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Instructor’s Solutions Manual for

Introductory Statistics: A Problem Solving Approach

Second Edition

Stephen Kokoska

Bloomsburg University

1

Chapter 9: Hypothesis Tests Based on a

Single Sample

Section 9.1: The Parts of a Hypothesis Test

and Choosing the Alternative Hypothesis

9.1 True

9.2 False

9.3 False

9.4 Null hypothesis; alternative hypothesis, teststatistic, rejection region

9.5 There is evidence to suggest the alternativehypothesis is true. There is no evidence to suggestthe alternative hypothesis is true.

9.6 a. Valid, null hypothesis b. Invalid, p̂ is astatistic. c. Invalid, s is a statistic. d. Invalid, x isa statistic. e. Valid, alternative hypothesisf. Valid, alternative hypothesis g. Invalid, x̃ is astatistic. h. Valid, null hypothesis

9.7 a. Valid, null hypothesis b. Invalid, IQR is astatistic. c. Valid, alternative hypothesisd. Invalid, y is a statistic. e. Invalid, Q3 is astatistic. f. Valid, alternative hypothesis g. Valid,alternative hypothesis h. Valid, null hypothesis

9.8 a. Valid b. Invalid. The alternative hypothesisshould be Ha: µ > 9.7. c. Invalid. The alternativehypothesis should involve 98.6, for example, Ha:σ2 6= 98.6. d. Valid

9.9 a. Valid b. Valid c. Invalid. The nullhypothesis should be H0: p = 0.32. However, thereis often an implied inequality in the null hypothesis.d. Invalid. Both hypotheses involve a statistic.They should be H0: µ = 78.5 and Ha: µ 6= 78.5.

9.10 a. Valid b. Valid c. Invalid. The nullhypothesis should be stated so that µ (a parameter)equals a single value. d. Valid.

9.11 a. Not permissible. We never accept the nullhypothesis. We fail to reject the null hypothesis.b. Permissible c. Permissible d. Permissiblee. Permissible f. Permissible

9.12 Let µ = the population mean cumulative SATscore. H0: µ = 1498, Ha: µ > 1498

9.13 Let p = the population proportion ofhouseholds that have at least one telescope.H0: p = 0.11, Ha: p > 0.11

9.14 Let µ = the population mean number of acresburned during wildfires.H0: µ = 17,060, Ha: µ < 17,060

9.15 Let µ = the population mean diameter of sanddollars found on Delray Beach. H0: µ = 4.25,Ha: µ < 4.25

9.16 (a) is appropriate. The software company islooking for evidence that the mean age is greaterthan 25.

9.17 H0: σ2 = 32, Ha: σ2 < 32

9.18 H0: p = 0.75, Ha: p > 0.75

9.19 H0: µ = 178 (minutes), Ha: µ < 178

9.20 (c) is appropriate. The bus company is lookingfor evidence that the true proportion of parents whofavor seat-belt installation is greater than 0.50.

9.21 H0: p = 0.65, Ha: p > 0.65

9.22 H0: µ̃ = 350, Ha: µ̃ > 350

9.23 H0: µ = 2000, Ha: µ < 2000

9.24 H0: σ = 7, Ha: σ < 7

9.25 H0: µ = 525, Ha: µ < 525

9.26 H0: p = 0.60, Ha: p > 0.60

9.27 H0: µ = 25, Ha: µ < 25

9.28 (c) H0: p = 0.80, Ha: p > 0.80 is appropriate.The City Council is looking for evidence that thetrue proportion of residents who favor the plan isgreater than 0.80.

9.29 H0: µ = 1925, Ha: µ < 1925

9.30 H0: µ̃ = 125.50, Ha: µ̃ < 125.50

9.31 H0: µ = 2666, Ha: µ > 2666

Section 9.2: Hypothesis Test Errors

9.32 a. Type I error b. Type II error

9.33 False

9.34 False

9.35 True

9.36 True

9.37 a. Type I error b. Correct decision c. TypeII error d. Type I error

9.38 a. Correct decision b. Type II errorc. Type II error d. Correct decision

9.39 a. Type I error b. Type II error c. Correctdecision d. Correct decision

9.40 a. β(11) > β(15). As the alternative value ofµ moves farther from the hypothesized value, theprobability of a type II error decreases. There is a

2 Chapter 9. Hypothesis Tests Based on a Single Sample

better chance of detecting the difference. b. Theprobability of a type II error decreases.

9.41 There is always a chance of making a mistakein any hypothesis test because we never look at theentire population, only a sample. In addition, thevalue of the parameter is unknown. So it is alwayspossible that we have incorrectly decided to rejectthe null hypothesis or failed to reject the nullhypothesis.

9.42 Since α and β are inversely related, if we setthe probability of a type I error (α) to a very smallvalue, then probability of a type II error (β) will bevery large.

9.43 a. H0: µ = 140,000, Ha: µ > 140,000b. Type I error: decide the mean is greater than140,000 when the true mean is 140,000 (or less).Type II error: decide the mean is 140,000 (or less)when the true mean is greater than 140,000.c. Drivers are more angry. Tolls did not have to beraised.d. Transportation officials are more angry. Theyreally need additional tolls to fund planned repairs.

9.44 a. A type I error would be deciding that themean age of the files is greater than 10 years when itis actually no more than 10 years. A type II errorwould be failing to realize the mean age of the filesis more than 10 years when it actually is.

b. A type II error is more serious from the warden’sperspective because in this case the files are reallyvery old and should be archived.

c. A type I error is more serious from a statesenator’s point of view because in this case the filesare really not that old and the money should not bespent to archive them.

9.45 a. H0: µ = 0.65, Ha: µ > 0.65

b. A type I error would be canceling the racebecause you decide that the mean current velocity ismore than 0.65 knots when it is actually less thanthat. A type II error would be failing to realize thatmean current velocity is too high and running therace when it should have been canceled.

c. A type II error is more serious for swimmersbecause it would result in swimming in dangerouscurrents.

d. A type I error is more serious for race organizersbecause it would result in unnecessarily cancelingthe race and losing a great deal of revenue and goodwill.

9.46 a. H0: p = 0.08, Ha: p < 0.08

b. A type II error is more serious for college officials

because, in this case, the new academic policy isreally working, but there is no evidence of that.

c. A type I error is more serious for studentsbecause, in this situation, the new academic policyis not working, but fewer students are showing uplate for exams. This probably means the new policywould remain in effect, but it really isn’t necessary.

9.47 a. A type I error would be deciding the meanpercentage of antioxidants in the bloodstream wasmore than 0.4 when it was actually less. A type IIerror would be failing to realize the meanpercentage of antioxidants in the bloodstream wasmore than 0.4.

b. A type II error would be more serious for theHershey Corporation because it would mean lostrevenue from its inability to advertise the highpercentages of antioxidants in its dark chocolatethat were actually present.

9.48 a. H0: p = 0.60, Ha: p > 0.60

b. A type I error would be deciding that more than60% of the town residents favor the variance whenthe true percentage is really 60% or less. A type IIerror would be failing to realize that more than 60%of the town residents favor the variance.

c. A type II error is more serious for the developerbecause, in this case, the residents are actually infavor of the extended structure, but the developerwill not be able to move forward with the project.

d. A type I error is more serious for the city councilmember because, in this case, they believe theresidents are in favor of the extended structure, butthey really aren’t.

9.49 a. H0: µ = 1200, Ha: µ < 1200

b. Type I error: decide µ < 1200 when the truemean is really 1200 (or greater). That is, decidethat the helmets do not meet the standard whenthey really do. Type II error: decide µ = 1200 (orgreater) when the true mean is really less than 1200.That is, decide the helmets meet the standard whenthey really do not.

9.50 a. H0: µ = 1367, Ha: µ > 1367

b. A type I error would be deciding the meanstamp duty for first-time home buyers is more than£1367 when it is, at most, £1367. A type II errorwould be failing to realize that the mean stamp dutyhas risen to more than £1367.

9.51 a. H0: p = 0.02, Ha: p > 0.02.

b. A type I error would be deciding that theproportion of people who book a room using amobile device is grater than 0.02 when the true

9.3. Hypothesis Tests Concerning a Population Mean When σ Is Known 3

proportion is really 0.02 or less. A type II errorwould be failing to recognize that more than 0.02 ofpeople book a room using a mobile device.

9.52 a. H0: σ2 = 15, Ha: σ2 < 15

b. A type I error would be deciding the variancewas less than 15 when the true population varianceis really 15 or more. A type II error would be failingto recognize that the population variance was lessthan 15.

c. A type I error is more serious for the NSFbecause it would commit more money for TMresearch when there is really no evidence TMdecreases brain activity. A type II error would bemore serious for TM researchers because this wouldmake it difficult for them to obtain more researchsupport (money) in the face of the lack of evidenceof decreased brain activity when TM really works!

9.53 a. H0: p = 0.25, Ha: p > 0.25

b. A type I error would be to conclude thepopulation proportion of homes with satellite dishesis more than 0.25 when it is actually 0.25 or less. Atype II error would be continuing to believe thepopulation proportion of homes with satellite dishesis, at most, 0.25 when it is actually more than 0.25.

c. β(0.27) > β(0.35). As the alternative value of pmoves farther from the hypothesized value, theprobability of a type II error decreases. There is abetter chance of detecting the difference.

9.54 a. H0: p = 0.15, Ha: p > 0.15.

b. A type I error would be to conclude that thetrue proportion of boats with safety violations ismore than 0.15 when it is actually 0.15 or less. Atype II error would be to conclude that theproportion of boats with safety violations is 0.15 (orless) when it is really more than 0.15.

c. As the true value of p approaches 0.15 from theright, the probability of a type I error becomessmaller.

9.55 a. H0: µ = 6400, Ha: µ > 6400 is the pair ofhypotheses that should be used because we care ifthe stress on the bridge is too great.

b. If you regularly drive over the bridge, you wantthe probability of a type II error (β) to be smallbecause you do not want the engineers to fail torealize the bridge is unsafe. In order to make βsmall, you will need to allow α to be the largervalue, 0.1.

