidenceproportion.

interval estimate of the population

= 400 and X : 25, construct a 99ohinterval estimate of the population

ne's

I

a. Construct a 99o/o confidence interval estimate of thepopulation proportion of households that would pur-chase the additional telephone line.

b. How would the manager in charge of promotional pro-grams concerning residential customers use the resultsin (a)?

8.27 According to the Center for Work-Life Policy, a sur-vey of 500 highly educated women who left careers forfamily reasons found that 660/o wanted to return to work(Extracted from A. M. Chaker and H. Stout, 'After YearsOff, Women Struggle to Revive Careers," The Wall StreetJournal, May 6, 2004, p. Al).a. Construct a95oh confidence interval for the population

proportion of highly educated women who left careersfor family reasons who want to return to work.

b. Interpret the interval in (a).

8.28 In a survey conducted for American Express, 21o/o ofsmall business owners indicated that they never check in withthe office when on vacation ("Snapshots," usatoday.com,April 18, 2006.). The article did not disclose the sample sizeused in the study.a. Suppose that the survey was based on 500 small business

owners. Construct a 95o/o confidence interval estimatefor the population proportion of small business ownerswho never check in with the office when on vacation.

b. Suppose that the survey was based on 1,000 small busi-ness owners. Construct a 95o/o confidence interval esti-mate for the population proportion of small business own-ers who never check in with the office when on vacation.

c. Discuss the effect of sample size on the confidenceinterval estimate.

8.29 The number of older consumers in the United Statesis growing, and they are becoming an even bigger eco-nomic force. Many feel overwhelmed when confrontedwith the task of selecting investments, banking services,health care providers, or phone service providers. A tele-phone survey of 1,900 older consumers found that 27ohsaid they didn't have enough time to be good money man-agers (Extracted from "Seniors Confused by FinancialChoices-Study," msnbc.com, May 6, 2004).a. Construct a95o/o confidence interval for the population

proportion of older consumers who don't think theyhave enough time to be good money managers.

b. Interpret the interval in (a).

8.30 A survey of 705 workers (USA Today Snapshots,March 21, 2006, p. lB) were asked how much they used

8.4: Determining Sample Size 299

the Internet at work. 423 said they used it within limits, and183 said that they did not use the Internet at work.a. Construct a95oh confidence interval for the proportion

of all workers who used the Internet within limits.b. Construct a 95o/o confidence interval for the proportion

of all workers who did not use the Internet at work.

8.31 When do Americans decide what to make for din-ner? An online survey (N. Hellmich, 'Americans Go forthe Quick Fix for Dinner," USA Today, February 14, 2005,p. 1B) indicated thatT4o/o of Americans decided either atthe last minute or that day. Suppose that the survey wasbased on 500 respondents.a. Construct a 95o/o confidence interval for the proportion

of Americans who decided what to make for dinnereither at the last minute or that day.

b. Construct a 99o/o confidence interval for the proportionof Americans who decided what to make for dinnereither at the last minute or that day.

c. Which interval is wider? Explain why this is true.

8.32 In a survey of 894 respondents with salaries below$100,000 per year, 367 indicated that the primary reasonfor staying on their job was interesting job responsibilities("What Is the Primary Reason for Staying on Your Job?"USA Today Snapshots, October 5,2005, p. lB).a. Construct a 95o/o confidence interval for the proportion

of all workers whose primary reason for staying on theirjob was interesting job responsibilities.

b. Interpret the interval constructed in (a).

8.33 A large number of companies are trying to reducethe cost of prescr ipt ion drug benef i ts by requir ingemployees to purchase drugs through a mandatory mail-order program. In a survey of 600 employers, 126 in-dicated that they either had a mandatory mail-orderprogram in place or were adopting one by the end of2004 (B. Martinez, "Forcing Employees to Buy Drugsvia Mail," The Wall Street Journal, February 18,2004,p . 1B) .a. Construct a95oh confidence interval for the population

proportion of employers who had a mandatory mail-orderprogram in place or were adopting one by the end of 2004.

b. Construct a99oh confidence interval for the populationproportion of employers who had a mandatory mail-orderprogram in place or were adopting one by the end of2004.

c. Interpret the intervals in (a) and (b).d. Discuss the effect on the confidence interval estimate

when you change the level ofconfidence.JC

seil-AItshe

8.4 DETERMINING SAMPLE SIZEIn each example of confidence interval estimation so far in this chapter, the sample size wasreported along with the results with little discussion with regard to the width of the result-ing confidence interval. In the business world, sample sizes are determined prior to data

300 CHAPTER EIGHT Confidence Interval Estimatron

collection to ensure that the confidence interval is narrow enough to be useful in makingdecisions. Determining the proper sample size is a complicated procedure, subject to theconstraints of budget, t ime, and the amount of acceptable sampling error. In the SaxonHome Improvement example, you want to estimate the mean dollar amount of the salesinvoices, you must determine in advance how large a sampling error to allow in estimatingthe population mean. You must also determine in advance the level of confidence (that is,90%,95%, or 99oh) to use in estimating the population pa'rameter.

Sample Size Determination for the Mean

To develop an equation for determining the appropriate sample size needed when constructinga confidence interval estimate of the mean, recall Equation (8.1) on page 287:

X +z o_T"\l n

is equal to half the width of the interval. Thisthe estimate that results from sampling error.

l ln this context, some

statisticians refer to e as

the "margin of error."

2You use Z instead of t

because, to determine the

critical value of t, you need

to know the sample size, but

you do not know it yet. For

most studies, the sample size

needed is large enough that

the standardized normal

distribution is a good

approximation of the

t distribution.

The amount added to or subtracted from Xquantity represents the amount of imprecision in

r . lThe sampl ing error . ' e . is def ined as

(8.4)

FIG

Micrwondetefor esaleJfor tlmpr

l=

lGh

See Ithis.

e = Z

Solving for r gives the sample size needed to construct the appropriate confidence intervalestimate for the mean. 'Appropriate" means that the resulting interval will have an acceptableamount of sampling error.

SAMPLE SIZE DETERMINATION FOR THE MEAN

The sample size, n, is equal to the product of the Z value squared and the variance,o, squared, divided by the square of the sampling erroq e,

o_T1 n

22o2n - " " } : -

Ie

To determine the sample size, you must know three factors:

1. The desired confidence level, which determines the value of Z, the critical value from thestandardized normal d is t r ibut ion2

2. The acceptable sampling error, e3. The standard deviation, o

In some business-to-business relationships that require estimation of important parame-ters, legal contracts specify acceptable levels of sampling error and the confidence levelrequired. For companies in the food or drug sectors, government regulations often specifysampling errors and confidence levels. In general, however, it is usually not easy to specify

Le-

/el

fv

ifv

FIGURE 8 .11

Microsoft Excelworksheet fordetermining sample sizetor est imating the meansales invoice amountfor the Saxon Homeimprovement Company

t-rt - - tulSee Section EB.4 to createmls.

8.4: Dctcrnr in ins Sarro lc Sizc 301

the two factors needed to determine the sample size. How can you determine the level of corr-fidence and sarnpling error? Typically, these questions are answered only by the subject mat-ter expert (that is, the individual most farnil iar with the variables under study). Although 95%is the most common confidence level used, if rnore confidence is desired" then 99%, rnight bemore appropriate; if less confidence is deemed acceptable, then 90%, uright be used. For thesampling error, you should think not of how much sampling error you would l ike to have (youreally do not want any error) but of how much you'can tolerate when drawing conclusionsfrom the data.

In addition to specifying the confidence level and the sarnpling error, you need an estimateof the standard deviation. Unfortunately, you rarely know the population standard deviation, o.In some instances, you can estimate the standard deviation from past data. In other situations,you can make an educated guess by taking into account the range and distribution ofthe vari-able. For example, if you assume a normal distribution, the range is approximately equal to 6o(that is, +3o around the mean) so that you estimate o as the range divided by 6. If you cannotestintate o in this way, you can conduct a small-scale study and estimate the standard deviatiorrf rorn the resul t iug dala.

To explore how to determine the sample size needed for estirnating the population mean,consider again the audit at Saxort Home lmprovement. In Section 8.2, yoLr selected a sample of100 sales invoices and constructed a 95o/o confidence interval estimate of the population rnearrsales invoice amount. How was this sample size detennined? Should you have selected a dif-ferent sample size'/

Suppose that, after consultation with company officials, you determine that a sarnplingerror of no more than +$5 is desi red, a long wi th 95ol conf idence. Past data indicate thatthe standard deviat ion of the sales arnount is approximately $25. Thus, e: $5, o - $25, andZ : | .96 ( for 95oh conf idence). Using Equat ion (8.4) ,

2 2 o 2 ( I . 9 6 ) 2 ( 2 5 ) 2

e - ( 5 ) "

= 96.04

Because the general rule is to slightly oversatisfy the criteria by rounding the sample sizeup to the next whole integer, you should select a sample of size 97. Thus, the sample of 100used on page 292 is close to what is necessary to satisfy the needs of the company, based on theestimated standard deviation, desired confidence level, and sampling error. Because the calcu-lated sample standard deviation is slightly higher than expecte4 $28.95 compared to $25.00,the corrfidence interval is slightly wider tharr desired. Figure 8. I I i l lustrates a Microsoft Excelworksheet to determine the samnle size.

-NORTS[{V(l - 86}21-((89'B4)/85r2

-ROUr{DUP(810, 0}

I234q67ct

I101 1t z; ;I J


EXAMPLE 8 .5

Example 8.5 illustrates another application of determining thedeveloo a confidence interval estimate for the mean.

samole size needed to

DETERMINING THE SAMPLE SIZE FOR THE MEAN

Returning to Example 8.3 on page 293, suppose.you want to estimate the population meanforce required to break the insulator to within +25 pounds with 95% confidence. On the basisofa study taken the previous year, you believe that the standard deviation is 100 pounds. Findthe sample size needed.

