Chapter 9

Statistical Inference:
Hypothesis Testing for Single Populations

LEARNING OBJECTIVES

The main objective of Chapter 9 is to help you to learn how to test hypotheses on single populations,
thereby enabling you to:

1. Understand the logic of hypothesis testing and know how to establish null and alternate
hypotheses.
2. Understand Type I and Type II errors and know how to solve for Type II errors.
3. Know how to implement the HTAB system to test hypotheses.
4. Test hypotheses about a single population mean when  is known.
5. Test hypotheses about a single population mean when  is unknown.
6. Test hypotheses about a single population proportion.
7. Test hypotheses about a single population variance.

CHAPTER OUTLINE

9.1 Introduction to Hypothesis Testing

Types of Hypotheses
Research Hypotheses
Statistical Hypotheses
Substantive Hypotheses
Using the HTAB System to Test Hypotheses
Rejection and Non-rejection Regions
Type I and Type II errors

9.2 Testing Hypotheses About a Population Mean Using the z Statistic ( known)
Using a Sample Standard Deviation
Testing the Mean with a Finite Population
Using the p-Value Method to Test Hypotheses
Using the Critical Value Method to Test Hypotheses
Using the Computer to Test Hypotheses about a Population Mean Using the z Statistic

9.3 Testing Hypotheses About a Population Mean Using the t Statistic ( unknown)
Using the Computer to Test Hypotheses about a Population Mean Using the t Test

9.4 Testing Hypotheses About a Proportion

Using the Computer to Test Hypotheses about a Population Proportion

9.5 Testing Hypotheses About a Variance

9.6 Solving for Type II Errors

Some Observations About Type II Errors
Operating Characteristic and Power Curves
Effect of Increasing Sample Size on the Rejection Limits

151
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KEY WORDS

alpha ( ) one-tailed test

alternative hypothesis operating characteristic curve (OC)
beta ( ) p-value method
critical value power
critical value method power curve
hypothesis rejection region
hypothesis testing research hypothesis
level of significance statistical hypothesis
nonrejection region substantive result
null hypothesis two-tailed test
observed significance level Type I error
observed value Type II error

STUDY QUESTIONS

1. The first step in testing a hypothesis is to establish a(n) _______________ hypothesis and a(n)
_______________ hypothesis.

2. In testing hypotheses, the researcher initially assumes that the _______________ hypothesis is
true.

3. The region of the distribution in hypothesis testing in which the null hypothesis is rejected is
called the _______________ region.

4. The rejection and acceptance regions are divided by a point called the _______________ value.

5. The portion of the distribution which is not in the rejection region is called the _______________
region.

6. The probability of committing a Type I error is called _______________.

7. Another name for alpha is _______________ _______________ _______________.

8. When a true null hypothesis is rejected, the researcher has committed a _______________ error.

9. When a researcher fails to reject a false null hypothesis, a _______________ error has been
committed.

10. The probability of committing a Type II error is represented by _______________.

11. Power is equal to _______________.

12. Whenever hypotheses are established such that the alternative hypothesis is directional, then the
researcher is conducting a _______________-tailed test.

13. A _______________-tailed test is nondirectional.

14. If in testing hypotheses, the researcher uses a method in which the probability of the observed
statistic is compared to alpha to reach a decision, the researcher is using the _______________.
 Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 153

15. Suppose Ho: µ = 95 and Ha: µ  95. If the sample size is 50 and  = .05, the critical value of z is
________.

16. Suppose Ho: µ = 2.36 and Ha: µ < 2.36. If the sample size is 64 and  = .01, the critical value of z
is ________.

17. Suppose Ho: µ = 24.8 and Ha: µ  24.8. If the sample size is 49 and  = .10, the critical value of z
is ________.

18. Suppose a researcher is testing a null hypothesis that µ = 61. A random sample of n = 38 is taken
resulting in x = 63 and  = 8.76. The observed z value is _______________.

