StudyMaterial_FormulaeEtc

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Figure 1: Different Statistical Tests for different scenarios

This is a quick compilation of relevant formulae for the end trimester syl-labus. For each section, the shaded items, e.g. P. 406 Ex. 8.2 , refer toexercises from Statistics for Management by Levin and Rubin, that youshould be familiar with.

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1 Mean, Standard Deviation

Combined Arithmetic Mean of k sets of data, with the ith set having nielements

x =

ki=1

nixi

ki=1

ni

For data x1, x2, . . . , xn, the variance is defined as

2 =1

n

n

i=1

(xi

x)2

It is also written as

2 =1

n

ni=1

xi2 x2

In real life situations, whenever data is obtained as a sample, the varianceof the sample is

s2 =

ni=1

(xi x2)

n 1 =

ni=1

x2i nx2

n 1

Population Standard Deviation = 2Sample Standard Deviation =

s2 = s

2 Hypothesis Testing - One Sample Tests

x =

n

z =x

x

P. 406 Ex. 8.2 .

Q 8.2, 8-12 A grocery store has specially packaged oranges and has claimed a bagof oranges will yield 2.5 quarts of juice. After randomly selecting 42bags, a stocker found the average juice production per bag to be 2.2

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quarts. Historically, we know the population standard deviation is

0.2 quart. Using this sample and a decision criterion of 2.5 standarderrors, could we conclude the stores claims are correct?

A 8.2, 8-12 H0 : = 2.5H1 : < 2.5n = 42, x = 2.2, = 0.2xn

= 2.22.50.242

= -9.72

zobs < -2.5 (2.5 standard errors left of H0)So reject H0. Stores claims are not correct.

.

Confidence Interval: H0 zcritxOne tailed test of means P. 422 Ex. 8.4 .

Q 8.4, 8-28 Generally Electric has developed a new bulb whose design specifica-tions call for a light output of 960 lumens compared to an earlier modelthat produced only 750 lumens. The companys data indicate that thestandard deviation of light output for this type of bulb is 18.4 lumens.From a sample of 20 new bulbs, the testing committee found an aver-age light output of 954 lumens per bulb. At a 0.05 significance level,

can Generally Electric conclude that its new bulb is producing thespecified 960 lumen output?

A 8.4, 8-28 H0 : = 960H1 : = 960n = 20, x = 954, = 18.4, = 0.05xn

= 95496018.420

= -1.4583

zcrit = 1.96Since -1.96 < zobs = 1.4583 < +1.96, we do not reject H0Bulb is producing the required 960 lumen.

.

Power P. 426 Ex. 8.5

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2.1 Hypothesis Test for Proportion

p =

pH0qH0

n

z =ppH0

p

P. 431 Ex. 8.6 .

Q 8.6, 8-40 Feronetics specializes in the use of gene-splicing techniques to producenew pharmaceutical compounds. It has recently developed a nasal

spray containing interferon, which it believes will limit the transmis-sion of the common cold within families. In the general population,15.1 percent of all individuals will catch a rhinovirus-caused cold onceanother family member contracts such a cold. The interferon spraywas tested on 180 people, one of whose family members subsequentlycontracted a rhinovirus-caused cold. Only 17 of the test subjects de-veloped similar colds.

(a) At a significance level of 0.05, should Feronetics conclude thatthe new spray effectively reduces transmission of colds?

(b) What should it consider at = 0.02?

A 8.6, 8-40 H0 : pH0 = 0.151H1 : pH0 < 0.151n = 180, p = 17180 = 0.0944, = 0.05

p =

pH0qH0n

=

0.1510.849180 = 0.0266

p pH0p

= 0.09440.1510.0266 = -2.1209

(a) zcrit at 0.05 = 1.645Since zobs = 2.1209 < -1.645, we reject H0.The new spray effectively reduces transmission of cold.

(b) zcrit at 0.02 = 2.055Since zobs =

2.1209 < -2.055, we still reject H0.