9.56 a. H0: p = 0.019, Ha: p < 0.019

b. A type I error would be to conclude that thefailure rate has decreased when it is still 0.019 (or

greater). A type II error would be to conclude thatthe failure rate has not changed (decreased) but it isreally less than 0.019.

c. β(0.010) < β(0.015). As the alternative value ofp moves farther from the hypothesized value, theprobability of a type II error decreases. There is abetter chance of detecting the difference.

d. I would want α = 0.01, the smaller type I error.If we determine that the new implant really doesreduce the failure rate, I would like very strongevidence and a smaller chance of making this error.

Section 9.3: Hypothesis Tests Concerning a

Population Mean When σ Is Known

9.57 False

9.58 True

9.59 False

9.60 True

9.61 True

9.62 a. Z = X−17015/

√

38

b. i. Z ≤ −2.3263 ii. Z ≤ −1.96iii. Z ≤ −1.6449 iv. Z ≤ −1.2816v. Z ≤ −3.0902 vi. Z ≤ −3.7190

9.63 a. Z = X−45.615/

√

16

b. i. Z ≥ 2.3263 ii. Z ≥ 1.96 iii. Z ≥ 1.6449iv. Z ≥ 1.2816 v. Z ≥ 2.5758 vi. Z ≥ 3.2905

9.64 a. Z = X−(−11)

4.5/√

21

b. i. |Z| ≥ 2.5758 ii. |Z| ≥ 1.2816 iii. |Z| ≥ 1.96iv. |Z| ≥ 1.6449 v. |Z| ≥ 3.2905 vi. |Z| ≥ 3.7190

9.65 a. α = 0.05 b. α = 0.005 c. α = 0.02d. α = 0.01 e. α = 0.001 f. α = 0.0001

9.66 a. α = 0.05 b. α = 0.10 c. α = 0.005d. α = 0.001 e. α = 0.20 f. α = 0.02

9.67 a. α = 0.0001 b. α = 0.20 c. α = 0.01d. α = 0.05 e. α = 0.0005 f. α = 0.002

9.68 a. H0: µ = 212; Ha: µ > 212;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 2.3263

b. We assume the underlying population is normaland the population standard deviation (σ) is known.

c. z = 213.5−2122.88/

√

25= 2.6042 ≥ 2.3263. There is

evidence to suggest that the population mean isgreater than 212.

9.69 a. H0: µ = 3.14; Ha: µ < 3.14;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −3.0902


b. We assume the sample size is large and thepopulation standard deviation (σ) is known.

c. z = 1.63−3.146.8/

√

32= −1.2588. There is no evidence to

suggest the population mean is less than 3.14.

9.70 a. H0: µ = 365.25; Ha: µ 6= 365.25;

TS: Z = X−µ0

σ/√

n; RR: |Z| ≥ 1.96


c. z = 360−365.2522.3/

√

48= 1.6311. There is no evidence to

suggest the population mean is different from365.25.

9.71 a. If the test is right-tailed with α = 0.05, therejection region should be RR: Z ≥ 1.6449.

b. The numerator in the test statistic is reversed. Itshould be X − µ0.

c. We never accept the null hypothesis, and wecannot prove the null hypothesis is true. We simplyfail to find evidence in favor of the alternativehypothesis.

d. The null hypothesis is always stated with anequal sign.

e. The probabilities of type I and type II errors areinversely related. This does not mean they sum to 1.

9.72 H0: µ = 51,500; Ha: µ < 51,500;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −2.3263

z = 49,762−51,500

3750/√

38= −2.8570 ≤ −2.3263

There is evidence to suggest the mean income peryear of corporate communications workers hasdecreased.

9.73 Solution Trail

Keywords: Is there any evidence; decreased;σ = 65.Translation: Conduct a one-sided, left-tailed testabout a population mean.Concepts: Hypothesis test concerning a populationmean when σ is known.Vision: Use the template for a one-sided, left-tailedtest about µ. The underlying populationdistribution is unknown, but n is large and σ isknown. Determine the appropriate alternativehypothesis and the corresponding rejection region,find the value of the test statistic, and draw aconclusion.H0: µ = 335; Ha: µ < 335;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −1.6449

z = 325.65−33565/

√

40= −0.9098

There is no evidence to suggest that the mean levelof CO2 emissions has decreased.

9.74 H0: µ = 295; Ha: µ > 295;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 2.3263

z = 306.3−29552/

√

48= 1.5056

There is no evidence to suggest the mean length ofinternational calls has increased. Therefore, there isno evidence to suggest that the campaign wassuccessful.

9.75 H0: µ = 12.4; Ha: µ < 12.4;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −1.96

z = 12.3783−12.40.07/

√

12= −1.0722

There is no evidence to suggest the mean watertable is less than 12.4 feet.

9.76 a. H0: µ = 35; Ha: µ > 35;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 2.3263

z = 36.22−35√

5.7/√

41= 3.2720 ≥ 2.3263

There is evidence to suggest the LOA is greaterthan 35 feet.

b. The answer does not change if α = 0.10. Thecritical value in this case is 1.2816, which is evensmaller than 2.3263.

9.77 Solution Trail

Keywords: Is there any evidence; greater than;population standard deviation 8.6.Translation: Conduct a one-sided, right-tailed testabout a population mean; σ = 8.6.Concepts: Hypothesis test concerning a populationmean when σ is known.Vision: Use the template for a one-sided,right-tailed test about µ. The underlying populationdistribution is unknown, but n is large and σ isknown. Determine the appropriate alternativehypothesis and the corresponding rejection region,find the value of the test statistic, and draw aconclusion.H0: µ = 126.96; Ha: µ > 126.96;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 1.6449

z = 130.29−126.968.6/

√

34= 2.2578 ≥ 1.6449

There is evidence to suggest that the meancomposite indicator is greater than 126.96.

9.78 a. H0: µ = 2200; Ha: µ < 2200;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −1.6449

z = 2089−2200358/

√

37= −1.8860 ≤ −1.6449

There is evidence to suggest the mean caloric intake

9.3. Hypothesis Tests Concerning a Population Mean When σ Is Known 5

is below the daily energy requirement of 2200calories.

b. If we use α = 0.01, the rejection region is RR:Z ≤ −2.3263. In this case, there is no evidence tosuggest the mean caloric intake is below the dailyenergy requirement.

9.79 H0: µ = 21; Ha: µ > 21;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 2.5758

z = 22.4−212.7/

√

32= 2.9332 ≥ 2.5758

There is evidence to suggest the mean thickness isgreater than 21.

9.80 H0: µ = 13.91; Ha: µ > 13.91;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 2.3263

z = 17.8607−13.913.4/

√

14= 4.3477 ≥ 2.3263

There is overwhelming evidence to suggest that themean download speed for the United States isgreater than the world population mean.

9.81 H0: µ = 14; Ha: µ > 14;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 3.0902

z = 14.016−143.2/

√

50= 0.0354

There is no evidence to suggest the mean responsetime has increased.

9.82 a. H0: µ = 12; Ha: µ < 12;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −1.6449

z = 11.85−120.26/

√

12= −1.9985 ≤ −1.6449

There is evidence to suggest that the true meanimpact velocity is less than 12 m/s.

b. If α = 0.01, RR: Z ≤ −2.3263In this case we would not reject the null hypothesisbut would conclude there is no evidence to suggestthat the true mean impact velocity is less than 12m/s.

9.83 H0: µ = 68; Ha: µ > 68;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 1.6449

z = 72.35−6815.5/

√

26= 1.4310

There is no evidence to suggest the mean locksmithservice charge is greater than $68.

9.84 a. H0: µ = 225; Ha: µ < 225;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −1.6449

z = 224.2−2253.6/

√

52= −1.6025

There is not enough evidence to suggest that themean amount of fuel in these cartridges is less thanthe advertised amount.


9.85 H0: µ = 5; Ha: µ < 5;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −2.053s7

z = 5.0733−50.95/

√

18= 0.3275

There is no evidence to suggest that the meanamount of protein is less than 5 grams.

9.86 H0: µ = 42; Ha: µ 6= 42;

TS: Z = X−µ0

σ/√

n; RR: |Z| ≥ 1.96

z = 43.22−427.6/

√

75= 1.3902

There is no evidence to suggest that the true meanheight of mailboxes in Des Moines is different from42 inches.

9.87 H0: µ = 88; Ha: µ < 88;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −2.3263

z = 86.5−883.4/

√

35= −2.61 ≤ −2.3263

There is evidence to suggest that the mean iodineconcentration is less than 88.

9.88 a. P(

X−507.5/

√

25≥ 2.3263

)= 0.01

⇒ P(X ≥ 53.49) = 0.01

β(54) = P(X < 53.49) = P(

Z < 53.49−547.5/

√

25

)

= P(Z < −0.34) = 0.3668

b. β(55) = P(

Z < 53.49−557.5/

√

25

)

= P(Z < −1.01) = 0.1570

β(56) = P(

Z < 53.49−567.5/

√

25

)

= P(Z < −1.67) = 0.0471

c. P(

X−507.5/

√

25≥ 1.96

)= 0.025

⇒ P(X ≥ 52.94) = 0.025

β(54) = P(X < 52.94) = P(

Z < 52.94−547.5/

√

25

)

= P(Z < −0.71) = 0.2399

d. β(55) = P(

Z < 52.94−557.5/

√

25

)

= P(Z < −1.37) = 0.0848

β(56) = P(

Z < 52.94−567.5/

√

25

)

= P(Z < −2.04) = 0.0207

9.89 a. H0: µ = 0.23; Ha: µ < 0.23;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −2.3263

z = 0.1638−0.230.23/

√

26= −4.8189 ≤ −2.3263

There is evidence to suggest that the mean level ofHC emissions is less than 0.23.


b. A type I error is more important to the companybecause in this case it will be building the biodieselfuel plant when the biodiesel fuel does not reallydecrease the mean level of HC emissions. Therefore,the company would want to use a very smallsignificance level.