SOLUTION Using Equation (8.4) on page 300 and e : 25, o : 100, and Z :

confidence,for 95%

z2o2n = -' -

e 2

(1 .e6)2( loo)2

Therefore, you should select a sample size of 62 insulators because the general rule for deter-mining sample size is to always round up to the next integer value in order to slightly oversat-isfy the criteria desired.An actual sampling error slightly larger than 25 will result if the sample standard deviation cal-culated in this sample of 62 is greater than 100 and slightly smaller if the sample standard devi-ation is less than 100.

Sample Size Determination for the Proportion

So far in this section, you have learned how to determine the sample size needed for estthe population mean. Now suppose that you want to determine the sample size necessary forestimating the proportion of sales invoices at Saxon Home Improvement that contain errors.

To determine the sample size needed to estimate a population proportion, fi, you usemethod similar to the method for a population mean. Recall that in developing the samplefor a confidence interval for the mean, the sampling error is defined by

When estimating a proportion, you replace o with 1G(1 - " ) Thus, the sampling error is

Solving for n, you have the sample size necessary to develop a confidence interval estimatea proportion.

SAMPLE SIZE DETERMINATION FOR THE PROPORTION

The sample size r is equal to the Zvalne squared times the population proportion, r, timesminus the population proportion, n, divided by the square of the sampling error, e,

Q'2

Z2n(l - n), =

" ,

o-T" ln

e = Z

n ( l - n )

n

(8.s)


To determine the sample size, you must know three factors:

1. The desired confidence level, which determines the value of Z, the critical value from thestandardized normal distribution

2. The acceptable sampling error, e3. The population proportion, n

In practice, selecting these quantities requires some planning. Once you determine thedesired level of confidence, you can find the appropriate Z value from the standardized normaldistribution. The sampling error, e, indicates the amount of error that you are willing to toleratein estimating the population proportion. The third quantity, n, is actually the population paru-meter that you want to estimate! Thus, how do you state a value for what you are taking a sam-ple in order to determine?

Here you have two alternatives. In many situations, you may have past information or rele-vant experience that provide an educated estimate of n. Or, if you do not have past informationor relevant experience, you can try to provide a value for a that would never underestimate thesample size needed. Referring to Equation (8.5), you can see that the quantity n(l - ru)appears in the numerator. Thus, you need to determine the value of n that will make the quan-tity n(l - n) as large as possible. When n : 0.5, the product n(l - n) achieves its maximumresult. To show this, several values of n, along with the accompanying products of a(l - n), areas follows

When n : 0.9, then n(l - n): (0.9X0.1) : 0.09When n: 0.7, then n( l - n) : (0.7X0.3) :0.21When n : 0.5, then n( l - n): (0.5X0.5) :0.25

When n : 0.3, then n( l - n): (0.3X0.7) :0.21

When n :0 .1 , then n(1 - n ) : (0 .1 ) (0 .9 ) :0 .09

Therefore, when you have no prior knowledge or estimate of the population proportion, n,you should use 7r: 0.5 for determining the sample size. This produces the largest possible sam-ple size and results in the highest possible cost of sampling. Using ru : 0.5 may overestimate thesample size needed because you use the actual sample proportion in developing the confidenceinterval. You will get a confidence interval narrower than originally intended if the actual sam-ple proportion is different from 0.5. The increased precision comes at the cost of spendingmore time and money for an increased sample size.

Returning to the Saxon Home Improvement Using Statistics scenario, suppose that theauditing procedures require you to have 95%o confidence in estimating the population propor-tion of sales invoices with errors to within +0.07. The results from past months indicate thatthe largest proportion has been no more than 0.15. Thus, using Equation (8.5) on page 302 ande:0 .07 , n :0 .15 , andZ:1 .96 fo r 95% conf idence.

Z2n(l - n)

" 2(1.e6)2(0.1sxO.s5)

(0.07)"

= 99.96

Because the general rule is to round the sample size up to the next whole integer to slightlyoversatisfii the criteria, a sample size of 100 is needed. Thus, the sample size needed to satisf,ithe requirements of the company, based on the estimated proportion, desired confidence level,and sampling error, is equal to the sample size taken on page 297.The actual confidence inter-val is narrower than required because the sample proportion is 0. 10, while 0. l5 was used for nin Equation (8.5). Figure 8.12 shows a Microsoft Excel worksheet for determining sample size.


FIGURE 8.12

Microsoft Excelworksheet fordetermining samplesize for estimating theproportion of salesinvoices with errorsfor the Saxon Homelmprovement Companyt-lI \ N 7 TI - l l

E


A Bor thc Proportlon of InInor Salcr Invrlecl

Dd4 I|md. ofTru. ProDofdon 0.rl

unlllno Enor6 funcr Lrvrl7I lnlermedieto CalculalionsI ! Value 1t0 Celculated Samola Sizet l

t3 r Slr.

-lloRxsfirv(l - Bs]n]-(89^2. Bf. (1 - B4D/85^2

-ROUI{DUP(810,0)

Example 8.6 provides a second application of determining the sample sizethe population proportion.

Learning the Basics

@ 8.34 If you want to be 95%o confident of esti-lAsslsr I mating the population mean to within a sampling

error of +5 and the standard deviation is assumedto be 15, what sample size is required?

8.35 If you want to be 99o/o confident of esti-mating the population mean to within a samplingerror of +20 and the standard deviation is

assumed to be 100, what sample size is required?

8.36 If you want to be 99Yo confident of esti-mating the population proportion to within anerror of t0.04,what sample size is needed?

8.37 If you want to be 95o/o confident of esti-mating the population proportion to within anerror of +0.02 and there is historical evidence

that the population proportion is approximately 0.40, whatsample size is needed?


8.38 A survey is planned to determine the meanannual family medical expenses of employees ofa large company. The management of the com-

pany wishes to be 95oh confident that the sample mean iscorrect to within +$50 of the population mean annual fam-ily medical expenses. A previous study indicates that thestandard deviation is approximately $400.a. How large a sample size is necessary?b. If management wants to be correct to within +$25, what

sample size is necessary?

8.39 If the manager of a paint supply store wants to esti-mate the mean amount of paint in a l-gallon can to within

ffi

ffi

EXAMPLE 8.6 DETERMINING THE SAMPLE SIZE FOR THE POPULATION PROPORTION

You want to have 90% confidence of estimating the proportion of office workers whoto email within an hour to within +0.05. Because you have not previously undertaken suchstudy, there is no information available from past data. Determine the sample size needed.

SOLUTION Because no information is available from past data, assume that n: 0.50. UsiEquation (8.5) on page 302 and e :0.05, n : 0.50, and Z : 1.645 for 90% confidence,

(1.645)',(0.50X0.s0)

(0.05)2

= 270.6

Therefore, you need a sample of 27 | office workers to estimate the population proportionwithin +0.05 with 90% confidence.

with95% confidence and also assumes thatdeviation is 0.02 gallon. what sample size is

a quality control manager wants to estimate theof light bulbs to within +20 hours with 95% con-

also assumes that the population standard devi-100 hours, what sample size is needed?

If the inspection division of a county weights anddepartment wants to estimate the mean amount

soft-drink fill in 2-liter bottles to within +0.01 liter withconfidence and also assumes that the standard devia-is 0.05 liter, what sample size is needed?

8.42 A consumer group wants to estimate themean electric bill for the month of July for sin-gle-family homes in a large city. Based on studiesin other cities, the standard deviation is assumed

be $25. The group wants to estimate the mean bill forto within +$5 with 99% confidence.

What sample size is needed?lf 95% confidence is desired, what sample size isnecessarv?

An advertising agency that serves a major radio sta-wants to estimate the mean amount of time that the sta-audience spends listening to the radio daily. From past

ies, the standard deviation is estimated as 45 minutes.What sample size is needed if the agency wants to be90% confident of being correct to within +5 minutes?If 99% confidence is desired, what sample size isnecessary?

A growing niche in the restaurant business isbreakfast, lunch, and brunch. Chains in this

include Le Peep, Good Egg, Eggs & I, First Watch,Eggs Up Grill. The mean per-person check for First

is approximately $7, and the mean per-person checkEggs Up Grill is $6.50. (Extracted from J. Hayes,

ition Heats Up as Breakfast Concepts Eye Growth,"b Restaurant News, April 24,2006, pp. 8, 66.)

Assuming a standard deviation of $2.00, what samplesize is needed to estimate the mean per-person check forGood Egg to within $0.25 with 95% confidence?Assuming a standard deviation of $2.50, what samplesize is needed to estimate the mean per-person check forGood Egg to within $0.25 with 95% confidence?Assuming a standard deviation of $3.00, what samplesize is needed to estimate the mean per-person check forGood Egg to within $0.25 with 95% confidence?Discuss the effect of variation on selecting the samplesize needed.

The U.S. Department of Transportation defines anine flight as being "on time" if it lands less than 15 min-


eized reservation system. Cancelled and diverted flightsare counted as late. A study of the l0 largest U.S. domes-tic airlines found Southwest Airlines to have the lowestproportion of late arrivals, at 0.1577 (Extracted fromN. Tsikriktsis and J. Heineke, "The Impact of ProcessVariation on Customer Dissatisfaction: Evidence from theU S. Domesti c Airline lndustry," Dec isio n Sciences, Winter2004,35(l),pp.129-142). Suppose you were asked to per-form a follow-up study for Southwest Airlines in order toupdate the estimated proportion of late arrivals. What sam-ple size would you use in order to estimate the populationproportion to within an error ofa. +0.06 with95oh confidence?b. +0.04 with95o/o confidence?c. +0.02 with95o/o confidence?

8.46 In 2005, 34oh of workers reported that their jobswere more difficult, with more stress, and 37%o reportedthat they worry about retiring comfortably. (Extracted fromS. Armour, " Money Worries Hinder Job Performance,"USA Today, October 5,2005, p. Dl). Consider a follow-upstudy to be conducted in the near future.a. What sample size is needed to estimate the population

proportion of workers who reported that their jobs weremore difficult, with more stress, to within +0.02 with95% confidence?

b. What sample size is needed to estimate the populationproportion of workers who worried about retiring com-fortably to within +0.02 with 95% confidence?

c. Compare the results of (a) and (b). Explain why theseresults differ.

d. If you were to design the follow-up study, would you useone sample and ask the respondents both questions, orwould you select two separate samples? Explain therationale behind your decision.

