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#### Elementary Statistics Picturing The World 6th Edition By Larson -Test Bank

Ch. 6 Confidence Intervals6.1 Confidence Intervals for the Mean (Large Samples)1 Find a Critical ValueMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.1) Find the critical value zc that corresponds to a 95% confidence level.A) ±1.96 B) ±2.575 C) ±2.33 D) ±1.6452 Find the Margin of ErrorMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.2) Determine the sampling error if the grade point averages for 10 randomly selected students from a class of 125students has a mean of x = 1.8. Assume the grade point average of the 125 students has a mean ofμ = 2.4.A) 0.6 B) 2.1 C) -0.6 D) 1.53) A random sample of 120 students has a test score average with a standard deviation of 9.2. Find the margin oferror if c = 0.98.A) 1.96 B) 0.18 C) 0.84 D) 0.824) A random sample of 150 students has a grade point average with a standard deviation of 0.78. Find the marginof error if c = 0.98.A) 0.15 B) 0.08 C) 0.11 D) 0.125) A random sample of 40 students has a mean annual earnings of $3120 and a standard deviation of $677. Findthe margin of error if c = 0.95.A) $210 B) $77 C) $2891 D) $73 Construct and Interpret Confidence Intervals for the Population MeanMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.6) A random sample of 150 students has a grade point average with a mean of 2.86 and with a standard deviationof 0.78. Construct the confidence interval for the population mean, μ, if c = 0.98.A) (2.71, 3.01) B) (2.51, 3.53) C) (2.43, 3.79) D) (2.31, 3.88)7) A random sample of 40 students has a test score with x = 81.5 and s = 10.2. Construct the confidence intervalfor the population mean, μ if c = 0.90.A) (78.8, 84.2) B) (51.8, 92.3) C) (66.3, 89.1) D) (71.8, 93.5)8) A random sample of 40 students has a mean annual earnings of $3120 and a standard deviation of $677.Construct the confidence interval for the population mean, μ if c = 0.95.A) ($2910, $3330) B) ($210, $110) C) ($4812, $5342) D) ($1987, $2346)9) A random sample of 56 fluorescent light bulbs has a mean life of 645 hours with a standard deviation of 31hours. Construct a 95% confidence interval for the population mean.A) (636.9, 653.1) B) (539.6, 551.2) C) (112.0, 118.9) D) (712.0, 768.0)Page 118Copyright © 2012 Pearson Education, Inc.10) A group of 49 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8.Construct a 98% confidence interval for the population mean.A) (21.1, 23.7) B) (20.3, 24.5) C) (19.8, 25.1) D) (18.8, 26.3)11) A group of 40 bowlers showed that their average score was 192 with a standard deviation of 8. Find the 95%confidence interval of the mean score of all bowlers.A) (189.5, 194.5) B) (186.5, 197.5) C) (188.5, 195.6) D) (187.3, 196.1)12) In a random sample of 60 computers, the mean repair cost was $150 with a standard deviation of $36.Construct a 90% confidence interval for the population mean.A) ($142, $158) B) ($138, $162) C) ($141, $159) D) ($537, $654)SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.13) In a random sample of 60 computers, the mean repair cost was $150 with a standard deviation of $36.a) Construct the 99% confidence interval for the population mean repair cost.b) If the level of confidence was lowered to 95%, what will be the effect on the confidence interval?MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.14) In a recent study of 42 eighth graders, the mean number of hours per week that they watched television was19.6 with a standard deviation of 5.8 hours. Find the 98% confidence interval for the population mean.A) (17.5, 21.7) B) (14.1, 23.2) C) (18.3, 20.9) D) (19.1, 20.4)SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.15) In a recent study of 54 eighth graders, the mean number of hours per week that they watched television was19.5 with a standard deviation of 5.1 hours.a) Find the 98% confidence interval of the mean.b) If the standard deviation is doubled to 10.2, what will be the effect on the confidence interval?MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.16) In a sample of 10 randomly selected women, it was found that their mean height was 63.4 inches. Fromprevious studies, it is assumed that the standard deviation σ is 2.4 and that the population of heightmeasurements is normally distributed. Construct the 95% confidence interval for the population mean.A) (61.9, 64.9) B) (58.1, 67.3) C) (59.7, 66.5) D) (60.8, 65.4)SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.17) In a sample of 10 randomly selected women, it was found that their mean height was 63.4 inches. Fromprevious studies, it is assumed that the standard deviation, σ, is 2.4 inches and that the population of heightmeasurements is normally distributed.a) Construct the 99% confidence interval for the population mean height of women.b) If the sample size was doubled to 20 women, what will be the effect on the confidence interval?Page 119Copyright © 2012 Pearson Education, Inc.18) The numbers of advertisements seen or heard in one week for 30 randomly selected people in the United Statesare listed below. Construct a 95% confidence interval for the true mean number of advertisements.598 494 441 595 728 690 684 486 735 808481 298 135 846 764 317 649 732 582 677734 588 590 540 673 727 545 486 702 70319) The number of wins in a season for 32 randomly selected professional football teams are listed below.Construct a 90% confidence interval