9.90 a. H0: µ = 23.625; Ha: µ 6= 23.625;

TS: Z = X−µ0

σ/√

n; RR: |Z| ≥ 1.96

z = 23.7−23.6250.15/

√

10= 1.5811

There is no evidence to suggest that the mean is lessthan 23.625, that the assembly line should be shutdown.

b. P(

−1.96 ≤ X−23.6250.15/

√

10≤ 1.96

)= 0.025

⇒ P(23.532 ≤ X ≤ 23.718) = 0.025

9.91 a. H0: µ = 15.5; Ha: µ < 15.5;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −2.3263

z = 15.45−15.50.26/

√

250= −3.0407 ≤ −2.3263

There is evidence to suggest that the mean weight ofpretzel packages is less than 15.5 ounces.

b. The sample size is very large and σ is small.

9.92 a. H0: µ = 344; Ha: µ < 344;

TS: Z = X−µ0

σ/√

n; RR: Z ≤ −2.3263

z = 325.05−34430/

√

28= −3.3418 ≤ −2.3263

There is evidence to suggest the claim is false, thatthe mean depth is less than 344 meters.

b. P(

X−34430/

√

28≤ −2.3263

)= 0.01

⇒ P(X ≤ 330.8111) = 0.01

β(315) = P(X > 330.8111)

= P(

Z > 330.8111−31530/

√

28

)= P(Z > 2.78)

= 1 − P(Z ≤ 2.78) = 1 − 0.9974 = 0.0026

c. Illustration of the probability in part (b):

315 344330.81 x

d. β(310) = P(X > 330.8111)

= P(

Z > 330.8111−31030/

√

28

)= P(Z > 3.67)

= 1 − P(Z ≤ 3.67) = 1 − 0.9999 = 0.0001

9.93 a. H0: µ = 714; Ha: µ 6= 714;

TS: Z = X−µ0

σ/√

n; RR: |Z| ≥ 1.96

z = 601.2−714283/

√

16= −1.5943

There is no evidence to suggest the mean monthlywater usage is different from 714.

b. The standard deviation is very large.

9.94 a. H0: µ = 1250; Ha: µ > 1250;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 2.3263

z = 1305−1250155/

√

45= 2.3803 ≥ 2.3263

There is evidence to suggest that the populationmean square footage is larger than 1250 during asale.

b. P(

X−1250155/

√

45≥ 2.3263

)= 0.01

⇒ P(X ≥ 1303.752) = 0.01

β(1330) = P(X ≤ 1303.752)

= P(

Z ≤ 1303.752−1330155/

√

45

)

= P(Z < −1.14) = 0.1217

c. Illustration:

1250 13301303.8

β(1330) α

x

9.95 a. H0: µ = 12.0; Ha: µ > 12.0;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 2.3263

z = 12.3−12.01.25/

√

37= 1.4599

There is no evidence to suggest that the populationmean moisture content is greater than 12%.

b. P(

X−12.01.25/

√

37≥ 2.3263

)= 0.01

⇒ P(X ≥ 12.4781) = 0.01

β(12.2) = P(X < 12.4781)

= P(

Z < 12.4781−12.21.25/

√

37

)

= P(Z < 1.35) = 0.9120

9.96 a. P(

X−10503.7/

√

18≥ 2.5758

)= 0.005

⇒ P(X ≥ 1052.2464) = 0.005

P(

X−10503.7/

√

18≤ −2.5758

)= 0.005

⇒ P(X ≤ 1047.7536) = 0.005

Critical values: 1047.7536 and 1052.2464

9.4. p Values 7

b. H0: µ = 1050; Ha: µ 6= 1050;

TS: Z = X−µ0

σ/√

n; RR: |Z| ≥ 2.5758

z = 1049−10503.7/

√

18= 1.1467

There is no evidence to suggest that the populationmean is different from 1050. The assembly lineshould not be shut down. Alternatively, x = 1049 isnot in the rejection region in the X world.Therefore, this leads to the same conclusion.

9.97 a. H0: µ = 496; Ha; µ > 496;

TS: Z = X−µ0

σ/√

n; RR: Z > 2.3263

z = 498.42−49646.7/

√

33= 0.2977

There is no evidence to suggest that the true meanweight of passengers and gear is greater than 496.

b. P(

X−49646.7/

√

33≥ 2.3263

)= 0.01

⇒ P(X ≥ 514.9119) = 0.01

β(525) = P(X < 514.9119) = P(

Z < 514.9119−52546.7/

√

33

)

= P(Z < −1.24) = 0.1073

9.98 a. P(

X−80050/

√

25≤ −1.96

)= 0.025

⇒ P(X ≤ 780.4) = 0.025

β(775) = P(X > 780.4)

= P(

Z > 780.4−77550/

√

25

)

= P(Z > 0.54) = 1 − P(Z ≤ 0.54)

= 1 − 0.7054 = 0.2946

π(775) = 1 − 0.2946 = 0.7054

b. Table of power for various values of µa:

µa π(µa) µa π(µa)

735 1.0000 740 1.0000745 0.9998 750 0.9988755 0.9945 760 0.9793765 0.9382 770 0.8508775 0.7054 780 0.5160785 0.3228 790 0.1685795 0.0721 800 0.0250

c. Power curve:

æ æ æ æ æ ææ

æ

æ

æ

æ

æ

ææ

730 740 750 760 770 780 790 8000.0

0.2

0.4

0.6

0.8

1.0

π(µa)

µa

Section 9.4: p Values

9.99 False

9.100 True

9.101 True

9.102 True

9.103 Reject the null hypothesis

9.104 a. Do not reject H0 b. Reject H0 c. Donot reject H0 d. Do not reject H0 e. Reject H0

f. Do not reject H0

9.105 a. 0.0307 b. 0.0054 c. 0.1151 d. 0.2843e. 0.0000522 f. 0.8729

9.106 a. 0.0202 b. 0.0764 c. 0.0006 d. 0.2514e. 0.000002327 f. 0.5987

9.107 a. 2(0.0384) = 0.0768b. 2(0.0764) = 0.1528 c. 2(0.0049) = 0.0098d. 2(0.3557) = 0.7114 e. 2(0.0002) = 0.0004f. 2(0.1977) = 0.3954

9.108 a. p = 0.0764; do not reject H0

b. p = 0.0202; reject H0

c. p = 0.0801; reject H0

d. p = 0.0009; reject H0

e. p = 0.1230; do not reject H0

f. p = 0.0188; do not reject H0

9.109 a. p = 0.0059; reject H0

b. p = 0.0823; do not reject H0

c. p = 0.0113; do not reject H0

d. p = 0.5675; do not reject H0


f. p = 0.0030; do not reject H0

9.110 a. p = 0.2000; do not reject H0

b. p = 0.1671; do not reject H0

c. p = 0.0021; reject H0

d. p = 0.0068; do not reject H0


f. p = 0.0094; reject H0

9.111 H0: µ = 10; Ha: µ > 10; TS: Z = X−µ0

σ/√

n

z = 11.2458−102.66/

√

33= 2.69

p = P(Z ≥ 2.69) = 0.0036 ≤ 0.05

There is evidence to suggest the mean is greaterthan 10.

9.112 H0: µ = 52.1; Ha: µ > 52.1; TS: Z = X−µ0

σ/√

n

z = 57.09−52.116.7/

√

34= 1.74

p = P(Z ≥ 1.74) = 0.0407 ≤ 0.05

There is evidence to suggest the mean quarterbackrating has increased.


9.113 H0: µ = 1; Ha: µ 6= 1; TS: Z = X−µ0

σ/√

n

z = 1.025−10.04/

√

8= 1.77

p = 2P(Z ≥ 1.77) = 0.0771 > 0.01

There is no evidence to suggest that the true meanpercentage of 8h is different from 1.

9.114 a. H0: µ = 30; Ha: µ > 30;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 2.3263

z = 32.2−305.7/

√

58= 2.9394 ≥ 2.3263

There is evidence to suggest that the mean time tocomplete the safety checkup is more than 30minutes.

b. p = P(Z ≥ 2.94) = 0.0016 ≤ 0.01

9.115 H0: µ = 5.5; Ha: µ > 5.5; TS: Z = X−µ0

σ/√

n

z = 6.2375−5.51.3/

√

16= 2.27

p = P(Z ≥ 2.27) = 0.0116 ≤ 0.05 There is evidenceto suggest the mean time to reach the top is greaterthan 5.5 minutes.

9.116 H0: µ = 40; Ha: µ 6= 40; TS: Z = X−µ0

σ/√

n

z = 40.9125−402/

√

8= 1.29

p = 2P(Z ≥ 1.29) = 0.1969 > 0.05

There is no evidence to suggest that the populationmean weight of the blocks is different from 40pounds.

9.117 H0: µ = 8000; Ha: µ > 8000; TS: Z = X−µ0

σ/√

n

z = 8082.325−8000365/

√

40= 1.43

p = P(Z ≥ 1.43) = 0.0769 > 0.05

There is no evidence to suggest the mean is greaterthan 8000 feet.

9.118 H0: µ = 185; Ha: µ 6= 185; TS: Z = X−µ0

σ/√

n

z = 185.447−1852.7/

√

32= 0.94

p = 2P(Z ≥ 0.94) = 0.3491

There is no evidence to suggest that the true meanweight of recycled shingles is different from 185pounds per cubic foot.

9.119 H0: µ = 80; Ha: µ < 80; TS: Z = X−µ0

σ/√

n;

z = 78.6−803.45/

√

35= −2.40 p = P(Z ≤ 0.94) = 0.0082

There is evidence to suggest that the manufacturer’sclaim is false, that the mean rating of the woodfloors is less than 80.

9.120 a. H0: µ = 60; Ha: µ < 60; TS: Z = X−µ0

σ/√

n;

z = 54.5997−608.0/

√

40= −4.27 ⇒ p = 0.0000098

There is strong evidence to suggest that the meandrying time for highway paint is less than 60seconds.

b. Yes, there is very strong evidence to suggest thepaint dries in less than 60 seconds because the pvalue is so small.

c. p value illustration:

0-4.27 z

9.121 a. H0: µ = 1600; Ha: µ < 1600;

TS: Z = X−µ0

σ/√

n

z = 1595.6−160023/

√

25= −0.96

p = P(Z ≤ −0.96) = 0.1694 > 0.01

There is no evidence to suggest the mean is lessthan 1600.

b. p value illustration:

0-0.96 z

9.122 a. H0: µ = 85; Ha: µ < 85; TS: Z = X−µ0

σ/√

n;

z = 84.88−855.6/

√

60= −0.17

p = P(Z ≤ −0.17) = 0.4341

There is no evidence to suggest that the meandecibel level produced by smoke alarms in this cityis less than 85.

b. The smallest significance level at which thishypothesis test would be significant is the p value,0.4341.