8.47 What proportion of people hit snags with onlinetransactions? According to a poll conducted by HarrisInteractive, 89% hit snags with online transactions ("TopOnline Transaction Trouble," USA Today Snapshots, April4,2006,p. lD).a. To conduct a follow-up study that would provide 95%

confidence that the point estimate is correct to within+0.04 of the population proportion, how large a samplesize is required?

b. To conduct a follow-up study that would provide 99%confidence that the point estimate is correct to within+0.04 of the population proportion, how large a samplesize is required?

c. To conduct a follow-up study that would provide 95%confidence that the point estimate is correct to within+0.02 of the population proportion, how large a samplesize is required?

d. To conduct a follow-up study that would provide 99ohconfidence that the point estimate is correct to withinafter the scheduled time shown in the carrier's comput-


+0.02 of the population proportion, how large a samplesize is required?

e. Discuss the effects of changing the desired confidencelevel and the acceptable sampling error on sample sizerequirements.

8.48 A poll of 1,286 young adult cell phone users wasconducted in March 2006. These cell phone users, aged78-29, were actively engaged in multiple uses of their cellphones. The data suggest that 707 took still pictures withtheir phones, 604 played games, and 360 used the Internet(Extracted from "Poll: Cellphones Are Annoying butInvaluable," usatoday.com, April 3,2006). Construct a95% confidence interval estimate of the population propor-tion of young adults that used their cell phone toa. take still pictures.b. play games.c. use the Internet.

d. You have been asked to update the results ofthis study,Determine the sample size necessary to estimate thepopulation proportions in (a) through (c) to within +0.02with 95o/o confidence.

8.49 A study of 658 CEOs conducted by the ConferenceBoard reported that 250 stated that their company's great-est concern was sustained and steady top-line growth("CEOs' Greatest Concerns," USA Today Snapshots, May8,2006, p . 1D) .a. Construct a 95oh confidence interval for the proportion

of CEOs whose greatest concern was sustained andsteady top-line growth.

b. Interpret the interval constructed in (a).c. To conduct a follow-up study to estimate the population

proportion of CEOs whose greatest concern was sus-tained and steady top-line growth to within +0.01 with95% confidence, how many CEOs would you survey?

8.5 APPLICATIONS OF CONFIDENCE INTERVALESTIMATION IN AUDITINGThis chapter has focused on estimating either the population mean or the population propor-tion. In previous chapters, you have studied application to different business scenarios.Auditing is one of the areas in business that makes widespread use of probability samplingmethods in order to construct confidence interval estimates.

AUDITING

Auditing is the collection and evaluation of evidence about information relating to aneconomic entity, such as a sole business proprietor, a partnership, a corporation, or agovernment agency, in order to determine and report on how well the informationiorresponds to established criteria.

t,

Auditors rarely examine a complete population of information. Instead, they rely on esti-mation techniques based on the probability sampling methods you have studied in this text. Thefollowing list contains some of the reasons sampling is advantageous to examining the entirepopulation.

. Sampling is less t ime consuming.' Sampling is less costly.. Sampling provides results that are objective and defensible. Because the sample size is

based on demonstrable statistical principles, the audit is defensible before one's superionand in a court of law.

. Sampling provides an objective way of estimating the sample size in advance.' Sampling provides an estimate of the sampling error.' Sampling is often more accurate for drawing conclusions about large populations than

other methods. Examining large populations is time-consuming and therefore often subjectto more nonsampling error than statistical sampling.

. Sampling allows auditors to combine, and then evaluate collectively, samples collected bydifferent individuals.

r Sampling allows auditors to generalize their findings to the population with a known sam-pling error.

8.5: Applications of Confidence Interval Estimation inAuditing 307

Estimating the Population Total AmountIn auditing applications, you are often more interested in developing estimates of the popula-tion total amount than the population mean. Equation (8.6) shows how to estimate a popula-tion total amount.

ESTIMATING THE POPULATION TOTALThe point estimate for the population total is equal to the population size, N, times thesample mean.

Total = NX (8.6)

Equation (8.7) defines the confidence interval estimate for the population total.

CONFIDENCE INTERVAL ESTIMATE FOR THE TOTAL

(8.7)

To demonstrate the application of the confidence interval estimate for the population totalamount, return to the Saxon Home Improvement Using Statistics scenario on page 284. One ofthe auditing tasks is to estimate the total dollar amount of all sales invoices for the month. Ifthere are 5.000 invoices for that month and x = $ | 10.27. then using Equation (8.6).

ryX = (5,000X$l 10.27) = $551,350

If n : 100 and S: $28.95, then using Equation (8.7) with tnr: 7.9842 for 95o/o confidence,

rvx + uu --, ) ': .8: '" " ^ l n YN- l

,cNX + N( tn t l : -

" ln= 55r,350 t (s,000x1.e 84D?J-/t'999 too

' ./tOO 1 5,000 - I

N - n

N - 1

= 551,350 + 28,721.295(0.99005)

= 551,350 + 28,436

$522,914 < Population total < $579,786

Therefore, with95o/o confidence, you estimate that the total amount of sales invoices is between$522,914 and $579,786. Figure 8.13 shows a Microsoft Excel worksheet for these data.

8.13

Excelfor theinterva I

ofthe totalof all invoices

Saxon HomeCompany

Tolal Amounl of All Salos Invoiceta

3 Data4 Populodon She5 amph teln fio^2l6 Stmple Slze ilx

StmDlG Standard Devla$on 2SJlB fonffd**a Lsval 95t{s1B Intermediats Calcularioffit l rooulalion Total 551358.tXt2 'PC Factor 0.9s[13 itandard Enor of ths Toialtd eqrees of Freedom 9g1 5 Value I 54:l 6 r*srYal HafWidih m1x.7t1 718 Confidence lntcrval1S $tsffal Lowcr Llmlt 5,7,311Jf,

nlsrvrl Uoosr Llmlt 5797&t.72

*84 " 85-sQRr(Fi-B6)r{81-l}t-tB{. 87 " 812ySORTtB6l- 8 6 - t-illlvtt -m,814)-Bl5 * 813

-811 - 816-Sl1 + 316

E8.6 to create


Example 8.7 further illustrates the population total.

Difference EstimationAn auditor uses difference estimation when he or she believes that errors exist in a set ofand he or she wants to estimate the magnitude of the errors based only on a sample. Thelowing steps are used in difference estimation:

Determine the sample size required.Calculate the differences between the values reached during the audit and the originalues recorded. The difference in value i, denoted D,, is equal to 0 if the auditor finds thatoriginal value is correct, is a positive value when the audited value is larger than thenal value, and is negative when the audited value is smaller than the original value.

3. Compute the mean difference in the sample, D,by dividing the total differencebysample size, as shown in Equation (8.8).

MEAN DIFFERENCE

2o,D - t = t

n

where D, : Audited value - Original value

l .,

EXAMPLE 8 .7 DEVELOPING A CONFIDENCE INTERVAL ESTIMATEFOR THE POPULATION TOTAL

An auditor is faced with a population of 1,000 vouchers and wants to estimate the total valuethe population of vouchers. A sample of 50 vouchers is selected, with the following results:

Mean voucher amount (X ) = $1.076.39

Standard deviation (S) =$2'73.62

Construct a95o/o confidence interval estimate of the total amount for the population of

SOLUTION Using Equation (8.6) on page 307, the point estimate of the population total is

7try = (1,000)(1,07 6.39) = $1,076,390

From Equation (8.7) on page 307, a 95% confidence interval estimate of the populationamount is

( 1.000x 1,076.39) I (1.000x2.0 ogol2J3fr

= 1,076,390 + 77,762.878 (0.97517)

= 1,076,390 + 75,832

$1,000,558 < Population total < $1,152,222

Therefore, with95o/o confidence, you estimate that the total amount of the vouchers is$ 1,000,558 and $ l,l 52,222.

(8.8)

8.5: Applications of Confidence Interval Estimation in Auditing 309

4. Compute the standard deviation of the differences, Sr, as shown in Equation (8.9).Remember that any item that is not in error has a dilference value of 0.

STANDARD DEVIATION OF THE DIFFERENCE

S D =

S r - ^ : . ,

L (u i -D ri = l

n - l

5. Use Equation (8.10) to construct a confidence intervalthe population.

(8.e)

estimate of the total difference in

CONFIDENCE INTERVALESTIMATE FOR THE TOTAL DIFFERENCE

.s^ ffi;ND + N(tn) l#" , l - (s .10)'

4n \ l N - l

i,,- e o

D - t = t = ' " = 0 . 9 0n 100

The auditing procedures for Saxon Home Improvement require a 95o/o confidence intervalestimate of the difference between the actual dollar amounts on the sales invoices and theamounts entered into the integrated inventory and sales information system. Suppose that in asample of 100 sales invoices, you have 12 invoices in which the actual amount on the salesinvoice and the amount entered into the integrated inventory management and sales informa-tion system is different. These l2 differences (stored in the file[[@fift!) are

$9.03 $1 .41 $17 .32 $8.30 $s.21 $ r 0.80 $6.22 $5.63 $4.97 $7 .43 52.99 54.63

The other 88 invoices are not in error. Their differences are each 0. Thus,

numerator, there and3differences. Each

last88 are equal0.eF

S D =

n

\ tn, -D)'; - l

, - l

Sn = 2 '752

Using Equation (8.10), construct the95Vo confidence intervalence in the population of 5,000 sales invoices as follows:

(s,000)(0.90) r (s,000x1 .s84D2+

= 4 ,500 + 2 ,702.91

51,197.09 < Total difference < $'7,202.91

estimate for the total differ-

(9.03 - 0.9)2 + (7.47 - 0.9)2 + . . . + (0 - 0.9)2

3 10 CHAPTER EIGHT Confidence Interval Estimation

Thus, the auditor estimates with 95o/o confidence that the total difference between the salesinvoices, as determined during the audit and the amount originally entered into the account-ing system is between $1,797.09 and$7,202.91. Figure 8.14 shows an Excel worksheet forthese data.