19. Suppose a researcher is testing a null hypothesis that µ = 413. A random sample of n = 70 is
taken resulting in x = 405. The population standard deviation is 34. The observed z value is
_______________.

20. A researcher is testing a hypothesis of a single mean. The critical z value for  = .05 and a one-
tailed test is 1.645. The observed z value from sample data is 1.13. The decision made by the
researcher based on this information is to _______________ the null hypothesis.

21. A researcher is testing a hypothesis of a single mean. The critical z value for  = .05 and a two-
tailed test is ± 1.96. The observed z value from sample data is –1.85. The decision made by the
researcher based on this information is to _______________ the null hypothesis.

22. A researcher is testing a hypothesis of a single mean. The critical z value for  = .01 and a one-
tailed test is –2.33. The observed z value from sample data is –2.45. The decision made by the
researcher based on this information is to _______________ the null hypothesis.

23. A researcher has a theory that the average age of managers in a particular industry is over 35-
years-old. The null hypothesis to conduct a statistical test on this theory would be
_______________.

24. A company produces, among other things, a metal plate that is supposed to have a six inch hole
punched in the center. A quality control inspector is concerned that the machine which punches
the hole is "out-of-control". In an effort to test this, the inspector is going to gather a sample of
metal plates punched by the machine and measure the diameter of the hole. The alternative
hypothesis used to statistical test to determine if the machine is out-of-control is
_______________.

25. The following hypotheses are being tested:

Ho: µ = 4.6
Ha: µ  4.6

The value of alpha is .05. To test these hypotheses, a random sample of 22 items is selected
resulting in a sample mean of 4.1 with a sample standard deviation of 1.8. It can be assumed that
this measurement is normally distributed in the population. The degrees of freedom associated with
the t test used in this problem are _______________.

26. The critical t value for the problem presented in question 25 is _______________.

27. The problem presented in question 25 contains hypotheses which lead to a __________-tailed test.
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28. The observed value of t for the problem presented in question 25 is _______________.

29. Based on the results of the observed t value and the critical table t value, the researcher should
_______________ the null hypothesis in the problem presented in question 25.

30. It is believed that the average time to assemble a given product is less than 2 hours. To test this, a
sample of 18 assemblies is taken resulting in a sample mean of 1.91 hours with a sample standard
deviation of 0.73 hours. Suppose  = .01. If a hypothesis test is done on this problem, the table
value is ______________. The observed value is _________________. The decision is
________________.

31. A political scientist wants to statistically test the null hypothesis that her candidate for governor is
currently carrying at least 57% of the vote in the state. She has her assistants randomly sample
550 eligible voters in the state by telephone and only 300 declare that they support her candidate.
The observed z value for this problem is _______________.

32. Problem 31 is a _______________-tailed test.

33. Suppose that the value of alpha for problem 31 is .05. After comparing the observed value to the
critical value, the political scientist decided to _______________ the null hypothesis.

34. A company believes that it controls .27 of the total market share in the South for one of its products.
To test this belief, a random sample of 1150 purchases of this product in the South are contracted.
385 of the 1150 purchased this company's brand of the product. If a researcher wants to conduct
a statistical test for this problem, the alternative hypothesis would be _______________.

35. The observed value of z for problem 34 is _______________.

36. Problem 34 would result in a _______________-tailed test.

37. Suppose that a .01 value of alpha were used in problem 34. The critical value of z for the problem is
_______________.

38. Upon comparing the observed value of z to the critical value of z, it is determined to
_______________ the null hypothesis in problem 34.

39. A production process produces parts with a normal variance of 27.3. Engineers are concerned that
the process may now be producing parts with greater variance than that. To test this concern, a
sample of 9 newly produced parts is taken. The sample standard deviation is 5.93. Let  = .01. The
null hypothesis for this problem is _______________________.