9.123 a. H0: µ = 35,800; Ha: µ > 35,800;

TS: Z = X−µ0

σ/√

n

z = 35,849.95−35,800

315/√

40= 1.00

p = P(Z ≥ 1.00) = 0.1580 > 0.05

There is no evidence to suggest the mean is greaterthan 35,800.

9.5. Hypothesis Tests Concerning a Population Mean When σ Is Unknown 9

b. p value illustration:

0 1.00 z


Population Mean When σ Is Unknown

9.124 True

9.125 True

9.126 True

9.127 False

9.128 False

9.129 a. TS: T = X−µ0

S/√

n

b. i. T ≥ 3.3649 ii. T ≥ 2.0739 iii. T ≥ 1.7459iv. T ≥ 1.3125 v. T ≥ 4.2968 vi. T ≥ 6.4420

9.130 a. TS: T = X−µ0

S/√

n

b. i. T ≤ −2.6245 ii. T ≤ −4.5869iii. T ≤ −1.7247 iv. T ≤ −1.3195v. T ≤ −7.1732 vi. T ≤ −4.2340

9.131 a. TS: T = X−µ0

S/√

n

b. i. |T | ≥ 3.1058 ii. |T | ≥ 1.3304iii. |T | ≥ 2.0595 iv. |T | ≥ 1.7033 v. |T | ≥ 5.9588vi. |T | ≥ 2.39608

9.132 a. α = 0.025 b. α = 0.001 c. α = 0.01d. α = 0.001

9.133 a. α = 0.10 b. α = 0.001 c. α = 0.005d. α = 0.01

9.134 a. α = 0.1 b. α = 0.01 c. α = 0.002d. α = 0.0002

9.135 a. 0.01 ≤ p ≤ 0.025 b. 0.005 ≤ p ≤ 0.01c. p < 0.0001 d. 0.05 ≤ p ≤ 0.01

9.136 a. 0.025 ≤ p ≤ 0.05 b. p > 0.20c. 0.01 ≤ p ≤ 0.025 d. 0.0005 ≤ p ≤ 0.10

9.137 a. 0.05 ≤ p ≤ 0.10 b. 0.001 ≤ p ≤ 0.005c. p < 0.0001 d. 0.1 ≤ p ≤ 0.2

9.138 a. H0: µ = 1.618; Ha: µ < 1.618;

TS: T = X−µ0

S/√

n; RR: T ≤ −1.7291

b. t = 1.5−1.6180.45/

√

20= −1.1727

There is no evidence to suggest that the mean is less

than 1.618.

c. p = P(T ≤ −1.1727) = 0.1277

9.139 a. H0: µ = 57.71; Ha: µ > 57.71;

TS: T = X−µ0

S/√

n; RR: T ≥ 2.7638

b. t = 59.31−57.711.6037/

√

11= 3.3053 ≥ 2.7638

There is evidence to suggest the mean is greaterthan 57.71.

c. p = P(T ≥ 3.3053) = 0.0040

9.140 a. H0: µ = 9.96; Ha: µ 6= 9.96;

TS: T = X−µ0

S/√

n; RR: |T | ≥ 3.4210

b. t = 9.04−9.961.2/

√

28= −4.0568 ≤ −3.4210

There is evidence to suggest that the mean isdifferent from 9.96.

c. p = 2P(T ≤ −3.4210) = 0.0004

9.141 a. When the test statistic is T , the value of σ

is unknown. The test statistic should be T = X−µ0

S/√

n.

b. The sample size n = 25 should be used incomputing the value of the test statistic, not n − 1.

c. This is a two-sided test, so the rejection regionshould be |T | ≥ 2.0639.

d. If the value of the test statistic is t = 2.6732,then the p value is between 0.01 and 0.025.

9.142 Solution Trail

Keywords: Is any evidence; mean weight; less than;s = 37.8; normal; random sample.Translation: Conduct a one-sided, left-tailed testabout µ.Concepts: Hypothesis test concerning a populationmean when σ is unknown.Vision: Use the template for a one-sided, left-tailedt test about µ. The underlying population isassumed to be normal, and σ is unknown.Determine the appropriate alternative hypothesisand the corresponding rejection region, find thevalue of the test statistic, and draw a conclusion.

H0: µ = 550; Ha: µ < 550;

TS: T = X−µ0

S/√

n; RR: T ≤ −1.7959

t = 535.7−55037.8/

√

12= −1.3105

p = P(T ≤ −1.3105) = 0.1084 > 0.05

There is no evidence to suggest that the meanweight of Atlantic bluefin tuna is less than 550pounds.

9.143 H0: µ = 303; Ha: µ > 303;

TS: T = X−µ0

S/√

n; RR: T ≥ 2.1448

t = 310.9333−30313.6772/

√

15= 2.2465 ≥ 2.1448


p = P(T ≥ 2.2465) = 0.0207 ≤ 0.025

There is evidence to suggest that the mean airbornetime is greater than 303.

9.144 H0: µ = 31.9; Ha: µ < 31.9;

TS: T = X−µ0

S/√

n; RR: T ≤ −2.4851

t = 30.088−31.94.433/

√

26= −2.0842

p = P(T ≤ −2.0842) = 0.0238 > 0.01

There is no evidence to suggest the mean is lessthan 31.9.


Keywords: Is there any evidence; increased;s = 15.59; normal; random sample.Translation: Conduct a one-sided, right-tailed testabout µ.Concepts: Hypothesis test concerning a populationmean when σ is unknown.Vision: Use the template for a one-sided,right-tailed t test about µ. The underlyingpopulation is assumed to be normal, and σ isunknown. Determine the appropriate alternativehypothesis and the corresponding rejection region,find the value of the test statistic, and draw aconclusion.

H0: µ = 64.98; Ha: µ > 64.98;

TS: T = X−µ0

S/√

n; RR: T ≥ 2.6503

t = 70.45−64.9815.59/

√

14= 1.3128 < 2.6503

p = P(T ≥ 1.3128) = 0.1060 > 0.01

There is no evidence to suggest that the mean finehas increased.

9.146 H0: µ = 40; Ha: µ > 40;

TS: T = X−µ0

S/√

n; RR: T ≥ 2.6503

t = 44.3929−403.384/

√

14= 4.8568 ≥ 2.6503

p = P(T ≥ 4.8568) = 0.0003 ≤ 0.01

There is strong evidence to suggest that the meannumber of hours worked during the last week bywhite-collar employees was greater than 40. Weassumed the underlying distribution is normal.

9.147 H0: µ = 23.1; Ha: µ > 23.1;

TS: T = X−µ0

S/√

n; RR: T ≥ 2.8965

t = 24.6−23.12.1/

√

9= 2.1429 < 2.8965

p = P(T ≥ 2.1429) = 0.0322 > 0.01

There is no evidence to suggest that the mean hotelroom occupation rate has increased. If the test issignificant, one cannot conclude the ad campaigncaused the increase.

9.148 H0: µ = 5.64; Ha: µ > 5.64;

TS: T = X−µ0

S/√

n; RR: T ≥ 1.7459

t = 5.869−5.641.16/

√

17= 0.8140 < 1.7459

p = P(T ≥ 0.8140) = 0.2138 > 0.05

There is no evidence to suggest that the mean crudeoil imported per day in China has increased.

9.149 H0: µ = 3.2; Ha: µ 6= 3.2;

TS: T = X−µ0

S/√

n; RR: |T | ≥ 2.0860

t = 3.1429−3.20.9453/

√

21= −0.2770; |t| < 2.0860

p = 2P(T ≤ −0.2770) = 0.7846 > 0.05

There is no evidence to suggest that the meanphysician density is different from 3.2.

9.150 a. H0: µ = 0.75; Ha: µ < 0.75;

TS: T = X−µ0

S/√

n; RR: T ≤ −2.4999

t = 0.6658−0.750.2461/

√

24= −1.6753 > −2.4999

p = P(T ≤ −1.6753) = 0.0537 > 0.01

There is no evidence to suggest that the mean timehas decreased.

b. 0.05 ≤ p ≤ 0.10

9.151 a. H0: µ = 9; Ha: µ > 9;

TS: T = X−µ0

S/√

n; RR: T ≥ 1.8595

t = 10.3022−94.587/

√

9= 0.8517 < 1.8595

p = P(T ≥ 0.8517) = 0.2096 > 0.05

There is no evidence to suggest that the mean floorarea is greater than 9 million square feet.

b. p > 0.20

9.152 H0: µ = 55.5; Ha: µ > 55.5;

TS: T = X−µ0

S/√

n; RR: T ≥ 1.7959

t = 70.3417−55.552.8575/

√

12= 0.9727 < 1.7959

p = P(T ≥ 0.9727) = 0.1758 > 0.05

There is no evidence to suggest the true meanspending on Internet advertising has increased.

9.153 a. H0: µ = 1381; Ha: µ 6= 1381;

TS: T = X−µ0

S/√

n; RR: |T | ≥ 2.1199

t = 1857−1381786/

√

17= 2.4969; |t| ≥ 2.1199

p = 2P(T ≥ 2.4969) = 0.0238 ≤ 0.05

There is evidence to suggest that the true mean lossas a result of a residential break-in has changedfrom $1381.

b. 0.02 ≤ p ≤ 0.05

9.5. Hypothesis Tests Concerning a Population Mean When σ Is Unknown 11


-2.4969 2.49690 t

9.154 a. H0: µ = 8.85; Ha: µ > 8.85;

TS: T = X−µ0

S/√

n; RR: T ≥ 1.7531

t = 10.9131−8.853.1559/

√

16= 2.6149 ≥ 1.7531

p = P(T ≥ 2.6149) = 0.0098 ≤ 0.05

There is evidence to suggest that the mean amountbet on Major League baseball games has increased.

b. 0.005 ≤ p ≤ 0.01

9.155 a. H0: µ = 159,350; Ha: µ > 159,350;

TS: T = X−µ0

S/√

n; RR: T ≥ 2.7638

t = 163288−1593508792/

√

11= 1.4855 < 2.7638

p = P(T ≥ 1.4855) = 0.0841 > 0.01

There is no evidence to suggest that the meanmethane output per well per day has increased.

b. 0.05 ≤ p ≤ 0.10.