FIGURE 8.14

Microsoft Excelworksheet for the totadifference between theinvoice amounts foundduring audit and theamounts entered intothe accounting systemfor the Saxon Homelmprovement Company

-SUl{olficrencod).t!!AAl-80/85-Ba 'B !0-SARTGl6)-saRT{(Bl -85}/(8r - ll}-F4 - 812'8lrySQRT(Fq-B!t - |-Tll{\t(l - 86, 815}-815 'B l l

-811 .817-B1l + 817


In the previous example, all 12 differences are positive because the actual amount on thesales invoice is more than the amount entered into the accounting system. In some circum-stances, you could have negative errors. Example 8.8 illustrates such a situation.

EXAMPLE 8 .8 DIFFERENCE EST]MATION

Returning to Example 8.7 on page 308, suppose that 14 vouchers in the sample of 50 voucherscontain errors. The values of the 14 errors are listed below and stored in the file !@@.Observe that two differences are nesative:

$75.41 $38.97 $108.54 -$37.18 $62.7s $l18.32 -$88.84

$r27.74 $s5.42 $39.03 $29.4r $47.99 $28.73 $84.05

Construct a 950/o confidence interval estimate of the total difference in the population of 1,000vouchers.

SOLUTION For these data,

S D =

tr ,: ?i 6e0.34D - ' = ' = = 1 3 . 8 0 6 8

n 5 0

) to, - D)'j = I

n - l

(75 .4r - 13 .8068)2 + (38 .97 - 13 .8068)2 + ' . '+ (0 - 13 .8068)2

otal Dlfforcnce In Acfual lnd Enter.d

= 37.427

-]

iIIj

8.5: Applications of Confidence Interval Estinration in Auditing 3 I 1

Using Equation (8. l0) on page 309, construct the confidence interval estirnate for the total dif-ference in the population as follows:

( 1,000x l 3.8068) +( 1,000)(2.0096 ) lg

= 13,806.8 + 10,372.4

$3,434.40 < Total difference < 524,179.20

Therefore, with 95o/o confidence, you estimate that thevouchers is between 53.434.40 and524.179.20.

d i f ference in the populat ion of

One-Sided Confidence Interval Estimation of theRate of Noncompliance with Internal Controls

Organizations use internal control mechanisms to ensure that individuals act in accordancewith company guidelines. For example, Saxon Home Improvement requires that an authorizedwarehouse-removal slip be completed before goods are removed from the warehouse. Duringthe monthly audit of the company, the auditing team is charged with the task of estimating theproportion of times goods were removed without proper authorization. This is referred to as therate of nonc'ompliance with the internal c'ontrol. To estimate the rate of noncompliance, audi-tors take a random sample of sales invoices and determine how often merchandise was shippedwithout an authorized warehouse-removal slip. The auditors then compare their results with apreviously established tolerable exception rate, which is the maximum allowable proportion ofitems in the population not in compliance. When estimating the rate of noncompliance, it iscommonplace to use a one-sided confidence interval. That is, the auditors estimate an upperbound on the rate ofnoncompliance. Equation (8.1 I ) defines a one-sided confidence intervalfor a proportion.

ONE-SIDED CONFIDENCE INTERVAL FOR A PROPORTION

Upperbound - p+Z (8.11)

where Z : the value corresponding to a cumulative area of ( I - cr) from the standardizednormal distribution (that is, a right-hand tailprobability of a).

Ifthe tolerable exception rate is higher than the upper bound" the auditor concludes that thecompany is in compliance with the internal control. If the upper bound is higher than the toler-able exception rate, the auditor concludes that the control noncompliance rate is too high. Theauditor may then request a larger sample.

Suppose that in the monthly audit, you select 400 sales invoices from a population of10,000 invoices. In the sample of 400 sales invoices,20 are in violation of the internal control.If the tolerable exception rate for this internal control is 6%, what should you conclude? Use a95o/o level of confidence.

The one-sided confidence interval is cornputed usingp :201400 - 0.05 and Z: 1.645.Using Equat ion (8.1 I ) ,

p( t - p )

n

= 0 .05 + 1 .645(0 .0109X0.98) = 0 .05 + 0 .0176 = 0 .0676


Learning the Basics

8.50 A sample of 25 is selected from a population of 500items. The sample mean is 25.7, and the sample standarddeviation is 7.8. Construct a99oh confidence interval esti-mate of the population total.

8.51 Suppose that a sample of 200 (see the file

@ft@Q is selected from a population of 10,000 items.Of these, l0 items are found to have errors of the follow-ing amounts:

13.76 42.87 34.65 I 1.09 14.5422.87 25.52 9.81 10.03 15.49

Construct a 95o/o confidence interval estimate of the totaldifference in the population.

8.52 If p: 0.04, n : 300, and N: 5,000, calculate theupper bound for a one-sided confidence interval estimateof the population proportion, 7t, using the following levelsofconfidence:a.90o/o b. 95% c. 99o/o

Thus, you have95o/o confidence that the rate of noncompliance is less than 6.76%. Becausetolerable exception rate is 6Vo, the rate of noncompliance may be too high for this internaltrol. In other words, it is possible that the noncompliance rate for the population is higherthe rate deemed tolerable. Therefore, you should request alarger sample.

In many cases, the auditor is able to conclude that the rate of noncompliance with thepany's internal controls is acceptable. Example 8.9 illustrates such an occurrence.


8.53 A stationery store wants to estimate the total

value of the 1,000 greeting cards it has in its inventory.Construct a95%o confidence interval estimate of the popu-lation total value ofall greeting cards that are in inventoryif a random sample of 100 greeting cards indicates a meanvalue of $2.55 and a standard deviation of $0.44.

8.54 The personnel department of a large cor-poration employing 3,000 workers wants to esti-mate the family dental expenses of its employees

to determine the feasibility of providing a dental insuranceplan. A random sample of 10 employees reveals the follow-ing family dental expenses (in dollars) for the precedingyear (see the !!S@ file):

l l0 362 246 85 510 208 173 425 316 r79

Construct a 90oh confidence interval estimate of the totalfamily dental expenses for all employees in the precedingyeat.

EXAMPLE 8 .9 ESTIMATING THE RATE OF NONCOMPLIANCE

A large electronics firm writes I million checks ayear. An internal control policy for thepany is that the authorization to sign each check is granted only after an invoice has beentialed by an accounts payable supervisor. The company's tolerable exception rate for thistrol is 4%. If control deviations are found in 8 of the 400 invoices sampled, what shouldauditor do? To solve this. use a95Yo level ofconfidence.

SOLUTION The auditor constructs a95Yo one-sided confidence interval for theinvoices in noncompliance and compares this to the tolerable exception rate. Using(8. 1 1 ) on page 3 I l, p : 81 400 : 0.02, and Z : 1.645 for 95olo confidence,

= 0.02 + 1.645

= 0.02 + 1.645(0.007X0.9998) = 0.02 + 0.01 l5 = 0.0315

The auditor concludes with95% confidence that the rate of noncompliance is less than 3.Because this is less than the tolerable exception rate, the auditor concludes that the icontrol compliance is adequate. In other words, the auditor is more thang5oh confidentrate of noncompliance is less than 4%.

0.02(l - 0.02)

8.6: Confidence Interval Estimation and Ethical Issues 3 l3

A branch ofa chain oflaree electronics stores is con-ing an end-of-month inventory of the merchandise in

There were 1,546 items in inventory at that time. Aof 50 items was randomlv selected and an audit

conducted, with the following results:

Value of Merchandise

X = 5252.28 ,S: $93.67

a 95% confidence interval estimate of the totalofthe merchandise in inventory at the end of the month.

A customer in the wholesale sarment trade is oftento a discount for a cash payment for goods. Theof discount varies by vendor. A sample of 150 itemsfrom apopulation of4,000 invoices at the end ofa

of time (see the!@@file) revealed that in 13the customer failed to take the discount to which he

she was entitled. The amounts (in dollars) of the 13 dis-that were not taken were as follows:

t5.32 97.36 230.63 r04.18 84.92 132.7688.32 47.81 89.01t2 26.55 129.43

a 99oh confidence interval estimate of the popu-total amount of discounts not taken.

Econe Dresses is a small company that manufactures's dresses for sale to specialty stores. It has 1,200

items, and the historical cost is recorded on ain, first out (FIFO) basis. In the past, approximatelyof the inventory items were incorrectly priced.

; any misstatements were usually not significant.of 120 items was selected (see the @$ file),

the historical cost of each item was compared with thevalue. The results indicated that l5 items differed

ir historical costs and audited values. These valuesas follows:

8.6

Sample Historical AuditedNumber Cost ($) Value ($)

Sample Historical AuditedNumber Cost ($) Value ($)

5 2619 8 7

17 201t8 t2l2 8 3 1 535 4t l43 24951 216

240105276l l 02983562tl305

60 21 2r073 140 r5286 t29 tlz95 340 2t696 341 402

107 135 97ll9 228 220

Construct a 95oh confidence interval estimate of the totalpopulation difference in the historical cost and auditedvalue.

8.58 Tom and Brent's Alpine Outfitters conduct an annualaudit ofits financial records. An internal control policy forthe company is that a check can be issued only after theaccounts payable manager initials the invoice. The tolera-ble exception rate for this internal control is 0.04. Duringan audit, a sample of 300 invoices is examined from a pop-ulation of 10,000 invoices, and 1l invoices are found toviolate the internal control.a. Calculate the upper bound for a 95oh one-sided confi-

dence interval estimate for the rate of noncompliance.b. Based on (a), what should the auditor conclude?

8.59 An internal control policy for Rhonda's OnlineFashion Accessories requires a quality assurance checkbefore a shipment is made. The tolerable exception ratefor this internal control is 0.05. During an audit, 500 ship-ping records were sampled from a population of 5,000shipping records, and 12 were found that violated theinternal control.a. Calculate the upper bound for a 95o/o one-sided confi-

dence interval estimate for the rate of noncompliance.b. Based on (a), what should the auditor conclude?