for the true mean number of wins in a season.9 9 9 8 10 9 7 211 10 6 4 11 9 8 812 10 7 5 12 6 4 312 9 9 7 10 7 7 54 Determine the Minimum Sample SizeMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.20) The standard IQ test has a mean of 101 and a standard deviation of 16. We want to be 98% certain that we arewithin 4 IQ points of the true mean. Determine the required sample size.A) 87 B) 10 C) 188 D) 121) A nurse at a local hospital is interested in estimating the birth weight of infants. How large a sample must sheselect if she desires to be 99% confident that the true mean is within 2 ounces of the sample mean? Thestandard deviation of the birth weights is known to be 7 ounces.A) 82 B) 81 C) 10 D) 922) In order to set rates, an insurance company is trying to estimate the number of sick days that full time workersat an auto repair shop take per year. A previous study indicated that the standard deviation was 2.8 days. Howlarge a sample must be selected if the company wants to be 95% confident that the true mean differs from thesample mean by no more than 1 day?A) 31 B) 141 C) 512 D) 102423) In order to efficiently bid on a contract, a contractor wants to be 95% confident that his error is less than twohours in estimating the average time it takes to install tile flooring. Previous contracts indicate that thestandard deviation is 4.5 hours. How large a sample must be selected?A) 20 B) 4 C) 5 D) 1924) In order to fairly set flat rates for auto mechanics, a shop foreman needs to estimate the average time it takes toreplace a fuel pump in a car. How large a sample must he select if he wants to be 99% confident that the trueaverage time is within 15 minutes of the sample average? Assume the standard deviation of all times is 30minutes.A) 27 B) 26 C) 6 D) 5Page 120Copyright © 2012 Pearson Education, Inc.SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.25) In order to set rates, an insurance company is trying to estimate the number of sick days that full time workersat an auto repair shop take per year. A previous study indicated that the standard deviation was 2.8 days. a)How large a sample must be selected if the company wants to be 90% confident that the true mean differs fromthe sample mean by no more than 1 day? b) Repeat part (a) using a 95% confidence interval. Which level ofconfidence requires a larger sample size? Explain.5 Determine the Finite Population Correction FactorSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.Provide an appropriate response.26) There were 800 math instructors at a mathematics convention. Forty instructors were randomly selected andgiven an IQ test. The scores produced a mean of 130 with a standard deviation of 10. Find a 95% confidenceinterval for the mean of the 800 instructors. Use the finite population correction factor.6 ConceptsSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.Provide an appropriate response.27) A random sample of 200 high school seniors is given the SAT-V test. The mean score for this sample is x = 493.What can you say about the mean score μ of all high school seniors?28) The grade point averages for 10 randomly selected students in a statistics class with 125 students are listedbelow. What can you say about the mean score μ of all 125 students?2.1 3.1 2.0 3.9 3.5 3.7 2.8 1.9 2.5 2.229) A certain confidence in interval is 9.75 < μ < 11.05. Find the sample mean x and the error of estimate E.30) Given the same sample statistics, which level of confidence will produce the narrowest confidence interval:75%, 85%, 90%, or 95%? Explain your reasoning.31) The grade point averages for 10 randomly selected students in a statistics class with 125 students are listedbelow.2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8What is the effect on the width of the confidence interval if the sample size is increased to 20? Explain yourreasoning.6.2 Confidence Intervals for the Mean (Small Samples)1 Find a Critical ValueMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.1) Find the critical value, tc for c = 0.99 and n = 10.A) 3.250 B) 3.169 C) 2.262 D) 1.8332) Find the critical value, tc, for c = 0.95 and n = 16.A) 2.131 B) 2.120 C) 2.602 D) 2.947Page 121Copyright © 2012 Pearson Education, Inc.3) Find the critical value, tc, for c = 0.90 and n = 15.A) 1.761 B) 1.753 C) 2.145 D) 2.6242 Construct and Interpret Confidence Intervals for the Population MeanMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.4) Find the value of E, the margin of error, for c = 0.90, n = 16 and s = 2.5.A) 1.1 B) 0.27 C) 0.84 D) 0.215) Find the value of E, the margin of error, for c = 0.99, n = 10 and s = 3.2.A) 3.29 B) 2.85 C) 1.04 D) 3.216) Find the value of E, the margin of error, for c = 0.95, n = 15 and s = 5.2.A) 2.88 B) 2.96 C) 2.36 D) 0.747) In a random sample of 28 families, the average weekly food expense was $95.60 with a standard deviation of$22.50. Determine whether a normal distribution or a t-distribution should be used or whether neither of thesecan be used to construct a confidence interval. Assume the distribution of weekly food expenses is normallyshaped.A) Use the t-distribution.B) Use normal distribution.C) Cannot use normal distribution or t-distribution.8) For a sample of 20 IQ scores the mean score is 105.8. The standard deviation, σ, is 15. Determine whether anormal distribution or a t-distribution should be used or whether neither of these can be used to construct aconfidence interval. Assume