40. The critical table value of 2 for problem 39 is ______________________.

41. The observed value of chi-square in problem 39 is ______________________.

42. The decision reached for problem 39 is _______________.

43. The null hypothesis for a test is H0:  = 30. A one-tailed test is being conducted. After taking a
sample of 49 items and computing a mean and standard deviation, it is decided to fail to reject the
null hypothesis. Let  = .05. Suppose the population standard deviation is .63. If the null
hypothesis is not true and if the true alternative hypothesis is 29.6, the value of beta is
_______________________.
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44. Suppose the alternative mean in problem 43 is really 29.9, the value of beta is
_______________________.

45. Plotting the power values against the various values of the alternative hypotheses produces a
______________________ curve.

46. Plotting the values of against various values of the alternative hypothesis produces a
_____________________ curve.
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ANSWERS TO STUDY QUESTIONS

1. Null, Alternative 24.   6”

2. Null 25. 21
3. Rejection 26. + 2.08
4. Critical 27. Two
5. Nonrejection Region 28. – 1.30

6. Alpha 29. Fail to Reject

7. Level of Significance 30. – 2.567, – 0.52, Fail to Reject

8. Type I 31. – 1.16

9. Type II 32. One

10. Beta 33. Fail to Reject

11. 1 –  34. p  .27

12. One 35. 4.95

13. Two 36. Two

14. p-value 37. + 2.575

15. 1.96 38. Reject

16. – 2.33 39. H0: 2 = 27.3

17. + 1.645 40. 20.0902

18. 1.41 41. 10.3047

19. – 1.97 42. Fail to Reject the Null Hypothesis

20. Fail to Reject 43. .0026

21. Fail to Reject 44. .7019

22. Reject 45. Power

23.  = 35 46. Operating Characteristic

 Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 157

SOLUTIONS TO ODD-NUMBERED PROBLEMS IN CHAPTER 9

9.1 a) Ho: µ = 25
Ha: µ  25

x = 28.1 n = 57  = 8.46  = .01

For two-tail, /2 = .005 zc = 2.575

x 28.1  25
z=  = 2.77
 8.46
n 57

observed z = 2.77 > zc = 2.575

Reject the null hypothesis

b) from Table A.5, inside area between z = 0 and z = 2.77 is .4972

p-value = .5000 – .4972 = .0028


Since the p-value of .0028 is less than = .005, the decision is to:
2
Reject the null hypothesis

c) critical mean values:

xc  
zc =
s
n

x c  25
± 2.575 =
8.46
57

x c = 25 ± 2.885

x c = 27.885 (upper value)

x c = 22.115 (lower value)

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9.3 a) Ho: µ = 1,200

Ha: µ > 1,200

x = 1,215 n = 113  = 100  = .10

For one-tail,  = .10 zc = 1.28

x 1,215  1,200

z =  = 1.59
 100
n 113

observed z = 1.59 > zc = 1.28

Reject the null hypothesis

b) Probability > observed z = 1.59 is .0559 which is less than  = .10.

Reject the null hypothesis.

c) Critical mean value:

xc  
zc =

n

x c  1,200
1.28 =
100
113

x c = 1,200 + 12.04

Since calculated x = 1,215 which is greater than the critical x = 1212.04, reject the null
hypothesis.

9.5 H0:  = $424.20

Ha:   $424.20

x = $432.69 n = 54  = $33.90  = .05

2-tailed test, /2 = .025 z.025 = + 1.96

x 432.69  424.20

z =  = 1.84
 33.90
n 54

Since the observed z = 1.85 < z.025 = 1.96, the decision is to fail to reject the null hypothesis.
 Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 159

9.7 H0:  = 5
Ha:   5

x = 5.0611 n = 42  = 0.2803  = .10

2-tailed test, /2 = .05 z.05 = + 1.645

x 5.0611  5
z =  = 1.41
 0.2803
n 42

Since the observed z = 1.41 < z.05 = 1.645, the decision is to fail to reject the null hypothesis.