CONFIDENCE INTERVAL ESTIMATION AND ETHICAL ISSUESEthical issues relating to the selection of samples and the inferences that accompany them canoccur in several ways. The major ethical issue relates to whether confidence interval estimatesare provided along with the sample statistics. To provide a sample statistic without also includ-ing the confidence interval limits (typically set at 95o/o),the sample size used, and an interpre-tation of the meaning of the confidence interval in terms that a layperson can understand raisesethical issues. Failure to include a confidence interval estimate might mislead the user of theresults into thinking that the point estimate is all that is needed to predict the population char-acteristic with certainty. Thus, it is important that you indicate the interval estimate in a promi-nent place in any written communication, along with a simple explanation of the meaning ofthe confidence interval. In addition, you should highlight the sample size.

One of the most common areas where ethical issues concerning estimation occurs is in thepublication of the results of political polls. Often, the results of the polls are highlighted on the


front page of the newspapeq and the sampling error involved along with the methodology usedis printed on the page where the article is typically continued, often in the middle of the news-paper. To ensure an ethical presentation ofstatistical results, the confidence levels, sample size,and confidence limits should be made available for all surveys and other statistical studies.

8.7 o (CD-ROM Topicl ESTIMATION AND SAMPLE SlzEDETERMINATION FOR FINITE POPULATIONSIn this section, confidence intervals are developed and the sample size is determined for situa-tions in which sampling is done without replacement from a finite population. For further dis-cussion, seeE@EEEEUon the Student CD-ROM that accompanies this book.

This chapter discusses confidence intervals for estimatingthe characteristics of a population, along with how you candetermine the necessary sample size. You learned how anaccountant at Saxon Home Improvement can use the sample

data from an audit to estimate important population parame-ters such as the total dollar amount on invoices and the pro-portion of shipments made without the proper authorization.Table 8.3 provides a list oftopics covered in this chapter.

Type ofDataT A B L E 8 . 3

Summary of Topicsin Chapter B

Tlpe ofAnalysis Numerical Categorical

Confidence interval fora population parameter

To determine what equation to use for a particular situ-ation, you need to ask several questions:

. Are you developing a confidence interval or are youdetermining sample size?

. Do you have a numerical variable or do you have a cate-gorical variable?

Confidence Interval for the Mean (o Known)

Confidence interval estimatefor the mean (Sections 8.1and 8.2)

Confidence intervalestimate for the total andthe mean difference(Section 8.5)

Confidence interval estimatefor the proportion (Section 8.3).

One-sided confidence intervalestimate for the proportion(Section 8.5)

X <X+z* (8.1)

. Ifyou have a numerical variable, do you know the popu-lation standard deviation? If you do, use the normal dis-tribution. If you do not, use the t distribution.

The next four chapters develop a hypothesis-testingapproach to making decisions about population parameten.

Confidence Interval for the Mean (o Unknown)

,sT,

s.sX - t n - t - - r < t t < X + t n - ' #

tln .ln

o-T=\ ln

X+Z

- ' * ,\ ln

p(r - p)n

Confidence Interval Estimate for the ProportionL .

P = zl!!!:-!)\ n

Chapter Review Problems 315

Sample Size Determination for the Mean

22o2ll -- ----=-

e '

Sample Size Determination for the Proportion

Z2n( l - n)

e -

(8.4)

(8.s)

Checking Your Understanding8.60 Why can you never really have 100% confidence ofconectly estimating the population characteristic of interest?

1 When do you use the / distribution to develop theinterval estimate for the mean?

Why is it true that for a given sample size, n. anin confidence is achieved by widening (and mak-

less precise) the confidence interval?

Under what circumstances

p -Z <1 r< p+z

auditing 306confidence interval estimate 284critical value 287degrees offreedom 290

(8.3)

differenceestimation 308level ofconfidence 287one-sided confidence interval 3l Ipoint estimate 284

sampling error 300Student's I distribution 290total amount 307

the national average of 900. After four weeks, the samplestores stabilize at a mean customer count of 974 and a stan-dard deviation of 96. This increase seems like a substantialamount to you, but it also seems like a pretty small sample.Is there some way to get a feel for what the mean per-storecount in all the stores will be if you cut coffee prices nation-wide? Do you think reducing coffee prices is a good strat-egy for increasing the mean customer count?

8.67 Companies are spending more time screening appli-cants than in the past. A study of 102 recruiters conducttldby ExecuNet found that 77 did Internet research on candi-dates. (Extracted from P. Kitchen, "Don't Let Any 'Digital

Dirt'BuryYour Job Prospects," Nousday, August 21,2005,p. A5e).a. Construct a 95Vo confidence interval estimate of the

population proportion of recruiters who do Internetresearch on candidates.

b. Based on (a), is it correct to conclude that more than70o/o ofrecruiters do Internet research on candidates?

c. Suppose that the study uses a sample size of 400recruiters and 302 did Internet research on candidates.Construct a 95o/o confidence interval estimate of thepopulation proportion of recruiters who do Internetresearch on candidates.

d. Based on (c), is it correct to conclude that more than700/o of recrtiters do Internet research on candidates?

e. Discuss the effect of sample size on your answers to(a) through (d).

8.68 High-fructose corn syrup (HFCS) was created in the1970s and is used todav in a wide varietv of foods and

idence interval instead of ado you use a one-sidedtwo-sided confidence

When would you want to estimatethe populationinstead of the population mean?

How does difference estimation differ from estima-of the mean?

the ConceptsYouwork in the corporate office for a nationwide

store franchise that operates nearly 10,000The per-store daily customer count has been steady

900 for some time (that is, the mean number of customersa store in one day is 900). To increase the customer count,franchise is considering cutting coffee prices by approx-

half. The l2-ounce size will now be $.59 instead of, and the 16-ounce size will be $.69 instead of $1.19.with this reduction in price, the franchise will have agross margin on coffee. To test the new initiative, theise has reduced coffee prices in a sample of 34 stores,customer counts have been running almost exactly at

p(r - p)n


beverages. HFCS is cheaper than sugar and is about 75%sweeter than sucrose. Some researchers think that HFCS islinked to the growing obesity problem in the United States(Extracted from P. Lempert, "War of the Sugars,"Progressive Grocer, April 15, 2006,p.20). The followingconsumer views are from a nationwide survev of 1.1l4responses:

Views on HFCS

Yes No

Are you concerned about consuming HFCS? 80o/o 20o/oDo you think HFCS should be bannedin food sold to schools? 88% 12%Do you think HFCS should be bannedinallfoods? 56% 44%

Construct a95o/o confidence interval estimate of the popu-lation proportion of people whoa. are concerned about consuming HFCS.b. think HFCS should be banned in food sold to schools.c. think HFCS should be banned in all foods.d. You are in charge of a follow-up survey. Determine the

sample size necessary to estimate the proportions in (a)through (c) to within +0.02 with 95olo confidence.

8.69 Starwood Hotels conducted a survey of 401 topexecutives who play golf (Extracted from D. Jones, "ManyCEOs Bend the Rules (of Golf)," USA Today, June 26,2002). Among the results were the following:

. 329 cheat at golf.

. 329 hate others who cheat at golf.

. 289 believe business and golfbehavior parallel.

. 80 would let a client win to get business.

. 40 would call in sick to play golf.Construct a 95o/o confidence interval estimate for each ofthese questions. Based on these results, what conclusionscan you reach about CEOs'attitudes toward golfl

8.70 A market researcher for a consumer electronicscompany wants to study the television viewing habits ofresidents of a particular area. A random sample of 40respondents is selected, and each respondent is instructedto keep a detailed record of all television viewing in a par-ticular week. The results are as follows:

. Viewing t ime per week: X = 15.3 hours, . t : 3.8hours.

. 27 respondents watch the evening news on at least 3weeknights.

a. Construct a 95o/o confidence interval estimate for themean amount of television watched per week in thiscity.

b. Construct a 95oh confidence interval estimate for thepopulation proportion who watch the evening news onat least 3 weeknights per week.

Suppose that the market researcher wants to takeanother survey in a different city. Answer these questions:c. What sample size is required to be 95o/o confident of

estimating the population mean to within *2 hours andassumes that the population standard deviation is equalto 5 hours?

d. What sample size is needed to be 95o/o confident ofbeing within +0.035 of the population proportion whowatch the evening news on at least 3 weeknights if noprevious estimate is available?

e. Based on (c) and (d), what sample size should the mar-ket researcher select if a single survey is being con-ducted?

8.71 The real estate assessor for a county governmentwants to study various characteristics of single-familyhouses in the county. A random sample of 70 housesreveals the following

. Heated area of the houses (in square feet): X = 1,759,^s: 380.

. 42 houses have central air-conditioning.a. Construct a 99o/o confidence interval estimate of the

population mean heated area of the houses.b. Construct a 95o/o confidence interval estimate of the

population proportion of houses that have central air-conditioning.

8.72 The personnel director of a large corporation wishesto study absenteeism among clerical workers at the corporation's central office during the year. A random sample of25 clerical workers reveals the following:

. Absenteeism: X = 9.7 days, S: 4.0 days.

. l2 clerical workers were absent more than l0 days.a. Construct a 95o/o confidence interval estimate of t

mean number of absences for clerical workers durithe year.

b. Construct a 95o/o confidence interval estimate of thepopulation proportion of clerical workers absentthan 10 days during the year.Suppose that the personnel director also wishes to take

survey in a branch office. Answer these questions:c. What sample size is needed to have 95% confidence i

estimating the population mean to within +1.5 daysthe population standard deviation is 4.5 days?

d. What sample size is needed to have 90% confidenceestimating the population proportion to within +0.075no previous estimate is available?

e. Based on (c) and (d), what sample size is needed if agle survey is being conducted?

8.73 The market research director for Dotfy'sStore wants to study women's spending onA survey of the store's credit card holders is designed into estimate the proportion of women who purchase

primarily from Dotty's Department Store and theyearly amount that women spend on cosmetics. A pre-survev found that the standard deviation of the amount

spend on cosmetics in a year is approximately $ 18.sample size is needed to have 99o/o confidence of

imating the population mean to within +$5?

,-What sample size is needed to have 90% confidenceestimating the population proportion to within.045?

on the results in (a) and (b), how many of thecard holders should be sampled?