The new spray effectively reduces transmission of cold.

.

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2.2 Hypothesis Test for Means when is not known

x = n = st =

xH0x

P. 436 Ex. 8.7 .

Q 8.7, 8-48 The data-processing department at a large life insurance company hasinstalled new color video display terminals to replace the monochromeunits it previously used. The 95 operators trained to use the newmachines averaged 7.2 hours before achieving a satifactory level of

performance. Their sample variance was 16.2 squared hours. Long ex-perience with operators on the old monochrome terminals showed thatthey averaged 8.1 hours on the machines before their performanceswere satisfactory. At the 0.01 significance level, should the supervisorof the department conclude that the new terminals are easier to learnto operate?

A 8.7, 8-48 H0 : = 8.1H1 : < 8.1n = 95, x = 7.2, =

16.2, = 0.01

zobs =7.28.1

16.295

= -2.179

zcrit at 0.01 = 2.33Since -2.33 < zobs = 2.179 < +2.055, we do not reject H0.The new terminals are not significantly easier to operate.

.

3 Hypothesis Testing - Two Sample Tests

3.1 Tests for difference between Means

x1x2 =

21n1

+22n2

= s

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x1x2 =21

n1 +

22

n2

z =(x1 x2) (1 2)H0

x1x2

Large Sample Sizes P. 459 Ex. 9.2 .

Q 9.2, 9-4 A sample of 32 money-market mutual funds was chosen on January 1,1996, and the average annual rate of return over the past 30 days wasfound to be 3.23 percent, and the sample standard deviation was 0.51

percent. A year earlier, a sample of 38 money-market funds showed anaverage rate of return of 4.36 percent. Is it reasonable to conclude (at = 0.05) that the money-market interest rates declined during 1995?

A 9.2, 9-4 Jan 1, 1996 n2 = 32, x2 = 3.23%, s2 = 0.51Jan 1, 1995 n1 = 38, x1 = 4.36%, s1 = 0.84 = 0.05, H0 : 2 1 = 0, H1 : 2 1 < 0

zobs =x2 x121

n1+

22

n2

=3.23 4.360.842

38 +0.512

32

= 6.9155

zcrit (at = 0.05) is 1.645Since zobs = -6.9155 < zobs = -1.645, we reject H0Conclusion: Rates have declined.

.

Small Sample SizesPooled estimate of 2

s2p =(n1 1)s21 + (n2 1)s22

n1 + n2 2

x1x2 = sp

1n1+ 1n2

t =(x1 x2) (1 2)H0

x1x2

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t =x H0

xP. 473 Ex. 9.4 .

Q 9.4, 9-14 The data below are a random sample of 9 firms chosen from the Digestof Earnings Reports in The Wall Street Journal on February 6, 1992:

(a) Find the mean change in earnings per share between 1991 and1992

(b) Find the standard deviation of the change and the standard errorof the mean.

(c) Were the average earnings per share different in 1991 and 1992?Test at = 0.02.

Firm 1 2 3 4 5 6 7 8 9

1991 earnings 1.38 1.26 3.64 3.50 2.47 3.21 1.05 1.98 2.721992 earnings 2.48 1.50 4.59 3.06 2.11 2.80 1.59 0.92 0.47

A 9.4, 9-14 = 0.02, H0 : diff = 0, H1 : diff = 0The differences in earnings are as follows:

Firm 1 2 3 4 5 6 7 8 9

Difference in earnings 1.10 0.24 0.95 -0.44 -0.36 -0.41 0.54 -1.06 -2.25

x = 1.69, x2 = 9.1391, x = 0.18777, n = 9s =

x2 9x29 1 =

1.1027 = 1.05

(a) Mean change, x = -0.1878

(b) s = 1.05, sx =1.05

9= 0.35

(c) tobs =xH0diff

n

= 0.187770.35 = 0.536

tcrit(8df, = 0.02) = 2.896Since tobs = 0.536 < tcrit = 2.896, we do not reject H0Conclusion: the average earnings did not change significantly.