9.156 a. H0: µ = 15; Ha: µ < 15;

TS: T = X−µ0

S/√

n; RR: T ≤ −2.2281

t = 14.35−153.75/

√

11= −0.5749 > −2.2281

p = P(T ≤ −0.5749) = 0.2890 > 0.025

There is no evidence to suggest that the meansavings is less than $15.

b. With such a small sample size, due to naturalsampling variability it is reasonable to obtain asample mean of $14.35 if the population mean isreally $15.

c. p > 0.20

9.157 a. H0: µ = 13.5; Ha: µ > 13.5;

TS: T = X−µ0

S/√

n; RR: T ≥ 2.4999

t = 16.005−13.54.039/

√

24= 3.0384 ≥ 2.4999

p = P(T ≥ 3.0384) = 0.0029 ≤ 0.01

There is evidence to suggest that the mean length oflionfish in this area is greater than 13.5 inches.

b. 0.001 ≤ p ≤ 0.005

9.158 H0: µ = 2.5; Ha: µ > 2.5;

TS: T = X−µ0

S/√

n; RR: T ≥ 2.7181

t = 2.6267−2.50.3978/

√

12= 1.1030 < 2.7181

p = P(T ≥ 1.1030) = 0.1468 > 0.01

There is no evidence to suggest that the mean timeto finish this triathlon was more than 2.5 hours.

9.159 a. H0: µ = 4.75; Ha: µ < 4.75;

TS: T = X−µ0

S/√

n; RR: T ≤ −2.4121

t = 4.66−4.750.25/

√

46= −2.4416 ≤ −2.4121

p = P(T ≤ −2.4416) = 0.0093 ≤ 0.01

There is evidence to suggest that the populationmean width of littlenecks has decreased.

b. This test is statistically significant even thoughthe sample mean (4.66) is so close to the assumedpopulation mean (4.75) because the sample size isfairly large (n = 46) and the sample standarddeviation is quite small (s = 0.25) relative to thesample mean.

c. 0.005 ≤ p ≤ 0.01

9.160 a. H0: µ = 0; Ha: µ 6= 0;

TS: T = X−µ0

S/√

n; RR: |T | ≥ 2.8453

t = 1.3776−01.633/

√

21= 3.8658; |t| ≥ 2.8453

p = 2P(T ≥ 3.8658) = 0.0010

There is evidence to suggest that the mean PDSI isdifferent from 0.

b. This conclusion suggests that conditions in theEastern United States have been wetter thannormal. Farmers may not have to water as muchand reservoir levels may be high.

9.161 a. H0: µ = 1.0; Ha: µ > 1.0;

TS: T = X−µ0

S/√

n; RR: T ≥ 1.7033

t = 1.4429−1.01.1564/

√

28= 2.0264 ≥ 1.7033

p = P(T ≥ 2.0264) = 0.0264 ≤ 0.05

There is evidence to suggest that the true meanmagnitude is greater than 1.0.

b. 0.025 ≤ p ≤ 0.05

9.162 a. H0: µ = 1000; Ha: µ > 1000;

TS: T = X−µ0

S/√

n; RR: T ≥ 1.7033

t = 1003.6786−100020.7097/

√

28= 0.9399 < 1.7033

p = P(T ≥ 0.9399) = 0.1778 > 0.05

There is no evidence to suggest that the meanfluoride concentration per tube is greater than 1000ppmF.

b. 0.10 ≤ p ≤ 0.20


9.163 a. H0: µ = 65; Ha: µ > 65;

TS: T = X−µ0

S/√

n; RR: T ≥ 2.6503

t = 70.8571−6545.7045/

√

14= 0.4795 < 2.6503

p = P(T ≥ 0.4795) = 0.3198 > 0.01

There is no evidence to suggest that the meanamount of aspartame is greater than 65 mg.

b. p ≥ 0.20

9.164 a. H0: µ = 9000; Ha: µ 6= 9000;

TS: T = X−µ0

S/√

n; RR: |T | ≥ 3.1693

t = 8722−9000460.4/

√

11= −2.0027; |t| < 3.1693

p = 2P(T ≤ −2.0027) = 0.0731 > 0.01

There is no evidence to suggest that the true meanattendance is different from 9000.

b. 0.05 ≤ p ≤ 0.10

c. 8722−9000460.4/

√

n= −t0.005,n−1 ⇒ n = 22

The sample size would have to be at least 22 for thistest to be significant.

9.165 a. H0: λ = 4; Ha: λ < 4;

TS: X = the number of people locked out of theirrooms;

RR: X ≤ 0 (α ≈ 0.0183)

b. x = 2. There is no evidence to suggest the meannumber of people locked out of their rooms per dayis less than 4.

c. P(X ≤ 2) = 0.2381

9.166 a. H0: λ = 19; Ha: λ > 19;

TS: Z = x−λ0√

λ0; RR: Z ≥ 2.3263

b. Z = 24−19√

19= 1.1471 < 2.3263

There is no evidence to suggest that the meannumber of dog bites per day is greater than 19.

c. p = P(Z ≥ 1.471) = 0.1257

Section 9.6: Large-Sample Hypothesis Tests

Concerning a Population Proportion

9.167 True

9.168 False

9.169 False

9.170 False

9.171 The hypothesis test is not valid.

9.172 a. 276(0.3) = 82.8 ≥ 5, 276(0.7) = 193.2 ≥ 5The test is appropriate.

b. 1158(0.6) = 694.8 ≥ 5, 1158(0.4) = 463.2 ≥ 5The test is appropriate.

c. 645(0.03) = 19.35 ≥ 5, 645(0.97) = 625.65 ≥ 5The test is appropriate.

d. 159(0.97) = 154.23 ≥ 5; 159(0.03) = 4.77 < 5The test is not appropriate.

e. 322(0.38) = 122.36 ≥ 5; 322(0.62) = 199.64 ≥ 5The test is appropriate.

f. 443(0.82) = 363.26 ≥ 5; 443(0.18) = 79.74 ≥ 5The test is appropriate.

9.173

RR Value of TS Conclusion

a. Z ≥ 1.6449 0.3033 Do not reject H0

b. Z ≥ 1.2816 1.4882 Reject H0

c. Z ≥ 2.3263 2.3327 Reject H0

d. Z ≥ 1.9600 0.6872 Do not reject H0

e. Z ≥ 2.3263 0.3307 Do not reject H0

9.174


a. Z ≤ −2.3263 −0.7090 Do not reject H0

b. Z ≤ −1.6449 −1.4596 Do not reject H0

c. Z ≤ −1.9600 −2.9056 Reject H0

d. Z ≤ −3.0902 −3.3245 Reject H0

e. Z ≤ −1.6449 −1.1619 Do not reject H0

9.175


a. |Z| ≥ 2.2414 −2.0142 Do not reject H0.b. |Z| ≥ 2.3263 2.3451 Reject H0

c. |Z| ≥ 1.9600 −2.0294 Reject H0

d. |Z| ≥ 2.8070 −1.4025 Do not reject H0

e. |Z| ≥ 2.5758 2.6455 Reject H0

9.176

Value of TS p value Conclusion

a. 0.9796 0.1636 Do not reject H0

b. 1.8453 0.0325 Reject H0

c. 1.5216 0.0641 Do not reject H0

d. 2.8587 0.0021 Reject H0

e. 3.2703 0.0005 Reject H0

9.177


a. −2.0285 0.0213 Reject H0

b. 0.0574 0.5229 Do not reject H0

c. −0.7611 0.2233 Do not reject H0

d. −2.2166 0.0133 Reject H0

e. −2.6187 0.0044 Reject H0

9.6. Large-Sample Hypothesis Tests Concerning a Population Proportion 13

9.178


a. −1.5489 0.1214 Do not reject H0

b. 2.4086 0.0160 Reject H0

c. −3.3621 0.0008 Reject H0

d. 2.7775 0.0055 Reject H0

e. −2.7110 0.0067 Reject H0

9.179 a. n = 500, x = 16, p0 = 0.02

b. 500(0.02) = 10 ≥ 5; 500(0.98) = 490 ≥ 5The large sample test is appropriate.

c. H0: p = 0.02; Ha: p > 0.02;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≥ 1.6449

z = 0.032−0.02√(0.02)(0.98)

500

= 1.9166 ≥ 1.6449

There is evidence to suggest that the proportion ofchildren who eat the recommended number ofservings each day has increased.

d. p = P(Z ≥ 1.9166) = 0.0276

9.180 a. n = 500, x = 115, p0 = 0.20

b. 500(0.20) = 100 ≥ 5; 500(0.80) = 400 ≥ 5The large sample test is appropriate.

c. H0: p = 0.20; Ha: p 6= 0.20;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: |Z| ≥ 1.96

z = 0.23−0.20√(0.20)(0.80)

500

= 1.6771; |z| < 1.96

There is no evidence to suggest the proportion ofworkers in the retail industry who believe a boss hashurt their career is different from 0.20.

d. p = 2P(Z ≥ 1.6771) = 0.0935

9.181 a. n = 225, x = 189, p0 = 0.90

b. 225(0.90) = 202.5 ≥ 5; 225(0.10) = 22.5 ≥ 5The large-sample test is appropriate.

c. H0: p = 0.90; Ha: p 6= 0.90;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: |Z| ≥ 1.96

z = 0.84−0.90√(0.90)(0.10)

225

= −3.0 ≤ −1.96

There is evidence to suggest that the proportion ofhomeschooled children who attend college isdifferent from 0.90.

d. p = 2P(Z ≤ −3.0) = 0.0027

9.182 a. n = 130, x = 62, p0 = 0.52

b. 130(0.52) = 67.6 ≥ 5; 130(0.48) = 62.4 ≥ 5The large-sample test is appropriate.

c. H0: p = 0.52; Ha: p < 0.52;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −2.3263

z = 0.4769−0.52√(0.52)(0.48)

130

= −0.9831 > −2.3263

There is no evidence to suggest that the turnoverrate of rural doctors has decreased.

d. p = P(Z ≤ −0.9831) = 0.1628


Keywords: Is there any evidence; greater than;1250 women; 502 developed cancer; 38%.Translation: Conduct a one-sided, right-tailed testabout p; n = 1250; x = 502; p0 = 0.38.Concepts: Large-sample hypothesis test concerninga population proportion.Vision: Check the nonskewness criterion. Use thetemplate for a one-sided, right-tailed test about p.Use α = 0.01 to find the critical value, compute thevalue of the test statistic, and draw a conclusion.