that IQ scores are normally distributed.A) Use normal distribution.B) Use the t-distribution.C) Cannot use normal distribution or t-distribution.9) A random sample of 40 college students has a mean earnings of $3120 with a standard deviation of $677 overthe summer months. Determine whether a normal distribution or a t-distribution should be used or whetherneither of these can be used to construct a confidence interval.A) Use normal distribution.B) Use the t-distribution.C) Cannot use normal distribution or t-distribution.10) A random sample of 15 statistics textbooks has a mean price of $105 with a standard deviation of $30.25.Determine whether a normal distribution or a t-distribution should be used or whether neither of these can beused to construct a confidence interval. Assume the distribution of statistics textbook prices is not normallydistributed.A) Cannot use normal distribution or t-distribution.B) Use normal distribution.C) Use the t-distribution.11) Construct a 95% confidence interval for the population mean, μ. Assume the population has a normaldistribution. A sample of 20 college students had mean annual earnings of $3120 with a standard deviation of$677.A) ($2803, $3437) B) ($1324, $1567) C) ($2135, $2567) D) ($2657, $2891)Page 122Copyright © 2012 Pearson Education, Inc.12) Construct a 90% confidence interval for the population mean, μ. Assume the population has a normaldistribution. A sample of 15 randomly selected students has a grade point average of 2.86 with a standarddeviation of 0.78.A) (2.51, 3.21) B) (2.41, 3.42) C) (2.37, 3.56) D) (2.28, 3.66)13) Construct a 95% confidence interval for the population mean, μ. Assume the population has a normaldistribution. A sample of 25 randomly selected students has a mean test score of 81.5 with a standard deviationof 10.2.A) (77.29, 85.71) B) (56.12, 78.34) C) (66.35, 69.89) D) (87.12, 98.32)SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.14) Construct a 98% confidence interval for the population mean, μ. Assume the population has a normaldistribution. A random sample of 20 college students has mean annual earnings of $3180 with a standarddeviation of $673.MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.15) Construct a 95% confidence interval for the population mean, μ. Assume the population has a normaldistribution. A random sample of 16 fluorescent light bulbs has a mean life of 645 hours with a standarddeviation of 31 hours.A) (628.5, 661.5) B) (876.2, 981.5) C) (531.2, 612.9) D) (321.7, 365.8)16) Construct a 99% confidence interval for the population mean, μ. Assume the population has a normaldistribution. A group of 19 randomly selected students has a mean age of 22.4 years with a standard deviationof 3.8 years.A) (19.9, 24.9) B) (16.3, 26.9) C) (17.2, 23.6) D) (18.7, 24.1)17) Construct a 98% confidence interval for the population mean, μ. Assume the population has a normaldistribution. A study of 14 bowlers showed that their average score was 192 with a standard deviation of 8.A) (186.3, 197.7) B) (222.3, 256.1) C) (328.3, 386.9) D) (115.4, 158.8)SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.18) Construct a 95% confidence interval for the population mean, μ. Assume the population has a normaldistribution. In a random sample of 26 computers, the mean repair cost was $149 with a standard deviation of$39.MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.19) Construct a 90% confidence interval for the population mean, μ. Assume the population has a normaldistribution. In a recent study of 22 eighth graders, the mean number of hours per week that they watchedtelevision was 19.6 with a standard deviation of 5.8 hours.A) (17.47, 21.73) B) (18.63, 20.89) C) (5.87, 7.98) D) (19.62, 23.12)SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.20) a) Construct a 95% confidence interval for the population mean, μ. Assume the population has a normaldistribution. In a random sample of 26 computers, the mean repair cost was $129 with a standard deviation of$37.b) Suppose you did some research on repair costs for computers and found that the standard deviation isσ = 37. Use the normal distribution to construct a 95% confidence interval for the population mean, μ. Comparethe results.Page 123Copyright © 2012 Pearson Education, Inc.MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.21) A random sample of 10 parking meters in a beach community showed the following incomes for a day.Assume the incomes are normally distributed.$3.60 $4.50 $2.80 $6.30 $2.60 $5.20 $6.75 $4.25 $8.00 $3.00Find the 95% confidence interval for the true mean.A) ($3.39, $6.01) B) ($2.11, $5.34) C) ($4.81, $6.31) D) ($1.35, $2.85)22) The grade point averages for 10 randomly selected high school students are listed below. Assume the gradepoint averages are normally distributed.2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8Find a 98% confidence interval for the true mean.A) (1.55, 3.53) B) (0.67, 1.81) C) (2.12, 3.14) D) (3.11, 4.35)23) A local bank needs information concerning the checking account balances of its customers. A random sampleof 15 accounts was checked. The mean balance was $686.75 with a standard deviation of $256.20. Find a 98%confidence interval for the true mean. Assume that the account balances are normally distributed.A) ($513.17, $860.33) B) ($238.23, $326.41) C) ($326.21, $437.90) D) ($487.31, $563.80)SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.24) A manufacturer