.

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3.3 Tests for difference between proportions - Large Samples

p =

pq

n

p1p2 =

p1q1

n1+

p2q2

n2

p1p2 =

p1q1

n1+

p2q2n2

p1, q1, p2 and q2 are the sample proportions

p =n1p1 + n2p2

n1 + n2

p1p2 =

pq

1

n1+

1

n2

z =(p1 p2) (p1 p2)H0

p1p2

P. 482 Ex. 9.5 .

Q 9.5, 9-22 A coal-fired power plant is considering two different systems for pol-lution abatement. The first system has reduced the emissionof pollu-

tants to acceptable levels 68 percent of the time, as determined from200 air samples. The second, more expensive system has reduced theemission of pollutants to acceptable levels 76 percent of the time, as de-termined from 250 air samples. If the expensive system is significantlymore effective than the inexpensive system in reducing pollutants toacceptable levels, then the management of the power plant will installthe expensive system. Which system will be installed if managementuses a significance level of 0.02 in making its decision?

A 9.5, 9-22 p1 = 0.68, n1 = 200p2 = 0.76, n2 = 250 = 0.02, H0 : 2 1 = 0, H1 : 2 1 > 0

p2p1 =

p1q1

n1+

p2q2

n2= 0.0426

zobs =(p2 p1) (2 1)H0

p2p1= 1.8765

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zcrit (at = 0.02) = 2.055

Since zobs = 1.8765 < zcrit = 2.055, we do not reject H0Conclusion: the 2nd system is not significantly more effective

.

P. 488 Ex. 9.6

4 2 Tests

Chi-Square Tests

2 =(fobsfexp)2

fexp

where fobs = observed cell frequency, fexp = expected cell frequency2 degrees of freedom = (rows - 1)(columns - 1)Expected frequency of a cell, fexp =

RowTotalColumnTotalGrandTotal

2 Test of Independence : P. 581 Ex. 11.2 .

Q 11.2, 11-7, 11-8 An advertising firm is trying to determine the demographics for a newproduct. They have randomly selected 75 people in each of 5 differentage groups nand introduced the product to them. The results of thesurvey are given below:

Age Group

Future Activity 18-29 30-39 40-49 50-59 60-69Purchase Frequently 12 18 17 22 32

Seldom Purchase 18 25 29 24 30Never Purchase 45 32 29 29 13

Develop a table of observed and expected frequencies for this prob-lem.

11-8 (a) Calculate the sample 2 value

11-8 (b) State the null and alternate hypotheses

11-8 (c) If the level of significance is 0.01, should the null hypothesis be

rejected?A 11.2, 11-7, 11-8 The table of observed and expected frequencies is

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Age Group Row

Future Activity 18-29 30-39 40-49 50-59 60-69 MarginalPurchase Frequently O 12 18 17 22 32 101

E (RTCTTot.

) 20.2 20.2 20.2 20.2 20.2

(O E)2 67.24 4.84 10.24 3.24 139.24(OE)2

E3.33 0.24 0.51 0.16 6.89

Seldom Purchase O 18 25 29 24 30 126

E (RTCTTot.

) 25.2 25.2 25.2 25.2 25.2

(O E)2 51.84 0.04 14.44 1.44 23.04(OE)2

E2.06 0.00 0.57 0.06 0.91

Never Purchase O 45 32 29 29 13 148

E (RTCTTot.

) 29.6 29.6 29.6 29.6 29.6

(O E)2 237.16 5.76 0.36 0.36 275.56(OE)2

E8.01 0.19 0.01 0.01 9.31

Column Marginals O 75 75 75 75 75 375

2crit,0.01,42=8df = 20.09; 2obs =

(OE)2E

= 32.36 null hypothesis is rejected; the purchasing habits is dependent onage group

.