9.9 Ho: µ = $4,292

Ha: µ < $4,292

x = $4,008 n = 55  = $386  = .01

For one-tailed test,  = .01, z.01 = –2.33

x $4,008  $4,292

z =  = –5.46
 $386
n 55

Since the observed z = –5.46 < z.01 = –2.33, the decision is to Reject the null hypothesis

The CEO could use this information as a way of discrediting the Runzheimer study and using her
own figures in recruiting people and in discussing relocation options. In such a case, this could be
a substantive finding. However, one must ask if the difference between $4,292 and $4,008 is
really an important difference in monthly rental expense. Certainly, Paris is expensive either way.
However, an almost $300 difference in monthly rental cost is a non trivial amount for most
people and therefore might be considered substantive.

9.11 n = 20 x = 16.45 s = 3.59 df = 20 – 1 = 19  = .05

Ho: µ = 16
Ha: µ  16

For two-tail test, /2 = .025, critical t.025,19 = ±2.093

x   16.45  16
t =  = 0.56
s 3.59
n 20

Observed t = 0.56 < t.025,19 = 2.093

The decision is to Fail to reject the null hypothesis

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9.13 n = 11 x = 1,235.36 s = 103.81 df = 11 – 1 = 10  = .05

Ho: µ = 1,160
Ha: µ > 1,160

For one-tail test,  = .05 critical t.05,10 = 1.812

x   1,236.36  1,160
t =  = 2.44
s 103.81
n 11

Observed t = 2.44 > t.05,10 = 1.812

The decision is to Reject the null hypothesis

9.15 n = 12 x = 1.85083 s = .02353 df = 12 – 1 = 11  = .10

H0: µ = 1.84
Ha: µ  1.84

For a two-tailed test, /2 = .05 critical t.05,11 = 1.796

x   1.85083  1.84
t =  = 1.59
s .02353
n 12

Since t = 1.59 < t11,.05 = 1.796,

The decision is to fail to reject the null hypothesis.

9.17 n = 19 x = $31.67 s = $1.29 df = 19 – 1 = 18  = .05

H0:  = $32.28
Ha:   $32.28

Two-tailed test, /2 = .025 t.025,18 = + 2.101

x   31.67  32.28
t =  = –2.06
s 1.29
n 19

The observed t = –2.06 > t.025,18 = –2.101,

The decision is to fail to reject the null hypothesis
 Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 161

9.19 Ho: p = .45

Ha: p > .45

n = 310 p̂ = .465  = .05

For one-tail,  = .05 z.05 = 1.645

pˆ  p .465  .45
z =  = 0.53
pq (.45)(.55)
n 310

observed z = 0.53 < z.05 = 1.645

The decision is to Fail to reject the null hypothesis

9.21 Ho: p = .29

Ha: p  .29

x 207
n = 740 x = 207 pˆ   = .28  = .05
n 740

For two-tail, /2 = .025 z.025 = ±1.96

pˆ  p .28  .29
z =  = –0.60
pq (.29)(.71)
n 740

observed z = –0.60 > zc = –1.96

The decision is to Fail to reject the null hypothesis

p-Value Method:

z = –0.60

from Table A.5, area = .2257

Area in tail = .5000 – .2257 = .2743

.2743 > .025

Again, the decision is to Fail to reject the null hypothesis

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Solving for critical values:

pˆ c  p
z =
pq
n

pˆ c  .29
±1.96 =
(.29)(. 71)
740

p̂c = .29 ± .033

.257 and .323

Sample p = p̂ = .28 not outside critical values in tails

Again, the decision is to Fail to reject the null hypothesis

9.23 Ho: p = .79

Ha: p < .79

n = 415 x = 303  = .01 z.01 = –2.33

x 303
pˆ   = .7301
n 415

pˆ  p .7301  .79
z =  = –3.00
pq (.79)(.21)
n 415

Since the observed z = –3.00 is less than z.01= –2.33, The decision is to reject the null
hypothesis.