receives an order for fluorescent light bulbs. The order requires that the bulbs have a mean lifespan of 500 hours. The manufacturer selects a random sample of 25 fluorescent light bulbs and finds that theyhave a mean life span of 495 hours with a standard deviation of 15 hours. Test to see if the manufacturer ismaking acceptable light bulbs. Use a 95% confidence level. Assume the data are normally distributed.25) A coffee machine is supposed to dispense 12 ounces of coffee in each cup. An inspector selects a randomsample of 40 cups of coffee and finds they have an average amount of 12.2 ounces with a standard deviation of0.3 ounce. Use a 99% confidence interval to test whether the machine is dispensing acceptable amounts ofcoffee.6.3 Confidence Intervals for Population Proportions1 Find a Point EstimateMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.1) When 435 college students were surveyed,120 said they own their car. Find a point estimate for p, thepopulation proportion of students who own their cars.A) 0.276 B) 0.724 C) 0.381 D) 0.2162) A survey of 100 fatal accidents showed that 12 were alcohol related. Find a point estimate for p, the populationproportion of accidents that were alcohol related.A) 0.12 B) 0.88 C) 0.136 D) 0.1073) A survey of 400 non-fatal accidents showed that 189 involved the use of a cell phone. Find a point estimate forp, the population proportion of non-fatal accidents that involved the use of a cell phone.A) 0.472 B) 0.527 C) 0.896 D) 0.321Page 124Copyright © 2012 Pearson Education, Inc.4) A survey of 250 homeless persons showed that 17 were veterans. Find a point estimate p, for the populationproportion of homeless persons who are veterans.A) 0.068 B) 0.932 C) 0.073 D) 0.0645) A survey of 2650 golfers showed that 379 of them are left-handed. Find a point estimate for p, the populationproportion of golfers that are left-handed.A) 0.143 B) 0.857 C) 0.167 D) 0.1252 Construct and Interpret Confidence Intervals for the Population ProportionMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.6) In a survey of 2480 golfers, 15% said they were left-handed. The surveyʹs margin of error was 3%. Construct aconfidence interval for the proportion of left-handed golfers.A) (0.12, 0.18) B) (0.18, 0.21) C) (0.12, 0.15) D) (0.11, 0.19)SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.7) The Federal Bureau of Labor Statistics surveyed 50,000 people and found the unemployment rate to be 5.8%.The margin of error was 0.2%. Construct a confidence interval for the unemployment rate.8) When 495 college students were surveyed, 150 said they own their car. Construct a 95% confidence interval forthe proportion of college students who say they own their cars.9) A survey of 300 fatal accidents showed that 123 were alcohol related. Construct a 98% confidence interval forthe proportion of fatal accidents that were alcohol related.10) A survey of 400 non-fatal accidents showed that 197 involved the use of a cell phone. Construct a 99%confidence interval for the proportion of fatal accidents that involved the use of a cell phone.MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.11) A survey of 280 homeless persons showed that 63 were veterans. Construct a 90% confidence interval for theproportion of homeless persons who are veterans.A) (0.184, 0.266) B) (0.176, 0.274) C) (0.167, 0.283) D) (0.161, 0.289)12) A survey of 2450 golfers showed that 281 of them are left-handed. Construct a 98% confidence interval for theproportion of golfers that are left-handed.A) (0.100, 0.130) B) (0.203, 0.293) C) (0.369, 0.451) D) (0.683, 0.712)SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.13) In a survey of 10 golfers, 2 were found to be left-handed. Is it practical to construct the 90% confidence intervalfor the population proportion, p? Explain.14) The USA Today claims that 44% of adults who access the Internet read the international news online. Youwant to check the accuracy of their claim by surveying a random sample of 120 adults who access the Internetand asking them if they read the international news online. Fifty-two adults responded ʺyes.ʺ Use a 95%confidence interval to test the newspaperʹs claim.Page 125Copyright © 2012 Pearson Education, Inc.3 Determine the Minimum Sample SizeMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.15) A researcher at a major hospital wishes to estimate the proportion of the adult population of the United Statesthat has high blood pressure. How large a sample is needed in order to be 95% confident that the sampleproportion will not differ from the true proportion by more than 4%?A) 601 B) 13 C) 1201 D) 42316) A pollster wishes to estimate the proportion of United States voters who favor capital punishment. How large asample is needed in order to be 95% confident that the sample proportion will not differ from the trueproportion by more than 5%?A) 385 B) 271 C) 10 D) 76917) A private opinion poll is conducted for a politician to determine what proportion of the population favorsdecriminalizing marijuana possession. How large a sample is needed in order to be 95% confident that thesample proportion will not differ from the true proportion by more than 2%?A) 2401 B) 1692 C) 4802 D) 2518) A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. How large a sample isneeded in order to be 98% confident that the sample proportion will not differ from the true proportion bymore than 2%? A previous study indicates