The branch manager of a nationwide bookstorewants to study characteristics of her store's cus-

She decides to focus on two variables: the amountspent by customers and whether the customers

consider purchasing educational DVDs relating topreparation exams, such as the GMAT, GRE, or

. The results from a sample of 70 customers are as

Amount spent: X = $28.52,,S: $11.39.28 customers stated that they would consider pur-chasins the educational DVDs.

a 95%o confidence interval estimate of themean amount spent in the bookstore.a 90% confidence interval estimate of the

Lon proportion of customers who would considerine educational DVDs.that the branch manaser of another store in the

wants to conduct a similar survey in his store.the following questions:sample size is needed to have 95% confidence ofing the population mean amount spent in his

to within +$2 if the standard deviation is assumed$10?sample size is needed to have 90% confidence ofing the population proportion who would con-

purchasing the educational DVDs to within,|

on your answers to (c) and (d), how large a sam-should the manager take?

branch manager of an outlet (Store l) of achain of pet supply stores wants to study char-

ics of her customers. In particular, she decides totwo variables: the amount of money spent by cus-

and whether the customers own only one dog, onlyor more than one dog and/or cat. The results from aof 70 customers are as follows:

t of money spent: X = $21.34, S:59.22.customers own only a dog.customers own only a cat.

Chapter Review Problems 317

a. Construct a 95oh confidence interval estimate of thepopulation mean amount spent in the pet supply store.

b. Construct a 90oh confidence interval estimate of thepopulation proportion of customers who own only acat.The branch manager of another outlet (Store 2) wishes

to conduct a similar survey in his store. The manager doesnot have access to the information generated by the man-ager of Store l. Answer the following questions:c. What sample size is needed to have 95% confidence of

estimating the population mean amount spent in hisstore to within +$1.50 if the standard deviation is $10?

d. What sample size is needed to have 90% confidence ofestimating the population proportion of customers whoown only a cat to within +0.045?

e. Based on your answers to (c) and (d), how large a sam-ple should the manager take?

8.76 The owner of a restaurant that serves continentalfood wants to study characteristics of his customers. Hedecides to focus on two variables: the amount of moneyspent by customers and whether customers order dessert.The results from a sample of 60 customers are asfollows:

. Amount spent: X = $38.54,S= $7.26.

. l8 customers purchased dessert.a. Construct a 95o/o confidence interval estimate of the

population mean amount spent per customer in therestaurant.

b. Construct a 90oh confidence interval estimate of thepopulation proportion of customers who purchasedessert.The owner of a competing restaurant wants to conduct a

similar survey in her restaurant. This owner does not haveaccess to the information of the owner of the first restau-rant. Answer the following questions:c. What sample size is needed to have 95% confidence of

estimating the population mean amount spent in herrestaurant to within +$1.50, assuming that the standarddeviation is $8?

d. What sample size is needed to have 90% confidence ofestimating the population proportion of customers whopurchase dessert to within +0.04?

e. Based on your answers to (c) and (d), how large a sam-ple should the owner take?

8.77 The manufacturer of "Ice Melt" claims its productwill melt snow and ice at temperatures as low as 0oFahrenheit. A representative for a large chain of hardwarestores is interested in testing this claim. The chain pur-chases a large shipment of 5-pound bags for distribution.The representative wants to know with 95% confidence,within +0.05, what proportion of bags of Ice Melt performthe job as claimed by the manufacturer.

's credit

The

mers own more than one dos and/or cat.


a. How many bags does the representative need to test?What assumption should be made concerning the popu-lation proportion? (This is called destructive testing;that is, the product being tested is destroyed by the testand is then unavailable to be sold.)

b. The representative tests 50 bags, and42 ofthem do thejob as claimed. Construct a 95o/o confidence intervalestimate for the population proportion that will do thejob as claimed.

c. How can the representative use the results of (b) todetermine whether to sell the Ice Melt product?

8.78 An auditor needs to estimate the percentage of times acompany fails to follow an internal control procedure. A sam-ple of 50 from a population of 1,000 items is selecte4 and in7 instances, the internal control procedure was not followed.a. Construct a90% one-sided confidence interval estimate

of the population proportion of items in which the inter-nal control procedure was not followed.

b. If the tolerable exception rate is 0.15, what should theauditor conclude?

8.79 An auditor for a government agency needs to evalu-ate payments for doctors'office visits paid by Medicare ina particular zip code during the month of June. A total of25,056 visits occurred during June in this area. The auditorwants to estimate the total amount paid by Medicare towithin +$5 with 95% confidence. On the basis of pastexperience, she believes that the standard deviation isapproximately $30.a. What sample size should she select?Using the sample size selected in (a), an audit is conductedwith the following results.

Amount of Reimbursement

X = $93.70 ,l = $34.55

In l2 of the office visits, an incorrect amount of reimburse-ment was provided. For the 12 office visits in which therewas an incorrect reimbursement, the differences betweenthe amount reimbursed and the amount that the auditordetermined should have been reimbursed were as follows(and are stored in the file@!@

$17 $2s $14 -$10 $20 $40 $3s $30 $28 $22 $15 $5

b. Construct a 90% confidence interval estimate of the pop-ulation proportion of reimbursements that contain errors.

c. Construct a 95o/o confidence interval estimate of thepopulation mean reimbursement per office visit.

d. Construct a 95%o confidence interval estimate of thepopulation total amount of reimbursements for this geo-graphic area in June.

e. Construct a 95%o confidence interval estimate of thetotal difference between the amount reimbursed and the

amount that the auditor determined should have beenreimbursed.

8.80 A home furnishings store that sells bedroom furni-ture is conducting an end-of-month inventory of the beds(mattress, bed spring, and frame) in stock. An auditor forthe store wants to estimate the mean value of the beds instock at that time. She wants to have 99% confidence thather estimate of the mean value is correct to within +$100.On the basis of past experience, she estimates that the stan-dard deviation ofthe value ofa bed is $200.a. What sample size should she select?b. Using the sample size selected in (a), an audit was con-

ducte4 with the following results:

X = $1,654.27 S = $184.62

Construct a 99o/o confidence interval estimate of the totalvalue of the beds in stock at the end of the month if therewere 258 beds in stock.

8.81 A quality characteristic of interest for a tea-bag-filling process is the weight of the tea in the individualbags. In this example, the label weight on the package indi-cates that the mean amount is 5.5 grams of tea in a bag. Ifthe bags are underfille4 two problems arise. First, cus-tomers may not be able to brew the tea to be as strong asthey wish. Second, the company may be in violation of thetruth-in-labeling laws. On the other hand, if the meanamount of tea in a bag exceeds the label weight, the com-pany is giving away product. Getting an exact amount oftea in a bag is problematic because of variation in the tem-perature and humidity inside the factory, differences in thedensity of the tea, and the extremely fast filling operationof the machine (approximately 170 bags per minute). Thefollowing data are the weights, in grams, of a sample of 50tea bags produced in one hour by a single machine (thedata are stored in the file S$!!@:

Weight of Tea Bags, in Grams

5.65 s.44 5.42

5.57 5.40 5.53

5.47 5.40 s.47

5.77 5.57 5.42

5.61 5.45 5.44

5.40 s.53 5.34 5.54 s.4s 5.52

5.54 5.s5 5.62 5.56 5.46 5.44

5.61 5.53 5.32 5.67 5.29 5.49

5.58 5.58 5.50 5.32 5.50 5.53

5.25 5.56 s.63 5.50 5.57 5.67

5.41

5.5Ii

5.55

a. Construct a 99o/o confidence interval estimate ofpopulation mean weight of the tea bags.

b. Is the company meeting the requirement set forth onlabel that the mean amount of tea in a bag is 5.5

8.82 A manufacturing company produces steelfor electrical equipment. The main component part ofhousing is a steel trough that is made out of a l4-gauge

I

ll

5

It is produced using a 250-ton progressive punch pressa wipe-down operation that puts two 90-degree formsflat steel to make the troush. The distance from one

of the form to the other is critical because of weather-in outdoor applications. The data (stored in the

from a sample of 49 troughs follows:

Width of Trough, in Inches

8.343 8.317 8.383 8.348 8.410 8.351 8.313 8.481 8.422

8.382 8.484 8.403 8.414 8.419 8.385 8.465 8.498 8.447

8.413 8.489 8.414 8.481 8.4t5 8.479 8.429 8.458 8.462

8.M 8.429 8.460 8.412 8.420 8.410 8.40s 8.323 8.420

8.447 8.405 8.439 8.411 8.427 8.420 8.498 8.409

Construct a 95o/o confidence interval estimate of themean width of the troughs.Interpret the interval developed in (a).

The manufacturer of Boston and Vermont asphaltknow that product weight is a major factor in the's perception of quality. The last stage of theline packages the shingles before they are placed

wooden pallets. Once a pallet is full (a pallet for mostholds l6 squares of shingles), it is weighed, and the

t is recorded. The data file [@ containsweight (in pounds) from a sample of 368 pallets of

shingles and 330 pallets of Vermont shingles.For the Boston shingles, construct a 95oh confidenceinterval estimate of the mean weight.For the Vermont shingles, construct a 95Vo confidenceinterval estimate of the mean weight.Evaluate whether the assumption needed for (a) and (b)has been seriously violated.Based on the results of(a) and (b), what conclusions canyou reach concerning the mean weight of the Bostonand Vermont shingles?

The manufacturer of Boston and Vermont asphaltprovides its customers with a 20-year warranty on

of its products. To determine whether a shingle will lastlong as the warranty period" acceleratedlife testing is con-

at the manufacturing plant. Accelerated-life testingthe shingle to the stresses it would be subject to in aof normal use via a laboratory experiment that takes

a few minutes to conduct. In this test, a shingle isscraped with a brush for a short period oftime,

and the shingle granules removed by the brushing areweighed (in grams). Shingles that experience low amountsof granule loss are expected to last longer in normal use thanshingles that experience high amounts of granule loss. In thissituation, a shingle should experience no more than 0.8grams of granule loss if it is expected to last the lengh of the

Chapter Review Problems 3I9

warranty period. The data file !@ contains a sampleof 170 measurements made on the company's Boston shin-gles and 140 measurements made on Vermont shingles.a. For the Boston shingles, construct a 95oh confidence

interval estimate of the mean granule loss.b. For the Vermont shingles, construct a 95o/o confidence

interval estimate of the mean granule loss.c. Evaluate whether the assumption needed for (a) and (b)

has been seriously violated.d. Based on the results of (a) and (b), what conclusions can

you reach concerning the mean granule loss of theBoston and Vermont shingles?