H0: p = 0.38; Ha: p > 0.38;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≥ 2.3263

z = 0.4016−0.38√(0.38)(0.62)

1250

= 1.5733 < 2.3263

p = P(Z ≥ 1.5733) = 0.0578 > 0.01

There is no evidence to suggest the proportion oftaller women who develop cancer is greater than0.38.

9.184 H0: p = 0.58; Ha: p > 0.58;

TS: Z = P̂ −p0√p0(1−p0)

n

;

z = 0.6226−0.58√(0.58)(0.42)

575

= 2.0701

p = P(Z ≥ 2.0171) = 0.0192 ≤ 0.05

There is evidence to suggest that the trueproportion of Missouri residents worried aboutviolations of privacy rights is greater than 0.58.

9.185 H0: p = 0.50; Ha: p > 0.50;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≥ 1.6449

z = 0.6333−0.50√(0.50)(0.50)

120

= 2.9212 ≥ 1.6449

p = P(Z ≥ 2.9212) = 0.0017 ≤ 0.05.

There is evidence to suggest that more than half ofall people have trouble sleeping during a full moon.


Keywords: Is there any evidence; greater than;1500 households; 235 too much in debt; 14%;random sample.Translation: Conduct a one-sided, right-tailed testabout p; n = 1500; x = 235; p0 = 0.14.Concepts: Large-sample hypothesis test concerninga population proportion.


Vision: Check the nonskewness criterion. Use thetemplate for a one-sided, right-tailed test about p.Use α = 0.05 to find the critical value, compute thevalue of the test statistic, and draw a conclusion.

H0: p = 0.14; Ha: p > 0.14;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≥ 1.6449

z = 0.1567−0.14√(0.14)(0.86)

1500

= 1.8603 ≥ 1.6449

p = P(Z ≥ 1.8603) = 0.0314 ≤ 0.05

There is evidence to suggest that the trueproportion of Canadian households that are toomuch in debt is greater than 0.14.

9.187 H0: p = 0.40; Ha: p < 0.40;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −2.3263

z = 0.3387−0.40√(0.40)(0.60)

375

= −2.4244 ≤ −2.3263

p = P(Z ≤ −2.4244) = 0.0077 ≤ 0.01

There is evidence to suggest that fewer than 40% ofall residents are satisfied with the local government.This politician should enter the race for mayor.

9.188 item H0: p = 0.02; Ha: p 6= 0.02;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: |Z| ≥ 1.96

z = 0.0185−0.02√(0.02)(0.98)

1300

= −0.3962; |z| < 1.96

p = 2P(Z ≤ −0.3962) = 0.6919 > 0.05

There is no evidence to suggest that the proportionof adults in this area who hear the Hum haschanged.

9.189 H0: p = 0.10; Ha: p < 0.10;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −1.6449

z = 0.09−0.10√(0.10)(0.90)

100

= −0.3333 > −1.6449

p = P(Z ≤ −03333) = 0.3694 > 0.05

There is no evidence to suggest that the proportionof rent-controlled apartments is less than 0.10.

9.190 H0: p = 0.86; Ha: p < 0.86;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −2.3263

z = 0.83−0.86√(0.86)(0.14)

1000

= −2.7341 ≤ −2.3263

p = P(Z ≤ −2.7341) = 0.0031 ≤ 0.01

There is evidence to suggest that the proportion ofAmericans who disapprove of the job Congress isdoing has decreased.

9.191 H0: p = 1/3; Ha: p 6= 1/3;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: |Z| ≥ 1.96

z = 0.3582−0.3333√(0.3333)(0.6667)

656

= 1.3528; |z| < 1.96

p = 2P(Z ≥ 1.3528) = 0.1761 > 0.05.

There is no evidence to suggest that the proportionof Americans not using handwriting is different from1/3.

9.192 H0: p = 0.14; Ha: p < 0.14;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −1.6449

z = 0.125−0.14√(0.14)(0.86)

1200

= −1.4975 > −1.6449

p = P(Z ≤ −1.4975) = 0.0671 > 0.05

There is no evidence to suggest that the proportionof students participating in these programs hasdecreased.

9.193 a. H0: p = 0.95; Ha: p < 0.95;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −1.6449

z = 0.915−0.95√(0.95)(0.05)

200

= −2.2711 ≤ −1.6449

There is evidence to suggest that the proportion ofall tractor batteries that last at least three years isless than 0.95.

b. p = P(Z ≤ −2.2711) = 0.0116 ≤ 0.05

c. Since there is evidence to suggest that less than95% of all batteries last at least three years, thecompany should not implement the new warranty.

9.194 a. H0: p = 0.45; Ha: p > 0.45;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≥ 3.0902

z = 0.5275−0.45√(0.45)(0.55)

400

= 3.1156 ≥ 3.0902

There is evidence to suggest that the proportion ofpeople who complete the puzzle within the allottedtime is greater than 0.45.

b. p = P(Z ≥ 3.1156) = 0.0009


0 3.1156 z

9.195 a. H0: p = 0.23; Ha: p < 0.23;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −1.6449

9.7. Hypothesis Tests Concerning a Population Variance or Standard Deviation 15

z = 0.1816−0.23√(0.23)(0.77)

347

= −2.1443 ≤ −1.6449

There is evidence to suggest that the proportion offoreign purchases of U.S. homes by Canadians is lessthan 0.23.

b. p = P(Z ≤ −2.1443) = 0.0160 ≤ 0.05.

c. Critical value and p value illustration:

-2.1443 -1.6449 0 z

9.196 a. H0: p = 0.43; Ha: p < 0.43;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −2.3263

z = 0.38−0.43√(0.43)(0.57)

1000

= −3.1937 ≤ −2.3263

p = P(Z ≤ −3.1937) = 0.0007 ≤ 0.01

There is evidence to suggest that the proportion ofworking Americans who do not seek medicalassistance when needed has decreased.

b. P

(P̂ −0.43√(0.43)(0.57)

1000

≤ −2.3263

)= 0.01

⇒ P(P̂ ≤ 0.3936) = 0.01

β(0.38) = P(P̂ ≥ 0.3936) = P

(Z ≥ 0.3936−0.38√

(0.38)(0.62)1000

)

= P(Z ≥ 0.8847) = 0.1882

c. Needed critical value is 1.2816:

P(Z ≥ 1.2816) = 0.10

Solve: 0.3936−pa√pa(1−pa)

1000

= 1.2816

⇒ (0.3936 − pa)2 = (1.2816)2 (pa)(1−pa)1000

⇒ 0.39362 − 0.7872pa + pa = 1.28162

1000 (pa − p2a)

⇒(

1 + 1.28162

1000

)p2

a −(

0.7872 − 1.28162

1000

)pa

+ 0.39362 = 0

⇒ pa = 0.3740

d. pc = −2.3263√

90.43)(0.57)n + 0.43

Solve for n: pc−0.35√(0.35)(0.65)

n

= 1.96

⇒ n = 680.29

Since n must be an integer, n ≥ 681.

9.197 a. H0: p = 0.51; Ha: p < 0.51;

TS: X = the number of students who admit tocheating out of the 25.

RR: X ≤ 7 (α ≈ 0.0170)

b. x = 9 > 7. There is no evidence to suggest thatthe true proportion of Long Island students whoadmit to cheating is less than 0.51.

c. Assuming X ∼ B(25, 0.51):p = P(X ≤ 9) = 0.0964


Population Variance or Standard Deviation

9.198 False

9.199 False

9.200 True

9.201 False

9.202 True

9.203 a. X2 = (n−1)S2

σ20

b. i. X2 ≥ 19.6751 ii. X2 ≥ 31.5264iii. X2 ≥ 40.2894 iv. X2 ≥ 37.9159v. X2 ≥ 20.5150 vi. X2 ≥ 42.5793

9.204 a. X2 = (n−1)S2

σ20

b. i. X2 ≤ 3.9416 ii. X2 ≤ 5.2260iii. X2 ≤ 12.5622 iv. X2 ≤ 17.2919v. X2 ≤ 20.0719 vi. X2 ≤ 0.9893

9.205 a. X2 = (n−1)S2

σ20

b. i. X2 ≤ 23.6543 or X2 ≥ 58.1201ii. X2 ≤ 13.7867 or X2 ≥ 53.6720 iii. X2 ≤ 5.8957or X2 ≥ 49.0108 iv. X2 ≤ 5.2293 or X2 ≥ 30.5779v. X2 ≤ 10.3909 or X2 ≥ 56.8923 vi. X2 ≤ 1.0636or X2 ≥ 7.7794

9.206 a. α = 0.05 b. α = 0.005 c. α = 0.005d. α = 0.0005

9.207 a. α = 0.0005 b. α = 0.01 c. α = 0.025d. α = 0.005

9.208 a. α = 0.01 b. α = 0.02 c. α = 0.001d. α = 0.05

9.209 a. 0.01 ≤ p ≤ 0.025 b. 0.05 ≤ p ≤ 0.10c. 0.001 ≤ p ≤ 0.005 d. p ≤ 0.0001

9.210 a. 0.0001 ≤ p ≤ 0.001 b. p ≤ 0.0001c. 0.005 ≤ p ≤ 0.01 d. 0.025 ≤ p ≤ 0.05

9.211 a. 0.02 ≤ p ≤ 0.05 b. 0.0002 ≤ p ≤ 0.001c. 0.05 ≤ p ≤ 0.10 d. 0.001 ≤ p ≤ 0.002

9.212 a. H0: σ2 = 16.7; Ha: σ2 > 16.7;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 37.5662

b. χ2 = (20)(28)16.7 = 33.5329 < 37.5662


There is no evidence to suggest that the populationvariance is greater than 16.7.

c. 0.025 ≤ p ≤ 0.05p value illustration:

0 18 33.5329 x

9.213 a. H0: σ2 = 36.8; Ha: σ2 6= 36.8;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≤ 6.2621 or X2 ≥ 27.4884

b. χ2 = (15)(105.8633)36.8 = 43.1508 ≥ 27.4884

p = 2P(X2 ≥ 43.1508) = 0.0003 ≤ 0.05

There is evidence to suggest the population varianceis different from 36.8.

c. 0.0002 ≤ p ≤ 0.01

9.214 a. H0: σ2 = 75.6; Ha: σ2 < 75.6;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≤ 17.2616

b. χ2 = (39)(48.5)75.6 = 25.0198 > 17.2616

There is no evidence to suggest that the populationvariance is less than 75.6.

c. 0.025 ≤ p ≤ 0.05

9.215 H0: σ2 = 0.25; Ha: σ2 > 0.25;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 48.6024

χ2 = (34)(0.558)2

0.25 = 42.3455 < 48.6024

p = P(X2 ≥ 42.3455) = 0.1541 > 0.05

There is no evidence to suggest that the populationvariance is larger than 0.25.