that the proportion of left-handed golfers is 8%.A) 999 B) 707 C) 1086 D) 1719) A researcher wishes to estimate the number of households with two cars. How large a sample is needed inorder to be 98% confident that the sample proportion will not differ from the true proportion by more than 5%?A previous study indicates that the proportion of households with two cars is 19%.A) 335 B) 237 C) 413 D) 8SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.20) A state highway patrol official wishes to estimate the number of drivers that exceed the speed limit traveling acertain road.a) How large a sample is needed in order to be 95% confident that the sample proportion will not differ fromthe true proportion by more than 2%?b) Repeat part (a) assuming previous studies found that 70% of drivers on this road exceeded the speed limit.6.4 Confidence Intervals for Variance and Standard Deviation1 Find Critical ValuesMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.1) Find the critical values, X 2R andX 2L , for c = 0.95 and n = 12.A) 3.816 and 21.920 B) 3.053 and 24.725 C) 4.575 and 26.757 D) 2.603 and 19.6752) Find the critical values, X 2R andX 2L , for c = 0.90 and n = 15.A) 6.571 and 23.685 B) 4.075 and 31.319 C) 4.660 and 29.131 D) 5.629 and 26.119Page 126Copyright © 2012 Pearson Education, Inc.3) Find the critical values, X 2R andX 2L , for c = 0.98 and n = 20.A) 7.633 and 36.191 B) 6.844 and 27.204 C) 8.907 and 38.582 D) 10.117 and 32.8524) Find the critical values, X 2R andX 2L , for c = 0.99 and n = 10.A) 1.735 and 23.587 B) 2.156 and 25.188 C) 2.088 and 21.666 D) 2.558 and 23.2092 Construct Confidence Intervals for the Population Variance and Population Standard DeviationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Assume the sample is taken from a normally distributed population and construct the indicated confidence interval.5) Construct a 95% confidence interval for the population standard deviation σ of a random sample of 15 menwho have a mean weight of 165.2 pounds with a standard deviation of 13.5 pounds. Assume the population isnormally distributed.A) (9.9, 21.3) B) (97.7, 453.3) C) (2.7, 5.8) D) (10.4, 19.7)6) A random sample of 16 men have a mean height of 67.5 inches and a standard deviation of 1.5 inches.Construct a 99% confidence interval for the population standard deviation, σ.A) (1.0, 2.7) B) (1.0, 2.8) C) (0.8, 2.2) D) (1.1, 2.5)7) A random sample of 20 women have a mean height of 62.5 inches and a standard deviation of 3.2 inches.Construct a 98% confidence interval for the population variance, σ2.A) (5.4, 25.5) B) (2.3, 5.0) C) (1.7, 8.0) D) (5.7, 26.8)8) The heights (in inches) of 20 randomly selected adult males are listed below. Construct a 99% confidenceinterval for the variance, σ2.70 72 71 70 69 73 69 68 70 7167 71 70 74 69 68 71 71 71 72A) (1.47, 8.27) B) (21.61, 69.06) C) (1.35, 8.43) D) (2.16, 71.06)9) The grade point averages for 10 randomly selected students are listed below. Construct a 90% confidenceinterval for the population standard deviation, σ.2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8A) (0.81, 1.83) B) (0.32, 0.85) C) (0.53, 1.01) D) (1.10, 2.01)10) The mean replacement time for a random sample of 12 microwave ovens is 8.6 years with a standard deviationof 4.2 years. Construct the 98% confidence interval for the population variance, σ2.A) (7.8, 63.6) B) (2.8, 8.0) C) (1.9, 15.1) D) (7.4, 54.3)11) A student randomly selects 10 CDs at a store. The mean is $13.75 with a standard deviation of $1.50. Constructa 95% confidence interval for the population standard deviation, σ.A) ($1.03, $2.74) B) ($0.99, $2.50) C) ($1.06, $7.51) D) ($0.84, $2.24)Page 127Copyright © 2012 Pearson Education, Inc.12) The stem-and-leaf plot shows the test scores of 16 randomly selected students. Construct a 99% confidenceinterval for the population standard deviation.5678995 8 37 4 4 2 95 8 3 53 1 7A) (7.61, 20.33) B) (57.97, 413.27) C) (62.18, 363.63) D) (7.89, 19.07)13) The dotplot shows the weights (in pounds) of 15 dogs selected randomly from those adopted out by an animalshelter last week. Construct a 98% confidence interval for the population variance.Weights25 30 35PoundsA) (3.03, 18.97) B) (1.74, 4.36) C) (3.38, 15.70) D) (2.89, 16.91)SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.14) A container of car oil is supposed to contain 1000 milliliters of oil. A quality control manager wants to be surethat the standard deviation of the oil containers is less than 20 milliliters. He randomly selects 10 cans of oilwith a mean of 997 milliliters and a standard deviation of 32 milliliters. Use these sample results to construct a95% confidence interval for the true value of σ. Does this confidence interval suggest that the variation in the oilcontainers is at an acceptable level?Page 128Copyright © 2012 Pearson Education, Inc.Ch. 6 Confidence IntervalsAnswer Key6.1 Confidence Intervals for the Mean (Large Samples)1 Find a Critical Value1) A2 Find the Margin of Error2) A3) A4) A5) A3 Construct and Interpret Confidence Intervals for the Population Mean6) A7) A8) A9) A10) A11) A12) A13) a) ($138, $162)b) A decrease in the level of confidence will decrease the width of the confidence interval.14) A15) a) (17.9, 21.1)b) An increase in the standard deviation will widen the confidence interval.16) A17) a) (61.4, 65.4)b) An increase in the sample size will decrease the width of the confidence interval.18) (543.8, 658. 