Report Writing Exercises8.85 Referring to the results in Problem 8.82 on page 318concerning the width of a steel trough, write a report thatsummarizes your conclusions.

Team Proiect8.86 Refer to the team project on page 73 (see the

Eil@@@ file). Construct all appropriate confidenceinterval estimates of the population characteristics of low-risk, average-risk, and high-risk mutual funds. Includethese estimates in a report to the vice president for researchat the financial investment service.

Student Survey Database8.87 Problem 1.27 on page 15 describes a survey of 50undergraduate students (see the file GEEEEffiE!E$.a. For these data, for each variable, construct a95%o confi-

dence interval estimate of the population characteristic.b. Write a report that summarizes your conclusions.

8.88 Problem 1.27 on page 15 describes a survey of 50undergraduate students (see the fileEEE@).a. Select a sample of 50 undergraduate students at your

school and conduct a similar survey for those students.b. For the data collected in (a), repeat (a) and (b) of

Problem 8.87.c. Compare the results of (b) to those of Problem 8.87.

8.89 Problem 1.28 on page 15 describes a survey of 50MBA students (see the fileEEElffiElldH).a. For these data, for each variable, construct a95o/o confi-

dence interval estimate of the population characteristic.b. Write a report that summarizes your conclusions.

8.90 Problem 1.28 on page 15 describes a survey of 50MBA students (see the file!@@l!@).a. Select a sample of 50 graduate students in your MBA

program and conduct a similar survey for those students.b. For the data collected in (a), repeat (a) and (b) of

Problem 8.89.c. Compare the results of (b) to those of Problem 8.89.

fsr')6

h"Ihrbrgshercl


Managing the Springville Herald

The marketing department has been considering ways toincrease the number of new subscriptions and increase therate of retention among customers who agreed to a trialsubscription. Following the suggestion of AssistantManager Lauren Alfonso, the department staff designed asurvey to help determine various characteristics of readersof the newspaper who were not home-delivery subscribers.The survey consists of the following 10 questions:

1. Do you or a member ofyour household ever purchasethe Springville Hera ld?( l ) Y e s ( 2 ) N o

[ f the respondent answers no,terminated.]

2. Do you receive the Springvilledelivery?( l )Yes (2 ) No

[f no, skip to question 4.]3. Do you receive the Springville Herald:

( I ) Monday-Saturday (2) Sunday only (3) Every day[f every day, skip to question 9.]

6 .

7 .

How often during the Monday-Saturday period doyou purchase the Springville Herald?(l) Every day (2) Most days (3) Occasionally or neverHow often do you purchase the Springville Herald onSundays?( l) Every Sunday (2) 2-3 Sundays per month(3) No more than once a monthWhere are you most likely to purchase the SpringvilleHerald?(l) Convenience store (2) Newsstand/candy store(3) Vending machine (4) Supermarket (5) OtherWoufd you consider subscribing to the SpringvilleHerald for a trial period if a discount were offered?(l) Yes (2) No[f no, skip to question 9.]The Springville Herald currently costs $0.50Monday-Saturday and $1.50 on Sunday, for a total of$4.50 per week. How much would you be willing to payper week to get home delivery for a 90-day tnal period?Do you read a daily newspaper other than theSpringville Herald?( l ) Y e s ( 2 ) N oAs an incentive for long-term subscribers, the news-paper is considering the possibility ofoflering a cardthat would provide discounts at certain restaurants inthe Springville area to all subscribers who pay inadvance for six months of home delivery. Would youwant to get such a card under the terms of this offer?( l )Yes (2 ) No

The group agreed to use a random-digit dialingmethod to poll 500 local households by telephone. Usingthis approach, the last four digits of a telephone number arerandomly selected to go with an area code and exchange(the first 6 digits of a l0-digit telephone number). Onlythose pairs of area codes and exchanges that were for theSpringville city area were used for this survey.

Of the 500 households selected" 94 households eitherrefused to participate, could not be contacted after repeatedattempts, or represented telephone numbers that were notin service. The summary results are as follows:

Households That Purchase the Springville Herald Frequencythe in terv iew is

Herald via homeYesNo

Households with Home Delivery

352J A

Frequency

YesNo

Type of Home Delivery Subscription

1 3 6216

Frequency4.

5 .

Monday-SaturdaySunday only7 days a week

Purchase Behavior of Nonsubscribersfor Monday-Saturday Editions

l 82593

Frequency

Every dayMost daysOccasionally or never

Purchase Behavior of Nonsubscribersfor Sunday Editions

789543

Frequency

8 .

Every Sunday2-3 Sundays a monthNo more than once a month

Nonsubscribers' Purchase Location

1 3 854)4

Frequency

9.Convenience storeNewsstand/candy storeVending machineSupermarketOther locations

Would Consider Trial Subscriptionif Offered a Discount

74952113I J

Frequency

YesNo

+0

170

t0 .

Rate Willing to Pay per Week (in Dollars) Data file EIjEEEfor a 90-Day Home-Delivery Trial Subscription

4.15 3.60 4.10 3.60 3.60 3.60 4.40 3.ts 4.00 3.75 4.003.25 3.75 3.30 3.75 3.65 4.00 4.r0 3.90 3.50 3.75 3.003.40 4.00 3.80 3.50 4.10 4.25 3.50 3.90 3.95 4.30 4.203.50 3.75 3.30 3.85 3.20 4.40 3.80 3.40 3.50 2.85 3.753.80 3.90

Read a Daily Newspaper Other Thanthe Springville Herald Frequency

r38214

Frequency

Rcf-crenccs 321

EXERCISESSH8.1 Some members of the marketing departrrent are

concerned about the random-digit dialing methodused to collect survey responses. Prepare a memo-randum that examines the following issues:. The advantages and disadvantages of using the

random-di git dialing rnethod.' Possib le a l ternat ive approaches for conduct-

ing the survey and their advantages and dis-advantages,

SH8.2 Analyze the resul ts of thehouseholds. Write a reportket ing impl icat ions of theSpringville Heruld.

YesNo

66286

YesNo

survey of Spr ingvi l lethat discusses the urar-survey results for the

Would Prepal' Six Months to Receivea Restaurant Discount Card

Web Case

Applt' your knov'ledge about confidenc'e intervctl estima-tion in this Weh Case. whic'h extends the OurCamptrs! WebCase.from Chapter 6.

Among its other features, the OurCampus! Web siteallows customers to purchase OurCampus! LifeStyles mer-chandise online. To handle payment processing, the man-agement of OurCampus! has contracted with the followingfirrns:. PayAFr iend (PAF): an onl ine payment system wi th

which customers and businesses such as OurCampus!register in order to exchange payments in a secure andconvenient lranner without the need for a credit card.

. Continental Banking Company (Conbanco): a process-ing serv ices provider that a l lows OurCampus! cus-tomers to pay for merchandise using nationally recog-nized credit cards issued by a financial institution.

To reduce costs, the management is considering elimi-nating one of these two payment systems. However,Virginia Duffy of the sales department suspects that cus-tomers use the two forms of payment in unequal numbers

and that customers d isplay d i f ferent buying behaviorswhen using the two forrns of payment. Therefore, shewould l ike to first determrne

a. the proportion of customers using PAF and the pro-portion of customers using a credit card to pay for theirpurchases.

b. the mean purchase amount when using PAF and themean purchase amount when using a credit card.

Assis t Ms. Duf fy by prepar i r rg an appropr iate analys isbased on a random sample of 50 transactions that she hasprepared and placed in an internal f i le on the OurCampus!Web site, www.prenhall.com/Springvil le/OurCampus_PymtSample.htm. Summarize your f ind ings and deter-mine whether Ms. Duffy's conjectures about OurCampus!customer purchasing behaviors are correct. If you want thesampling error to be no more than $3 when estimating themean purchase amount, is Ms. Duffy's sanrple largeenough to perform a valid analysis?

1. Cochran, W. G., Sumplingkc'hnique.r, 3rd ed. (NewYork:Wiley, 1977).

2. Fisher, R. A., and F. Yates, Sraristical Tables fi irBiological, Agricultural and Medic'al Researc'h,5th ed.(Edinburgh: Oliver & Boyd, 1957).

3. Kirk, R. E., ed., Statistical Issues: A Reader /br theBehavinraI Sc'ienc'es (Beln-ront, CA: Wadsworth, 1972).

4. Larsen, R. L., and M. L. Marx, An Introduction to Mqth-ematical Statistic's and lts Applic'ations, 4th ed. (UpperSaddle River, NJ: Prentice Hall, 2006).

5. Microsoft Excel 2007 (Redmond, WA: Microsoft Corp.,2007).

6. Snedecor, G. W., and W G. Cochran" Statisticul Methods,7th ed. (Arrres, IA: Iowa State University Press, 1980).

322 EXCEL coMpANIoN to chaDter 8

E8.1 COMPUTING THE CONFIDENCEINTERVAL ESTIMATE FOR THEMEAN (o KNOWN)

You compute the confidence interval estimate for the mean(o known) either by using the PHStat2 Estimate for theMean, sigma known procedure or by making entries in theg@@EEworkbook.

Usinq PHStat2 Estimatefor tFe Mean, Sigma Known

Select PHStat ) Confidence Intervals ) Estimate forthe Mean, sigma known. ln the Estimate for the Mean,sigma known dialog box (shown below), enter values forthe Population Standard Deviation and the ConfidenceLevel. Click one of the input options and make the requiredentries. Enter a tit le as the Title and click OK.

If you know the sarnple size and sample mean of yoursample, click Sample Statistics Known and enter thosevalues. Otherwise, click Sample Statistics Unknown andenter the cell range of your sample as the Sample CellRange.