9.25 Ho: p = .18

Ha: p > .18

n = 376 p̂ = .22  = .01

one-tailed test, z.01 = 2.33

pˆ  p .22  .18
z =  = 2.02
pq (.18)(.82)
n 376

Since the observed z = 2.02 is less than z.01= 2.33, The decision is to fail to reject the null
hypothesis. There is not enough evidence to declare that the proportion is greater than .18.
 Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 163

9.27 Ho: p = .47

Ha: p  .47

n = 67 x = 40  = .05 /2 = .025

For a two-tailed test, z.025 = +1.96

x 40
pˆ   = .597
n 67

pˆ  p .597  .47
z =  = 2.08
pq (.47)(.53)
n 67

Since the observed z = 2.08 is greater than z.025= 1.96, The decision is to reject the null
hypothesis.

9.29 H0: 2 = 14  = .05 /2 = .025 n = 12 df = 12 – 1 = 11 s2 = 30.0833

Ha: 2  14

2.025,11 = 21.92 2.975,11 = 3.81575

(12  1)(30.0833)
2 = = 23.64
14

Since 2 = 23.64 < 2.025,11 = 21.92, the decision is to reject the null hypothesis.

9.31 H0: 2 = 199,996,164  = .10 /2 = .05 n = 13 df =13 – 1 = 12

Ha: 2  199,996,164 2
s = 832,089,743.7

2.05,12 = 21.0261 2.95,12 = 5.22603

(13  1)(832,089,743.7)
2 = = 49.93
199,996,164

Since 2 = 49.93 > 2.05,12 = 21.0261, the decision is to reject the null
hypothesis. The variance has changed.

9.33 Ho: µ = 100

Ha: µ < 100

n = 48 µ = 99  = 14
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a)  = .10 z.10 = –1.28

xc  
zc =

n

x c  100
–1.28 =
14
48

x c = 97.4

xc   97.4  99
z = = = –0.79
 14
n 48

from Table A.5, area for z = –0.79 is .2852

 = .2852 + .5000 = .7852

b)  = .05 z.05 = –1.645

xc  
zc =

n

x c  100
–1.645 =
14
48

x c = 96.68

xc   96.68  99
z = = = –1.15
 14
n 48

from Table A.5, area for z = –1.15 is .3749

 = .3749 + .5000 = .8749

 Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 165

c)  = .01 z.01 = –2.33

xc  
zc =

n

x c  100
–2.33 =
14
48

x c = 95.29

xc   95.29  99
z = = = –1.84
 14
n 48

from Table A.5, area for z = –1.84 is .4671

 = .4671 + .5000 = .9671

d) As gets smaller (other variables remaining constant), beta gets larger. Decreasing the probability
of committing a Type I error increases the probability of committing a Type II error if other
variables are held constant.

9.35 Ho: µ = 50
Ha: µ  50

µa = 53 n = 35 =7  = .01

Since this is two-tailed, /2 = .005 z.005 = ±2.575

xc  
zc =

n

x c  50
±2.575 =
7
35

x c = 50 ± 3.05

46.95 and 53.05

xc   53.05  53
z = = = 0.04
 7
n 35
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from Table A.5 for z = 0.04, area = .0160

Other end:

xc   46.9  53
z = = = –5.11
 7
n 35

Area associated with z = –5.11 is .5000

 = .5000 + .0160 = .5160

9.37 n = 58 x = 45.1  = 8.7  = .05 /2 = .025

H0: µ = 44
Ha: µ  44 z.025 = ± 1.96

45.1  44
z = = 0.96
8.7
58

Since z = 0.96 < zc = 1.96, the decision is to fail to reject the null hypothesis.

x c  44
+ 1.96 =
8 .7
58

± 2.239 = x c – 44

x c = 46.239 and 41.761

For 45 years:

46.29  45
z = = 1.08
8.7
58

from Table A.5, the area for z = 1.08 is .3599

 = .5000 + .3599 = .8599

Power = 1 –  = 1 – .8599 = .1401

 Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 167

For 46 years:

46.239  46
z = = 0.21
8 .7
58

From Table A.5, the area for z = 0.21 is .0832

 = .5000 + .0832 = .5832

Power = 1 –  = 1 – .5832 = .4168

For 47 years:

46.9  47
z = = –0.67
8 .7
58

From Table A.5, the area for z = –0.67 is .2486

 = .5000 – .2486 = .2514

Power = 1 –  = 1 – .2514 = .7486

For 48 years:

46.248  48
z = = 1.54
8.7
58

From Table A.5, the area for z = 1.54 is .4382

 = .5000 – .4382 = .0618

Power = 1 –  = 1 – .0618 = .9382

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9.39
1) Ho: µ = 36
Ha: µ  36

x
2) z =

n
3)  = .01

4) two-tailed test, /2 = .005, z.005 = + 2.575

If the observed value of z is greater than 2.575 or less than –2.575, the decision will be to reject
the null hypothesis.

5) n = 63, x = 38.4,  = 5.93

x 38.4  36
6) z = = = 3.21
 5.93
n 63

7) Since the observed value of z = 3.21 is greater than z.005 = 2.575, the decision is
to reject the null hypothesis.

8) The mean is likely to be greater than 36.

9.41

a. 1) Ho: p = .28
Ha: p > .28

pˆ  p
2) z =
pq
n
3)  = .10
 Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 169

4) This is a one-tailed test, z.10 = 1.28. If the observed value of z is greater than
1.28, the decision will be to reject the null hypothesis.

5) n = 783 x = 230

230
pˆ  = .2937
783

.2937  .28
6) z = = 0.85
(.28)(. 72)
783

7) Since z = 0.85 is less than z.10 = 1.28, the decision is to fail to reject the null
hypothesis.

8) There is not enough evidence to declare that p is not .28.

b. 1) Ho: p = .61
Ha: p  .61

pˆ  p
2) z =
pq
n

3)  = .05

4) This is a two-tailed test, z.025 = + 1.96. If the observed value of z is greater than
1.96 or less than –1.96, then the decision will be to reject the null hypothesis.

5) n = 401 p̂ = .56

.56  .61
6) z = = –2.05
(.61)(. 39)
401

7) Since z = –2.05 is less than z.025 = –1.96, the decision is to reject the null
hypothesis.

8) The population proportion is not likely to be .61.

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9.43 a) H0: µ = 130

Ha: µ > 130

n = 75  = 12  = .01 z.01 = 2.33 µa = 135

Solving for x c:

xc  
zc =

n

x c  130
2.33 =
12
75

x c = 133.23

133.23  135
z = = –1.28
12
75

from table A.5, area for z = –1.28 is .3997

 = .5000 – .3997 = .1003

b) H0: p = .44
Ha: p < .44

n = 1095  = .05 pa = .42 z.05 = –1.645

pˆ c  p
zc =
pq
n

pˆ c  .44
–1.645 =
(.44)(.56)
1095

p̂c = .4153

.4153  .42
z = = –0.32
(.42)(. 58)
1095

from table A.5, area for z = –0.32 is .1255

 = .5000 + .1255 = .6255
 Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 171

9.45 x = 3.45 n = 64 2 = 1.31  = .05

Ho: µ = 3.3
Ha: µ  3.3

For two-tail, /2 = .025 zc = ±1.96

x 3.45  3.3

z = = = 1.05
 1.31
n 64

Since the observed z = 1.05 < zc = 1.96, the decision is to Fail to reject the null hypothesis.

9.47 H0: 2 = 16 n = 12  = .05 df = 12 – 1 = 11

Ha: 2 > 16

s = 0.4987864 ft. = 5.98544 in.

2.05,11 = 19.6751

(12  1)(5.98544) 2
 =
2
= 24.63
16

Since 2 = 24.63 > 2.05,11 = 19.6751, the decision is to reject the null hypothesis.