0)19) (7.2, 8.8)4 Determine the Minimum Sample Size20) A21) A22) A23) A24) A25) a) 22b) 31; A 95% confidence interval requires a larger sample than a 90% confidence interval because more information isneeded from the population to be 95% confident.5 Determine the Finite Population Correction Factor26) (127.0, 133.0)6 Concepts27) The sample mean x = 493 is the best estimator of the unknown population mean μ.28) The sample mean x = 2.77 is the best point estimate of the unknown population mean μ.29) Sample mean x = 10.40 and the error of estimate E = 0.65.30) The 75% level of confidence will produce the narrowest confidence interval. As the level of confidence decreases, zcdecreases, causing narrower intervals.31) The width of the interval will decrease. As n increases, E decreases because n is in the denominator of the formulafor E. So the intervals become narrower.6.2 Confidence Intervals for the Mean (Small Samples)1 Find a Critical Value1) A2) APage 129Copyright © 2012 Pearson Education, Inc.3) A2 Construct and Interpret Confidence Intervals for the Population Mean4) A5) A6) A7) A8) A9) A10) A11) A12) A13) A14) ($2798, $3562)15) A16) A17) A18) ($133.24, $164.76)19) A20) a) ($114.05, $143.95)b) ($114.78, $143.22); The t-confidence interval is wider.21) A22) A23) A24) (488.81, 501.19). Because the interval contains the desired life span of 500 hours, they are making good light bulbs.25) (12.1, 12.3) Because the interval does not contain the desired amount of 12 ounces, the machine is not workingproperly.6.3 Confidence Intervals for Population Proportions1 Find a Point Estimate1) A2) A3) A4) A5) A2 Construct and Interpret Confidence Intervals for the Population Proportion6) A7) (0.058, 0.060)8) (0.263, 0.344)9) (0.344, 0.476)10) (0.428, 0.557)11) A12) A13) It is not practical to find the confidence interval. It is necessary that np^ > 5 to insure that thedistribution of p ^ be normal. (np ^= 2)14) (0.345, 0.522) Because the interval contains the reported percentage of 44%, the newspaperʹs claim is accurate.3 Determine the Minimum Sample Size15) A16) A17) A18) A19) A20) a) 2401b) 2017Page 130Copyright © 2012 Pearson Education, Inc.6.4 Confidence Intervals for Variance and Standard Deviation1 Find Critical Values1) A2) A3) A4) A2 Construct Confidence Intervals for the Population Variance and Population Standard Deviation5) A6) A7) A8) A9) A10) A11) A12) A13) A14) The 95% confidence interval is (22.01, 58.42). Because this interval does not contain 20, the data suggest that thestandard deviation is not at an acceptable level.Page 131Copyright © 2012 Pearson Education, Inc. Ch. 7 Hypothesis Testing with One Sample7.1 Introduction to Hypothesis Testing1 State a Null and Alternative HypothesisSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.Provide an appropriate response.1) The mean age of bus drivers in Chicago is 48.5 years. Write the null and alternative hypotheses.2) The mean IQ of statistics teachers is greater than 110. Write the null and alternative hypotheses.3) The mean score for all NBA games during a particular season was less than 101 points per game. Write the nulland alternative hypotheses.4) A candidate for governor of a particular state claims to be favored by at least half of the voters. Write the nulland alternative hypotheses.5) The dean of a major university claims that the mean time for students to earn a Masterʹs degree is at most 4.9years. Write the null and alternative hypotheses.6) The buyer of a local hiking club store recommends against buying the new digital altimeters because they varymore than the old altimeters, which had a standard deviation of one yard. Write the null and alternativehypotheses.7) The mean age of bus drivers in Chicago is 53.7 years. State this claim mathematically. Write the null andalternative hypotheses. Identify which hypothesis is the claim.8) The mean IQ of statistics teachers is greater than 120. State this claim mathematically. Write the null andalternative hypotheses. Identify which hypothesis is the claim.9) The mean score for all NBA games during a particular season was less than 101 points per game. State thisclaim mathematically. Write the null and alternative hypotheses. Identify which hypothesis is the claim.10) A candidate for governor of a particular state claims to be favored by at least half of the voters. State this claimmathematically. Write the null and alternative hypotheses. Identify which hypothesis is the claim.11) The dean of a major university claims that the mean time for students to earn a Masterʹs degree is at most 3.3years. State this claim mathematically. Write the null and alternative hypotheses. Identify which hypothesis isthe claim.12) The buyer of a local hiking club store recommends against buying the new digital altimeters because they varymore than the old altimeters, which had a standard deviation of one yard. State this claim mathematically.Write the null and alternative hypotheses. Identify which hypothesis is the claim.2 Identify Whether to Use a One-tailed or Two-tailed TestMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.13) Given H0: p ≥ 80% and Ha: p < 80%, determine whether the hypothesis test is left-tailed, right-tailed, ortwo-tailed.A) left-tailed B) right-tailed C) two-tailedPage 132Copyright © 2012 Pearson Education, Inc.14) Given H0: μ ≤ 25 and Ha: μ > 25, determine whether the hypothesis test is left-tailed, right-tailed, ortwo-tailed.A) right-tailed