Keywords: Is there any evidence; populationvariance; greater than 5800; random sample.Translation: Conduct a one-sided, right-tailed testabout σ2.Concepts: Hypothesis test concerning a populationvariance.Vision: The underlying distribution is assumednormal. Use the template for a one-sided,right-tailed test about σ2. Use α = 0.01 to find thecritical value. Compute the value of the teststatistic and draw a conclusion.

H0: σ2 = 5800; Ha: σ2 > 5800;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 49.5879

χ2 = (29)(6211.0851)5800 = 31.0554 < 49.5879

p = P(X2 ≥ 31.0554) = 0.3628 > 0.01

There is no evidence to suggest that the populationvariance in technically recoverable shale gas isgreater than 5800 tcf2.

9.217 H0: σ2 = 324; Ha: σ2 > 324;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 24.7250

χ2 = (11)(21.56)2

324 = 15.7814 < 24.7250

p = P(X2 ≥ 15.7814) = 0.1494 > 0.01

There is no evidence to suggest that the populationvariance in stress is greater than 324. Therefore,there is no evidence to refute the manufacturer’sclaim.

9.218 H0: σ2 = 0.09; Ha: σ2 > 0.09;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 36.7807

χ2 = (22)(0.105)0.09 = 25.6667 < 36.7807

p = P(X2 ≥ 25.667) = 0.2663 > 0.025

There is no evidence to suggest that the populationvariance in the fillings of the cream puffs is greaterthan 0.09 ounces.


Keywords: Is there any evidence; variance; greaterthanTranslation: Conduct a one-sided, right tailed testabout σ2.Concepts: Hypothesis test concerning a populationvariance.Vision: The underlying population is assumednormal. Use the template for a one-sided,right-tailed test about σ2. Use α = 0.01 to find thecritical value. Compute the value of the teststatistic, and draw a conclusion.

H0: σ2 = 62.5; Ha: σ2 > 62.5;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 21.6660

χ2 = (9)(70.1)62.5 = 10.0944 < 21.6660

p = P(X2 ≥ 10.0944) = 0.3429 > 0.01.

There is no evidence to suggest that the variance isgreater than 62.5.

9.220 H0: σ2 = 0.04; Ha: σ2 > 0.04;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 42.9798

χ2 = (24)(0.224)2

0.04 = 35.7216 < 42.9798

p = P(X2 ≥ 35.7216) = 0.0584 > 0.01

There is no evidence to suggest that the populationvariance in die weight is greater than 0.04 g2.

9.7. Hypothesis Tests Concerning a Population Variance or Standard Deviation 17

9.221 H0: σ2 = 0.32; Ha: σ2 > 0.32;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 37.6525

χ2 = (25)(0.1170)0.09 = 32.4900 < 37.6525

p = P(X2 ≥ 32.49) = 0.1443 > 0.05

There is no evidence to suggest that the standarddeviation is greater than 0.3.

9.222 H0: σ2 = 0.57; Ha: σ2 6= 0.57;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≤ 5.0917 or X2 ≥ 37.9461

χ2 = (17)(0.34)2

0.57 = 3.4477 ≤ 5.0917

p = 2P(X2 ≤ 3.4477) = 0.0004 ≤ 0.05

There is evidence to suggest that the populationvariance is different from 0.57.

9.223 H0: σ2 = 22.5; Ha: σ2 < 22.5;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≤ 23.2686

χ2 = (36)(15.6)22.5 = 24.9600 > 23.2686

p = P(X2 ≤ 24.96) = 0.0832 > 0.05

There is no evidence to suggest that the variance inbull-riding time has decreased.

9.224 a. H0: σ2 = 10002; Ha: σ2 > 10002;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 36.4150

χ2 = (24)(1197.332)10002 = 34.4064 < 36.4150

p = P(X2 ≥ 34.4054) = 0.0777 > 0.05

There is no evidence to suggest that the standarddeviation in outpatient charges for this service isgreater than 1000.

b. 0.05 ≤ p ≤ 0.10p value illustration:

0 22 34.4054 x

9.225 a. H0: σ2 = 0.36; Ha: σ2 > 0.36;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 30.1435

χ2 = (19)(0.42)0.36 = 22.1667 < 30.1435

p = P(X2 ≥ 22.1667) = 0.2760 > 0.05

There is no evidence to suggest the populationvariance is greater than 0.36.

b. p > 0.10

9.226 a. H0: σ2 = 4002; Ha: σ2 > 4002;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 14.0671

χ2 = (7)(571.6782)4002 = 14.2982 ≥ 14.0671

p = P(X2 ≥ 14.2982) = 0.0461 ≤ 0.05

There is evidence to suggest that the standarddeviation in minimum wage is greater than 400euros.

b. 0.025 ≤ p ≤ 0.05

9.227 a. H0: σ2 = 230; Ha: σ2 > 230;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 38.8851

χ2 = (26)(194.0425)230 = 21.9352 < 38.8851

p = P(X2 ≥ 21.9352) = 0.6922 > 0.05

There is no evidence to suggest the variance is lessthan 230.

b. p > 0.10

9.228 a. H0: σ2 = 502; Ha: σ2 > 502;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 62.4281

χ2 = (39)(3245.0128)2500 = 50.6222 < 62.4281

p = P(X2 ≥ 50.6222) = 0.1006 > 0.01

There is no evidence to suggest that the populationvariance in height is greater than the company’sdesired value.

b. p ≥ 0.10

9.229 a. H0: σ2 = 2.752; Ha: σ2 > 2.752;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 32.6706

χ2 = (21)(14.5214)7.5625 = 40.3239 ≥ 32.6706

p = P(X2 ≥ 40.3239) = 0.0068 ≤ 0.05

There is evidence to suggest that the populationvariance in wingspan is greater than 2.752.

b. 0.005 ≤ p ≤ 0.01

9.230 a. H0: σ2 = 4; Ha: σ2 6= 4;

TS: X2 = (n−1)S2

σ20

;

RR: X2 ≤ 5.6972 or X2 ≥ 35.7185

χ2 = (17)(5.23)4 = 22.2275 < 35.7185

p = 2P(X2 ≥ 22.2275) = 0.3523

There is no evidence to suggest that the standarddeviation in the resolution for these goggles isdifferent from 2.

b. p ≥ 0.20

9.231 a. H0: σ2 = 49; Ha: σ2 6= 49;

TS: X2 = (n−1)S2

σ20

;

RR: X2 ≤ 9.5908 or X2 ≥ 34.1696

b. S2 ≤ 23.4975 or S2 ≥ 83.7155


c. 23.4975 < 56 < 83.7155; p = 0.5917 > 0.05.

There is no evidence to suggest the populationvariance in thickness is different from 49.

d. s2 = 15.6 ≤ 23.4975; p = 0.0034 ≤ 0.05

There is evidence to suggest the population variancein thickness is different from 49.

9.232 a. H0: σ2 = 1.56; Ha: σ2 6= 1.56;

TS: Z =S2

−σ20√

2σ20/

√

n−1; RR: |Z| ≥ 1.96

b. z = 2.2−1.56√

2(1.56)/√

47= 1.9888 ≥ 1.96

There is evidence to suggest the population variancein exchange rate is different from 1.56.

c. p = 2P(Z ≥ 1.9888) = 0.0467

d. H0: σ2 = 1.56; Ha: σ2 6= 1.56;

TS: X2 = (n−1)S2

σ20

;

RR: X2 ≤ 29.9562 or X2 ≥ 67.8206

χ2 = (47)(2.2)1.56 = 66.2821

There is no evidence to suggest the populationvariance in exchange rate is different from 1.56.

p = 2P(X2 ≥ 66.2821) = 0.0666

The conclusion and p value are different.

Chapter 9 Exercises

9.233 a. H0: µ = 1.6; Ha: µ 6= 1.6;

TS: Z = x−µ0

σ/√

n; RR: |Z| ≥ 2.5758

z = 1.78−1.60.5/

√

23= 1.7265; |z| < 2.5758

There is no evidence to suggest the population meanis different from 1.6; there is no evidence to suggestthe machine is malfunctioning.

b. p = 2P(Z ≥ 1.7265) = 0.0843

9.234 H0: µ = 4; Ha: µ 6= 4;

TS: Z = (x − µ0)/(σ/√

n); RR: |Z| ≥ 2.5758

a. z = 4.014−40.05/

√

40= 1.7709; |z| < 2.5758

p = 2P(Z ≥ 1.7709) = 0.0766 > 0.01

There is no evidence to suggest the population meanthickness is different from 4. The process should notbe stopped.

b. z = 3.979−40.05/

√

40= −2.6563 ≤ −2.5758

p = 2P(Z ≤ −2.6563) = 0.0079

There is evidence to suggest the population meanthickness is different from 4. The process should bestopped.