9.49 x = $26,650 n = 100  = $12,000

a) Ho: µ = $25,000
Ha: µ > $25,000  = .05

For one-tail,  = .05 z.05 = 1.645

x 26,650  25,000

z = = = 1.38
 12,000
n 100

Since the observed z = 1.38 < z.05 = 1.645, the decision is to fail to reject the null hypothesis.
172 Solutions Manual and Study Guide

b) µa = $30,000 zc = 1.645

Solving for x c:

xc  
zc =

n

( x c  25,000)
1.645 =
12,000
100

x c = 25,000 + 1,974 = 26,974

26,974  30,000
z = = –2.52
12,000
100

from Table A.5, the area for z = –2.52 is .4941

 = .5000 – .4941 = .0059

9.51 H0: p = .46

Ha: p > .46

x 66
n = 125 x = 66  = .05 pˆ   = .528
n 125

Using a one-tailed test, z.05 = 1.645

pˆ  p .528  .46
z =  = 1.53
pq (.46)(.54)
n 125

Since the observed value of z = 1.53 < z.05 = 1.645, the decision is to fail to reject the null
hypothesis.
 Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 173

Solving for p̂c :

pˆ c  p
zc =
pq
n

pˆ c  .46
1.645 =
(.46)(.54)
125

p̂c = .533

pˆ c  p a .533  .50
z =  = 0.74
pa  qa (.50)(. 50)
n 125

from Table A.5, the area for z = 0.74 is .2704

 = .5000 + .2704 = .7704

9.53 H0: p = .16

Ha: p > .16

x 84
n = 428 x = 84  = .01 pˆ   = .1963
n 428

For a one-tailed test, z.01 = 2.33

pˆ  p .1963  .16
z =  = 2.05
pq (.16)(.84)
n 428

Since the observed z = 2.05 < z.01 = 2.33, the decision is to fail to reject the null hypothesis.

The probability of committing a Type I error is .01.

174 Solutions Manual and Study Guide

Solving for p̂c :

pˆ c  p
zc =
pq
n

. pˆ c  .16
2.33 =
(.16)(.84)
428

p̂c = .2013

pˆ c  p a .2013  .21
z =  = –0.44
pa  qa (.21)(. 79)
n 428

from Table A.5, the area for z = –0.44 is .1700

 = .5000 – .1700 = .3300

9.55 H0: 2 = 16 n = 22 df = 22 –1 = 21 s=6  = .05

Ha: 2 > 16

2.05,21 = 32.6705

(22  1)(6) 2
2 = = 47.25
16

Since the observed 2 = 47.25 > 2.05,21 = 32.6705, the decision is to reject the null hypothesis.

9.57 a) Ho:  = 23.58

Ha:   23.58

n = 95 x = 22.83  = 5.11  = .05

Since this is a two-tailed test and using /2 = .025: z.025 = + 1.96

x 22.83  23.58

z = = = –1.43
 5.11
n 95

Since the observed z = –1.43 > z.025 = –1.96, the decision is to fail to reject the null hypothesis.
 Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 175

xc  
b) zc =

n

xc  23.58
+ 1.96 =
5.11
95

xc = 23.58 + 1.03

xc = 22.55, 24.61

for Ha:  = 22.30

xc   a 22.55  22.30
z = = = 0.48
 5.11
n 95

xc   a 24.61  22.30
z = = = 4.41
 5.11
n 95

from Table A.5, the areas for z = 0.48 and z = 4.41 are .1844 and .5000

 = .5000 – .1844 = .3156

The upper tail has no effect on .

9.59 The sample size is 22. x is 3.967 s = 0.866 df = 21

The test statistic is:

x
t =
s
n

The observed t = –2.34. The p-value is .015.

The results are statistical significant at  = .05.

The decision is to reject the null hypothesis.

176 Solutions Manual and Study Guide

9.61 H0:  = 2.51

Ha:  > 2.51

This is a one-tailed test. The sample mean is 2.555 which is more than the hypothesized value.
The observed t value is 1.51 with an associated p-value of .072 for a one-tailed test. Because the
p-value is greater than  = .05, the decision is to fail to reject the null hypothesis. There is not
enough evidence to conclude that beef prices are higher.