Using the CIE_SK WorksheetOpen to the CIE_SK worksheet of theworkbook. This worksheet uses the NORMSINV(P<Eand CONFIDENCE(l-conJidence level, population stan-dard deviution, sample siee) functions to compute theZ value and interval half-width for the Example 8.1 meanpaper length problem on page 288. To adapt this worksheetto other problems, enter the appropriate population stan-dard deviation, sample mean, sample size, and confidencelevel values in the tinted cells 84 throush 87 and enter anew t i t le in cel l A l .

E8.2 COMPUTING THE CONFIDENCEINTERVAL ESTIMATE FOR THEMEAN (o UNKNOWN)

You compute the confidence interval estimate for the mean(o unknown) either by using the PHStat2 Estimate for theMean, sigma unknown procedure or by making entries inthe EIEEEEEEEEIE wo rkbook.

Usinq PHStat2 Estimatefor tFe Mean, Sigma Unknown

Select PHStat ) Confidence Intervals ) Estimate forthe Mean, sigma unknown. In the Estimate for the Mean,sigma unknown dialog box (shown on page 323), enter aConfidence Level value, click one of the input options,and make the required entries. Enter a title as the Title andclick OK.

If you know the sample size, sample mean, and samplestandard deviation of your sample, click Sample StatisticsKnown and enter those values. Otherwise, click SampleStatistics Unknown and enter the cell range of your sam-ple as the Sample Cell Range.

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Using the CIE_SU WorksheetOpen to the clE_su worksheet of th.EIEEEE@IEworkbook. The worksheet (see Figure 8.6 on page 293) usesthe TINV( 1 -c o nJi d e n c e I eve I, d egre e s of fre e d o m) functionto determine the crit ical value from the r distribution andcompute the interval half-width for the Section 8.2 SaxonHome Improvement Company example. To adapt this work-sheet to other problems, change the sample statistics andconfidence level values in the tinted cells 84 throush 87and enter a new tit le in cell Al.

E8.3 COMPUTING THE CONFIDENCEINTERVAL ESTIMATE FOR THEPROPORTION

You compute the confidence interval estimate for the pro-portion either by using the PHStat2 Estimate for theProportion procedure or by making entries in theq[f[[fis[[!workbook.

Usinq PHStat2 Estimatefor tf,e Proportion

Select PHStat ) Confidence Intervals ) Estimate forthe Proportion. In the Estimate for the Proportion dialog

E8.4: Computing the Sanrple Size Needed for Estimating the Mean 323

box (shown below), enter values for the Sample Size, theNumber of Successes, and the Confidence Level. Enter atitle as the Title and click OK.

Using the CIE_P worksheet

Open to the CIE_P worksheet of the Q@EEEworkbook. The worksheet (see Figure 8. I 0 on page 297)uses the NORMSINV(P<,\') function to determine theZ value and uses the square root function to compute thestandard error ofthe proportion for the Section 8.3 SaxonHome Improvement Company exarnple. To adapt th isworksheet to other problerns. enter the appropriate samplesize, number of successes, and conf idence level va luesin the t in ted cel ls 84, 85, and 86 and enter a new t i t le inc e l l A l .

E8.4 COMPUTING THE SAMPLE S/,ZENEEDED FOR ESTIMATING THEMEAN

You compute the sample size needed for estimating the meaneither by using the PHStat2 Determination for the Meanprocedure or by making entries in the EEEEEEEIEEEEEworkbook.

Using PHStat2 Determinationfor the Mean

Select PHStat ) Sample Size ) Determination for theMean. In the Sample Size Determinat ion for the Meandialog box (shown on page 324), enter values for thePopulat ion Standard Deviat ion, the Sampl ing Error ,and the Confidence Level. Enter a tit le as the Title andcl ick OK.

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324 EXCEL coMPANIoN ro chanter 8

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Using the SampleSize*M Worksheet

Open to the SampleSize_M worksheet of the

f,!ffi!f@fl!![[! workbook. The worksheet (see Figure8.1 I on page 301) uses the NORMSINV@<$ function tocompute the Z value and uses the ROUNDUP(value) func-tion to round up the sample size needed to the next higherinteger for the Section 8.4 Saxon Home ImprovementCompany example. To adapt this worksheet to other prob-lems, enter the appropriate population standard deviation,sampling error, and confidence level values in the tintedcells 84" 85. and 86 and enter a new tit le in cell A l .

E8.5 COMPUTING THE SAMPLE SIZENEEDED FOR ESTIMATING THEPROPORTION

You compute the sample size needed for estimating theproportion either by using the PHStat2 Determination forthe Proportion procedure or by making entries in the

@workbook.

Using PHStat2 Determinationfor the Proportion

Select PHStat ) $n6p1e Size ) Determination for theProport ion. In the Sample Size Determinat ion for theProportion dialog box (shown at right), enter values forthe Estimate of True Proportion, the Sampling Error,and the Confidence Level. Enter a tit le as the Title andcl ick OK.

Using the SampleSize_P Worksheet

Open to the SampleSize_P worksheet of the

@file. The worksheet (see Figure8.12 on page 304) uses the NORMSINV and ROUNDUPfunctions, discussed in Section E8.3, for the Saxon HomeImprovement Company example in Section 8.3. To adaptthis worksheet to other problems, enter the appropriate esti-mate of true proportion, sampling error, and confidencelevel values in the tinted cells 84 throush 86 and enter anew tit le in cell A l.

E8.6 COMPUTING THE CONFIDENCEINTERVAL ESTIMATE FOR THEPOPULATION TOTAL

You cornpute the conf idence interval est imate for thepopulation total either by using the PHStat2 Estimate forthe Population Total procedure or by making entries in the

EIEEIE[IE *orkbook.

Using PHStat2 Estimate for thePopulation Total

Select PHStat ) Confidence Intervals ) Estimate forthe Population Total. In the Estimate for the PopulationTotal dialog box (shown on page 325), enter values forthePopulation Size and the Confidence Level.

Click one of the input options and make the requiredentries.

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Data

Estimatc of Trrr ProporHml

5anphrg Error:

Cmf*Jence Lwd:

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If you know the sample size, sample mean, and sampledeviation of your sample, click Sample Statisticsand enter those values. Otherwise, click Sample

Unknown and enter the cell range of your sam-as the Sample Cell Range. Enter a title as the Title andoK.

the CIE_T Worksheetto the CIE-T worksheet of the EIEIEEIEIE workbook.worksheet (see Figure 8.13 on page 307) uses the

l-conjidence level, degrees offreedom) function toine the critical value from the I distribution and the

half-width for the Section 8.5 Saxon HomeCompany population total example. To adapt

worksheet to other problems, enter the appropriateation size, sample mean, sample size, sample stan-deviation. and confidence level values in the tinted84 through 88 and enter a new title in cell A I .

7 COMPUTING THE CONFIDENCEINTERVAL ESTIMATE FOR THETOTAL DIFFERENCE

comoute the confidence interval estimate for the totalnce either by using the PHStat2 Estimate for theDifference orocedure or bv makins entries in the

E8.7: Computing the Confidence Interval Estimate for theTotal Difference 325

Usinq PHStat2 Estimatefor tEe Total Difference

Select PHStat ) Confidence Intervals ) Estimate forthe Total Difference. In the Estimate for the TotalDifference dialog box (shown below), enter values for theSample Size, the Population Size, and the ConfidenceLevel. Enter the cell range of the differences as theDifferences Cell Range. If the first cell in the column ofdifferences contains a label, click First cell contains label,enter a title as the Title. and click OK.

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Using the CIE_TD WorksheetOpen to the CIE_TD worksheet of thegtr@file. This worksheet (see Figure 8.14on page 310) uses the TINV(I-conJidence level, degreesof freedom) function to determine the critical value fromthe I distribution and the interval half-width for the Section8.5 Saxon Home Improvement Company total differenceexample. The worksheet also contains a calculation area in

cel l range D9:E16, as shown in F igure E8.1, that countsand sums the differences l isted on a DifferencesDataworksheet. Figure EB.2 illustrates the first 6 of the 13 rowsin the Difference Data worksheet).

-COUtlT(DlFerencoellalalAAl-85 - El1-SUt(0lfieronc.doata!B:B)-Er2 ' (41012-El3 + El f-El5/815

FIGURE E8.1

Calculations area in the Difference Data worksheet

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or standard deviation of difierencesof 0i6erences Nol = 0 12ofDifersnces=B 88

for Diferences Nol = 0 678forDif isrences=O 71.n

ofSouares 719,6664of Differences 7 .

workbook.

326 ExcEL coMPANIoN to chapter 8

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FIGURE E8.2

DifferencesData worksheet (partia l)

To adapt this worksheet to other problems, you need tochange both the CIE_TD and DifferencesData worksheet.In the CIE_TD worksheet, enter the appropriate populationsize, sample size, and confidence level values in the tintedcells 84 through 86 and enter a new title in cell Al. In theDiffererencesData worksheet, enter the differences in col-umn A. Then adjust column B by either copying down the

formula in cell Bl3 to all rows with difference data. ifhave more than 12 differences; or by deletingcolumn B formulas, if you have fewer than 12 di

E8.8 COMPUTING FINITEPOPULATION CORRECTIONFACTORS

The workbooks for confidence interval estimations ofmean and proportion and for computing the sampleneeded for estimating the mean or proportion include aworksheet that calculates the confidence interval estimatesample size, using a finite population conection factorSection 8.7 on the Student CD-ROM). (Open to thosesheets for further information.) If you use PHStat2, youinclude these computations by clicking the FiniteCorrection output option and entering the Populationbefore clicking OK in the appropriate dialog boxes.

-{A2 - gE_TDltB$10}^2-(/[3 . gE_riDltBtl0fz-0[4 - CIE_Tl]lf B$10]^ Z-!45 - qE_T,-DltBtl0l^2-(A5 - CtE-T:DttB$10r 2

chap 8

Documents

computing confidence

computing sample

proportion e8

population total e8

mean o unknown e8

interval estimatefor

total difference e8

estimating population