B) left-tailed C) two-tailed15) A researcher claims that 62% of voters favor gun control. Determine whether the hypothesis test for this claimis left-tailed, right-tailed, or two-tailed.A) two-tailed B) left-tailed C) right-tailed16) A brewery claims that the mean amount of beer in their bottles is at least 12 ounces. Determine whether thehypothesis test for this claim is left-tailed, right-tailed, or two-tailed.A) left-tailed B) right-tailed C) two-tailed17) A car maker claims that its new sub-compact car gets better than 47 miles per gallon on the highway.Determine whether the hypothesis test for this is left-tailed, right-tailed, or two-tailed.A) right-tailed B) left-tailed C) two-tailed18) The owner of a professional basketball team claims that the mean attendance at games is over 30,000 andtherefore the team needs a new arena. Determine whether the hypothesis test for this claim is left-tailed,right-tailed, or two-tailed.A) right-tailed B) left-tailed C) two-tailed19) An elementary school claims that the standard deviation in reading scores of its fourth grade students is lessthan 4.35. Determine whether the hypothesis test for this claim is left-tailed, right-tailed, or two-tailed.A) left-tailed B) right-tailed C) two-tailed3 Identify Type I and Type II ErrorsSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.Provide an appropriate response.20) The mean age of bus drivers in Chicago is 52.5 years. Identify the type I and type II errors for the hypothesistest of this claim.21) The mean IQ of statistics teachers is greater than 120. Identify the type I and type II errors for the hypothesistest of this claim.22) The mean score for all NBA games during a particular season was less than 105 points per game. Identify thetype I and type II errors for the hypothesis test of this claim.23) A candidate for governor of a certain state claims to be favored by at least half of the voters. Identify the type Iand type II errors for the hypothesis test of this claim.4 Interpret a Decision Based on the Results of a Statistical TestMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.24) The mean age of bus drivers in Chicago is 50.2 years. If a hypothesis test is performed, how should youinterpret a decision that rejects the null hypothesis?A) There is sufficient evidence to reject the claim μ = 50.2.B) There is not sufficient evidence to reject the claim μ = 50.2.C) There is sufficient evidence to support the claim μ = 50.2.D) There is not sufficient evidence to support the claim μ = 50.2.Page 133Copyright © 2012 Pearson Education, Inc.25) The mean age of bus drivers in Chicago is 59.3 years. If a hypothesis test is performed, how should youinterpret a decision that fails to reject the null hypothesis?A) There is not sufficient evidence to reject the claim μ = 59.3.B) There is sufficient evidence to reject the claim μ = 59.3.C) There is sufficient evidence to support the claim μ = 59.3.D) There is not sufficient evidence to support the claim μ = 59.3.26) The mean age of bus drivers in Chicago is greater than 57.8 years. If a hypothesis test is performed, how shouldyou interpret a decision that rejects the null hypothesis?A) There is sufficient evidence to support the claim μ > 57.8.B) There is sufficient evidence to reject the claim μ > 57.8.C) There is not sufficient evidence to reject the claim μ > 57.8.D) There is not sufficient evidence to support the claim μ > 57.8.27) The mean age of bus drivers in Chicago is greater than 47.6 years. If a hypothesis test is performed, how shouldyou interpret a decision that fails to reject the null hypothesis?A) There is not sufficient evidence to support the claim μ > 47.6.B) There is sufficient evidence to reject the claim μ > 47.6.C) There is not sufficient evidence to reject the claim μ > 47.6.D) There is sufficient evidence to support the claim μ > 47.6.28) The mean IQ of statistics teachers is greater than 160. If a hypothesis test is performed, how should youinterpret a decision that rejects the null hypothesis?A) There is sufficient evidence to support the claim μ > 160.B) There is sufficient evidence to reject the claim μ > 160.C) There is not sufficient evidence to reject the claim μ > 160.D) There is not sufficient evidence to support the claim μ > 160.29) The mean IQ of statistics teachers is greater than 150. If a hypothesis test is performed, how should youinterpret a decision that fails to reject the null hypothesis?A) There is not sufficient evidence to support the claim μ > 150.B) There is sufficient evidence to reject the claim μ > 150.C) There is not sufficient evidence to reject the claim μ > 150.D) There is sufficient evidence to support the claim μ > 150.30) The mean score for all NBA games during a particular season was less than 92 points per game. If a hypothesistest is performed, how should you interpret a decision that rejects the null hypothesis?A) There is sufficient evidence to support the claim μ < 92.B) There is sufficient evidence to reject the claim μ < 92.C) There is not sufficient evidence to reject the claim μ < 92.D) There is not sufficient evidence to support the claim μ < 92.31) The mean score for all NBA games during a particular season was less than 100 points per game. If ahypothesis test is performed, how should you interpret a decision that fails to reject the null hypothesis?A) There is not sufficient evidence to support the claim μ < 100.B) There is sufficient evidence to reject