2 Test of Goodness of fit: P. 587 Ex. 11.3 .

Q 11.3, 11-17 The computer coordinator of a business school believes the amount oftime a graduate student spends reading and writing e-mail each week-day is normally distributed with mean = 14 and standard deviation = 5. In order to examine this belief, the coordinatorcollected dataone Wednesday, recording the amount of time in minutes each gradu-ate student spent checking e-mail. Using a chi-square goodness-of-fittest on these data, what would you conclude about the distributionof e-mail times? (Use a 0.05 significance level and clearly state your

hypothesis.)(Hint: Use five equally probable intervals.)8.2 7.4 9.6 12.8 22.4 6.2 8.7 9.7 12.4 10.61.2 18.6 3.3 15.7 18.4 12.4 15.9 19.4 12.8 20.4

12.3 11.3 10.9 18.4 14.3 16.2 6.7 13.9 18.3 19.214.3 14.9 16.7 11.3 18.4 18.8 20.4 12.4 18.1 20.1

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A 11.3, 11-17 Following is the table of observed and expected frequencies

Observed z (upper X N(14,5) ExpectedClass Frequency class lim.) P(X


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Brand A B C D E

4.4 5.8 4.8 2.9 4.64.6 5.2 5.9 2.7 4.34.5 4.9 4.9 2.9 3.84.1 4.7 4.6 3.9 5.23.8 4.6 4.3 4.3 4.4

(a) Compute the mean number of hours of relief for each brand anddetermine the grand mean.

(b) Estimate the population variance using the between-column vari-ance.

(c) Estimate the population variance using the within-column vari-

ance computed from the variance within the samples.(d) Calculate the F ratio. At the 0.05 level of significance, do the

brands produce significantly different amounts of relief to peoplewith strong stomach acid?

A 11.4, 11-26 Computation table:Brand A B C D E

4.4 5.8 4.8 2.9 4.64.6 5.2 5.9 2.7 4.34.5 4.9 4.9 2.9 3.84.1 4.7 4.6 3.9 5.23.8 4.6 4.3 4.3 4.4

X 21.4 25.2 24.5 16.7 22.3X 4.28 5.04 4.90 3.34 4.46

Grand Mean X = 4.404

(A)

(Xi X)2 0.428 0.932 1.460 2.032 1.032SSB X X2 0.015376 0.404496 0.246016 1.132096 0.003136

(B) = n SS B = 5.SSB 0.07688 2.02248 1.23008 5.66048 0.015682b =

(B) / 4 = 2.2514

2w =

(A) / (25 - 5) = 0.2942Fobs (df 4, 20) =

2.25140.2942 = 7.6526

Fcrit (df 4, 20; 0.05 significance) = 2.8661;

Since Fobs > Fcrit reject the null hypothesis of equality of meansConclusion: the brands produce significantly different amounts of re-lief to people with strong stomach acid.

.

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6 Correlation and Regression

6.1 Estimation using Regression line

b =

XY nXY

X2 nXa = Y bX

Standard Error of the Regression Estimate

se =Y2 aY bXYn 2

P. 671 Ex. 12.2 .

Q 12.2, 12-18 A study by the Atlanta, Georgia, Department of Transportation onthe effect of bus-ticket prices on the number of passengers producedthe following results:

Ticket price (cents) 25 30 35 40 45 50 55 60Passengers per 100 miles 800 780 780 660 640 600 620 620

(a) Plot these data.(b) Develop the estimating equation that best describes these data.

(c) Predict the number of passengers per 100 miles if the ticket priceswere 50 cent. Use a 95 percent approximate prediction interval.

A 12.2, 12-18 From the data provided, we getX = 340;

Y = 5500

X2 = 15500;

Y2 = 3830800X = 42.5; Y = 687.5;

XY = 227200

b = XY nXYX

2

nX2

=65501050

=

6.238

a = Y bX = 687.5 + 6.238 42.5 = 952.615 Y = 952.615 - 6.238 X PASSENGERS= 952.615 6.238.TICKETPRICE

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PASSENGERSTicketPrice=50 = 640.715

se =

Y2 aY bXY

n 2 = sqrt8691.1

6= 38.06

95% prediction interval for number of passengers at a ticket price of50 cents is

PASSENGERSTicketPrice=50 se tcrit,0.05= 640.715 38.06 2.447 = 640.715 93.132

.