9.235 a. H0: µ = 1.0; Ha: µ 6= 1.0;

TS: Z = x−µ0

σ/√

n; RR: |Z| ≥ 1.96

z = 1.0317−1.00.1/

√

18= 1.3435; |z| < 1.96

There is no evidence to suggest the mean weight isdifferent from 1.

b. p = 2P(Z ≥ 1.3435) = 0.1791

9.236 a. H0: µ = 23; Ha: µ > 23;

TS: Z = (x − µ0)/(σ/√

n); RR: Z ≥ 2.3263

z = 24.6−232.28/

√

18= 2.9773 ≥ 2.3263

There is evidence to suggest the population meanwidth is greater than 23 meters.

b. p = P(Z ≥ 2.9773) = 0.0015

9.237 H0: µ = 13; Ha: µ 6= 13;

TS: T = X−µ0

S/√

n; RR: |T | ≥ 2.2622

a. t = 12.89−130.96/

√

10= −0.3623; |t| < 2.2622

p = 2P(T ≤ −0.3623) = 0.7255 > 0.05

There is no evidence to suggest the mean diameteris different from 13 mm. The process should not bestopped.

b. t = 13.04−130.045/

√

10= 2.8109; |t| ≥ 2.2622

p = 2P(T ≥ 2.8109) = 0.0203 ≤ 0.05

There is evidence to suggest the mean diameter isdifferent from 13 mm. The process should bestopped.

c. I would prefer a large significance level. Thiswould mean a smaller type II error, failing to realizethe mean diameter is different from 13, and less of achance of costly engine damage.

9.238 a. H0: µ = 8.0; Ha: µ < 8.0;

TS: T = X−µ0

S/√

n; RR: T ≤ −1.7459

t = 7.3941−8.01.5754/

√

17= −1.5857 > −1.7459

p = P(T ≤ −1.5857) = 0.0662 > 0.05

There is no evidence to suggest that the truepopulation mean time spent with each patient is lessthan 8 minutes.

b. 0.05 ≤ p ≤ 0.10

9.239 a. p0 = 0.20, n = 1500, p̂ = 0.23

b. (1500)(0.20) = 300 ≥ 5; (1500)(0.80) = 1200 ≥ 5

c. H0: p = 0.20; Ha: p > 0.20;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≥ 2.3263

z = 0.23−0.20√(0.20)(0.90)

1500

= 2.9047 ≥ 2.3263

There is evidence to suggest that the proportion ofcompanies that require this information is greaterthan 0.20.

d. p = P(Z ≥ 2.9047) = 0.0018 ≤ 0.01

Chapter 9 Exercises 19

9.240 a. H0: p = 0.75; Ha: p < 0.75;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −2.3263

z = 0.70−0.75√(0.75)(0.25)

560

= −2.7325 ≤ −2.3263

There is evidence to suggest that the trueproportion of residents who favor additional powerto tap phones is less than 0.65.

b. p = P(Z ≤ −2.7325) = 0.0031 ≤ 0.01

9.241 a. H0: p = 0.22; Ha: p 6= 0.22;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: |Z| ≥ 1.96

z = 0.1774−0.22√(0.22)(0.78)

310

= −1.8098; |z| < 1.96

There is no evidence to suggest that the proportionof Americans affected by the sequester has changed.

b. p = 2P(Z ≤ −1.8098) = 0.0703 > 0.05

9.242 H0: p = 0.15; Ha: p 6= 0.15;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: |Z| ≥ 1.96

z = 0.1857−0.15√(0.15)(0.85)

420

= 2.0498; |z| > 1.96

p = 2P(Z ≥ 2.0498) = 0.0404 ≤ 0.05

There is evidence to suggest that the trueproportion of Parkinson’s patients with a familyhistory of the disease is different from 0.15.

9.243 a. H0: p = 0.74; Ha: p < 0.74;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −1.6449

z = 0.6589−0.74√(0.74)(0.26)

129

= −2.0996 ≤ −1.6449

There is evidence to suggest that the proportion ofcrude-oil spills that involve less than five gallons hasdecreased.

b. Rejecting the null hypothesis suggests that morespills are of greater volume.

c. p = P(Z ≤ −2.0996) = 0.0179 ≤ 0.05

9.244 H0: σ2 = 0.0015; Ha: σ2 > 0.0015;

TS: X2 = (n − 1)S2/σ20 ; RR: X2 ≥ 23.6848

χ2 = (14)(0.0026)0.0015 = 24.2667 ≥ 23.6848

p = P(X2 ≥ 24.2667) = 0.0425 ≤ 0.05

There is evidence to suggest the population variancein diameter of viruses has increased.

9.245 H0: σ2 = 0.50; Ha: σ2 < 0.50;

TS: X2 = (n − 1)S2/σ20 ; RR: X2 ≤ 12.4426

χ2 = (20)(0.39)0.50 = 15.6000 > 12.4426

p = P(X2 ≤ 15.6) = 0.2589 > 0.10

There is no evidence to suggest the populationvariance in shrinkage is less than 0.50.

9.246 H0: σ2 = 625002; Ha: σ2 > 625002;

TS: X2 = (n − 1)S2/σ20 ; RR: X2 ≥ 67.9852

χ2 = (36)(65268)2

625002 = 39.2593 < 67.9852

p = P(X2 ≥ 39.2593) = 0.3259 > 0.001

There is no evidence to suggest the populationvariance in blood platelet count has increased.

9.247 a. H0: µ = 58; Ha: µ < 58;

TS: T = X−µ0

S/√

n; RR: T ≤ −1.7109

t = 55.84−585.0964/

√

25= −2.1191 ≤ −1.7109

p = P(T ≤ −2.1191) = 0.0223 ≤ 0.05

There is evidence to suggest that the mean policeresponse time has decreased.

b. 0.01 ≤ p ≤ 0.025

9.248 a. H0: µ = 1800; Ha: µ > 1800;

TS: Z = X−µ0

σ/√

n; RR: Z ≥ 1.6449

z = 1852−1800202/

√

36= 1.5446 < 1.6449

There is no evidence to suggest the population meanamount of ore extracted each day is greater than1800. There is no evidence to suggest the newmachinery has improved production.

b. P(Z ≥ 1.6449) = 0.05

⇒ P(

X−1800202/6 ≥ 1.6449

)= 0.05

⇒ P(X ≥ 1855.3783) = 0.05

β(1875) = P(X ≤ 1855.3783)

= P(Z ≤ −0.5828) = 0.2800

β(1925) = P(X ≤ 1855.3783)

= P(Z ≤ −2.068) = 0.0193

9.249 a. H0: µ = 1; Ha: µ > 1;

TS: T = X−µ0

S/√

n; RR: T ≥ 2.6810

t = 1.5923−10.6006/

√

13= 3.5555 ≥ 2.6810

p = P(T ≥ 3.5555) = 0.0020 ≤ 0.01

There is evidence to suggest that the meansignificant wave height is greater than 1 meter.b. 0.001 ≤ p ≤ 0.005

9.250 a. H0: µ = 40; Ha: µ < 40;

TS: T = X−µ0

S/√

n; RR: T ≤ −2.5083

t = 38.63−405.6/

√

23= −1.1733 > −2.5083

p = P(T ≤ −1.1733) = 0.1266 > 0.01

There is no evidence to suggest the population meanbrightness is less than 40.

b. 0.10 ≤ p ≤ 0.20


9.251 a. H0: µ = 4,500,000; Ha: µ > 4,500,000;

TS: T = X−µ0

S/√

n; RR: T ≥ 2.3060

t = 4,675,250−4,500,000

482,556/√

9= 1.0895 < 2.3060

p = P(T ≥ 1.0895) = 0.1538 > 0.025

There is no evidence to suggest that the meanamount of oil stored is above the safety level.

b. 0.10 ≤ p ≤ 0.20

9.252 a. H0: p = 0.60; Ha: p < 0.60;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −2.3263

z = 0.475−0.60√(0.60)(0.40)

120

= −2.7951 ≤ −2.3263

There is evidence to suggest the populationproportion of cast-iron pans with harmful bacteria isless than 0.60.

b. p = P(Z ≤ −2.7951) = 0.0026

c. This is probably not a random sample. Thepeople who brought their pans for testingself-selected.

9.253 a. H0: p = 0.92; Ha: p < 0.92;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≤ −2.3263

z = 0.9152−0.92√(0.92)(0.08)

5000

= −1.2511 > −2.3263

There is no evidence to suggest that the trueproportion of assisted-living patients who aresatisfied is less than 0.92.

b. p = P(Z ≤ −1.2511) = 0.1055

c. Critical value and p value illustration:

-1.2511-2.3263 0

0.1055

0.01

z

9.254 a. H0: µ = 277; Ha: µ < 277;

TS: T = X−µ0

S/√

n; RR: T ≤ −1.8125

t = 270.436−27710.1303/

√

11= −2.1490 ≤ −1.8125

There is evidence to suggest that the truepopulation mean capacity has decreased.

b. p = P(T ≤ −2.1490) = 0.0286 ≤ 0.05

9.255 a. H0: σ2 = 4; Ha: σ2 > 4;

TS: X2 = (n−1)S2

σ20

; RR: X2 ≥ 26.2962

χ2 = (16)(7.75)4 = 31.00 ≥ 26.2962

p = P(X2 ≥ 31.00) = 0.0135 ≤ 0.05

There is evidence to suggest that the populationvariance is greater than 4 pounds.

b. 0.01 ≤ p ≤ 0.025

9.256 a. H0: p = 0.046; Ha: p > 0.046;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≥ 2.3263

z = 0.0673−0.046√(0.046)(0.954)

550

= 2.3815 ≥ 2.3263

There is evidence to suggest that the trueproportion of small businesses owned by women isgreater than 0.046.

b. p = P(Z ≥ 2.3815) = 0.0086 ≤ 0.01

9.257 H0: p = 0.24; Ha: p > 0.24;

TS: Z = P̂ −p0√p0(1−p0)

n

; RR: Z ≥ 1.6449

z = 0.2704−0.24√(0.24)(0.76)

355

= 1.3421 < 1.6449

p = P(Z ≥ 1.3421) = 0.0898 > 0.05

There is no evidence to suggest that more than 24%of all Liberty Mutual policy holders view padding asacceptable. The insurance company will not raiseautomobile rates in this case.

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