6.2 Correlation

Coefficient of Determination = r2

r2 =

XY nXY

(

X2 nX2)(Y2 nY2)Correlation Coefficient r =

r2

P. 685 Ex. 12.3 .

Q 12.3, 12-32 Zippy Cola is studying the effect of its latest advertising campaign.People chosen at random were called and asked how many cans ofZippy Cola they had bought in the past week and how many ZippyCola advertisements they had either read or seen in the past week.

X (number of ads) 3 7 4 2 0 4 1 2Y (cans purchased) 11 18 9 4 7 6 3 8

(a) Develop the estimating equation that best fits the data.

(b) Calculate the sample coefficient of determination and the samplecoefficient of correlation.

A 12.3, 12-32 From the data provided, the following summary statistics emerge:

X = 23; X2 = 99; X = 2.875Y = 66;

Y2 = 700; Y = 8.25

XY = 246

b =

XY nXYX2 nX2 =

56.25

32.875= 1.711

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a = Y = bX = 3.33

PURCHASES = 3.33 + 1.71 ADS

r2 =

XY nXY

(

X2 nX2)(Y2 nY2) =96.186

155.5

Coefficient of Determination, r2 = 0.6186Coefficient of Correlation, r =

0.6186 = 0.7865

r has a positive sign because b, the slope of the regression equation, is+ve

.

6.3 Inferences about the slope of the regression line

Standard Error of the Regression Slope, b

sb =se

X2 nX2

t =bBH0

sb

Confidence Interval of the regression coefficient = b tsbP. 691 Ex. 12.4 .

Q 12.4, 12-38 In 1969, a government health agency found that in a number of coun-ties, the relationship between smokers and heart-disease fatalities per100,000 population had a slope of 0.08. A recent study of 18 countiesproduced a slope of 0.147 and a standard error of the regression slopecoefficient of 0.032.

(a) Construct a 90 percent confidence interval estimate of the slope ofthe true regression line. Does the result from this study indicate

that the true slope has changed?(b) Construct a 99 percent confidence interval estimate of the slope of

the true regression line. Does the result from this study indicatethat the true slope has changed?

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A 12.4, 12-38 b = 0.08; b = 0.147; sb = 0.032; n = 18

t16df, 0.1 = 1.746; t16df, 0.01 = 2.921

(a) 90% confidence interval for b is b sbtcrit= 0.147 0.032 1.746 = 0.147 0.05587 = (0.091, 0.203)Since, 0.08, the earlier population slope, lies outside this range, weconclude that the slope has changed at a 90% level of significance

(b) 99% confidence interval for b is b sbtcrit= 0.147 0.032 2.921 = 0.147 0.09347 = (0.054, 0.240)Since, 0.08, the earlier population slope, lies inside this range,we conclude that the slope has not changed at a 99% level ofsignificance

.

7 Non Parametric Tests

7.1 Sign Test (Paired Data)

p =

pH0qH0

n

z =

p

pH0

p

7.2 Mann-Whitney U Test

U = n1n2 +n1(n1 + 1)

2R1

U =n1n2

2

U =

n1n2(n1 + n2 + 1)

12

z=

U

U

U

P. 809 Ex. 14.3, Problems 14-14, 14-17, 14-18, 14-20 .

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Q 14.3, 14-14 Test the hypothesis of no difference between the ages of male and fe-

male employees of a certain company using the Mann-Whitney U testfor the sample data. Use the 0.10 level of significance.

Males 31 25 38 33 42 40 44 26 43 35Females 44 30 34 47 35 32 35 47 48 34

A 14.3, 14-14 H0 : m = f; H1 : m = f; = 0.1The following rankings emerge from the given data

Age Gender Rank

25 M 126 M 230 F 3

31 M 432 F 533 M 634 F 7.534 F 7.535 M 1035 F 1035 F 1038 M 1240 M 1342 M 14

43 M 1544 M 16.544 F 16.547 F 18.547 F 18.548 F 20

nm = 10; Rm = 93.5nf = 10; Rf = 116.5

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U = nmnf + nm(nm + 1)

2Rm

= (10)(10) +(10)(11)

2 93.5

= 100 + 55 93.5= 61.5

Mean of the sampling distribution of U, U =nmnf

2 =(10)(10)

2 = 50

Standard error of the U statistic, U = nmnf(nm+nf+1)

12

U =

(10)(10)(10 + 10 + 1)

12

=

2100

12

=

175

= 13.23

zobs =UUU

= 61.55013.23 = 0.8692Since zL = 1.645 < zobs < zU, we conclude that there is no difference

between the ages. .

7.3 Kruskal-Wallis Test

K =12

n(n + 1)

kj=1

R2j

nj 3(n + 1)

where k = number of groupsK 2df(n 1)P. 809 Ex. 14.3, Problems 14-15, 14-16, 14-19 .

Q 14.3, 14-16 A mail-order gift company has the following sample data on dollarsales, separated according to how the order was paid. Test the hy-pothesis that there is no difference in the dollar amount of orders paid

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for by cash, by check, or by credit card. Use the Kruskal-Wallis test

with a 0.05 level of significance.Credit-card orders 78 64 75 45 82 69 60

Check orders 110 70 53 51 61 68Cash orders 90 68 70 54 74 65 59

A 14.3, 14-16 H0 : cc = ck = cs; H1: the means are not all equal; =0.05The following rankings emerge from the given data

Dollar amount Mode ofof Orders paid Payment Rank

45 CC 151 CK 2

53 CK 354 CS 459 CS 560 CC 661 CK 764 CC 865 CS 968 CK 10.568 CS 10.569 CC 1270 CS 13.5

70 CK 13.574 CS 1575 CC 1678 CC 1782 CC 1890 CS 19

110 CK 20

k = 3nCC = 7; RCC = 78nCK = 6; RCK = 56nCS = 7; RCS = 76

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K =12

n(n + 1)

kj=1

R2j

nj 3(n + 1)

=12

(20)(21)(

6084

7+

3136

6+

5776

7) 3(20 + 1)

= 0.02857(869.14 + 522.67 + 825.14) 63= 0.02857 2216.95 63= 63.34 63= 0.34

2

(2 df, significance 0.05) = 5.991Since K = 0.34 < 2crit = 5.991, we do not reject H0Conclusion: The average amounts paid by the three methods are not signif-icantly different. .

7.4 One Sample Run Test

r =2n1n2

n1 + n2+ 1

where n1 = sample size of one category

n2 = sample size of the other categoryr = number of runs

r =

2n1n2(2n1n2 n1 n2)(n1 + n2)

2(n1 + n2 1)

z =r r

r

P. 817 Ex. 14.4.

Q 14.4, 14-26 The NewsandClarion kept a record of the gender of people who calledthe circulation office to complain about delivery problems with theSunday paper. For a recent Sunday, these data were as follows:

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M,F,F,F,M,M,F,M,F,F,F,F,M,M,M,F,M,F,M,F,F,F,F,M,M,M,M,M

Using the 0.05 level of significance, test this sequence for random-ness. Is there anything about the nature of this problem that wouldcause you to believe that such a sequence would not be random?

A 14.4, 14-26 From the run sequence of geneders provided in the question:n1 = 14, n2 = 14, r = 13, = 0.05

r =2n1n2

n1 + n2+ 1

= 2 14 1414 + 14 + 1= 15

r =

2n1n2(2n1n2 n1 n2)(n1 + n2)

2(n1 + n2 1)

=

2.14.14.(2.14.14 28)

282.27

= sqrt14.13

27

= sqrt6.74= 2.6

zobs =r r

r

=13 15

2.6= 0.77

zcrit0.05 = 1.96Since zobs

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