the claim μ < 100.C) There is not sufficient evidence to reject the claim μ < 100.D) There is sufficient evidence to support the claim μ < 100.Page 134Copyright © 2012 Pearson Education, Inc.32) A candidate for governor of a certain state claims to be favored by at least half of the voters. If a hypothesis testis performed, how should you interpret a decision that rejects the null hypothesis?A) There is sufficient evidence to reject the claim ρ ≥ 0.5.B) There is not sufficient evidence to reject the claim ρ ≥ 0.5.C) There is sufficient evidence to support the claim ρ ≥ 0.5.D) There is not sufficient evidence to support the claim ρ ≥ 0.5.33) A candidate for governor of a certain state claims to be favored by at least half of the voters. If a hypothesis testis performed, how should you interpret a decision that fails to reject the null hypothesis?A) There is not sufficient evidence to reject the claim ρ ≥ 0.5.B) There is sufficient evidence to reject the claim ρ ≥ 0.5.C) There is sufficient evidence to support the claim ρ ≥ 0.5.D) There is not sufficient evidence to support the claim ρ ≥ 0.5.34) The dean of a major university claims that the mean time for students to earn a Masterʹs degree is at most 5.8years. If a hypothesis test is performed, how should you interpret a decision that rejects the null hypothesis?A) There is sufficient evidence to reject the claim μ ≤ 5.8.B) There is not sufficient evidence to reject the claim μ ≤ 5.8.C) There is sufficient evidence to support the claim μ ≤ 5.8.D) There is not sufficient evidence to support the claim μ ≤ 5.8.35) The dean of a major university claims that the mean time for students to earn a Masterʹs degree is at most 5.1years. If a hypothesis test is performed, how should you interpret a decision that fails to reject the nullhypothesis?A) There is not sufficient evidence to reject the claim μ ≤ 5.1.B) There is sufficient evidence to reject the claim μ ≤ 5.1.C) There is sufficient evidence to support the claim μ ≤ 5.1.D) There is not sufficient evidence to support the claim μ ≤ 5.1.5 Use Confidence Intervals to Make DecisionsMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.36) Given H0: μ ≤ 12, for which confidence interval should you reject H0?A) (13, 16) B) (11.5, 12.5) C) (10, 13)37) Given H0: p ≥ 0.45, for which confidence interval should you reject H0?A) (0.32, 0.40) B) (0.40, 0.50) C) (0.42, 0.47)6 ConceptsMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.38) Given H0: p = 0.85 and α = 0.10, which level of confidence should you use to test the claim?A) 90% B) 95% C) 99% D) 80%39) Given H0: μ ≥ 23.5 and α = 0.05, which level of confidence should you use to test the claim?A) 90% B) 99% C) 80% D) 95%Page 135Copyright © 2012 Pearson Education, Inc.7.2 Hypothesis Testing for the Mean (Large Samples)1 Find P-valuesMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.1) Suppose you are using α = 0.05 to test the claim that μ > 14 using a P-value. You are given the sample statisticsn = 50, x = 14.3, and s = 1.2. Find the P-value.A) 0.0384 B) 0.1321 C) 0.0128 D) 0.00122) Suppose you are using α = 0.05 to test the claim that μ ≠ 14 using a P-value. You are given the sample statisticsn = 35, x = 13.1, and s = 2.7. Find the P-value.A) 0.0488 B) 0.0591 C) 0.1003 D) 0.02443) Suppose you are using α = 0.01 to test the claim that μ ≤ 32 using a P-value. You are given the sample statisticsn = 40, x = 33.8, and s = 4.3. Find the P-value.A) 0.0040 B) 0.9960 C) 0.0211 D) 0.10304) Suppose you are using α = 0.01 to test the claim that μ = 1620 using a P-value. You are given the samplestatistics n = 35, x = 1590, and s = 82. Find the P-value.A) 0.0308 B) 0.0154 C) 0.3169 D) 0.00772 Test a Claim About a Mean Using P-valuesMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.5) Given H0: μ = 25, Ha: μ ≠ 25, and P = 0.034. Do you reject or fail to reject H0 at the 0.01 level of significance?A) fail to reject H0B) reject H0C) not sufficient information to decide6) Given H0: μ ≥ 18 and P = 0.070. Do you reject or fail to reject H0 at the 0.05 level of significance?A) fail to reject H0B) reject H0C) not sufficient information to decide7) Given Ha: μ > 85 and P = 0.007. Do you reject or fail to reject H0 at the 0.01 level of significance?A) reject H0B) fail to reject H0C) not sufficient information to decideSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.8) A fast food outlet claims that the mean waiting time in line is less than 3.4 minutes. A random sample of 60customers has a mean of 3.3 minutes with a standard deviation of 0.6 minute. If α = 0.05, test the fast foodoutletʹs claim.Page 136Copyright © 2012 Pearson Education, Inc.9) A local school district claims that the number of school days missed by its teachers due to illness is below thenational average of 5. A random sample of 40 teachers provided the data below. At α = 0.05, test the districtʹsclaim using P-values.0 3 6 3 3 5 4 1 3 57 3 1 2 3 3 2 4 1 62 5 2 8 3 1 2 5 4 11 1 2 1 5 7 5 4 9 33 Find Critical ValuesMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.10) Find the critical value for a right-tailed test with α = 0.01 and n = 75.A) 2.33 B) 2.575 C) 1.645 D) 1.9611) Find the critical value for a two-tailed test with α = 0.01 and n = 30.A) ±2.575 B) ±2.33 C) ±1.645 D) ±1.9612) Find the critical value for a left-tailed test with α = 0.05 and n = 48.A) -1.645 B) -2.33 C) -2.575 D) -1.9613) Find the critical value for a two-tailed test with α = 0.10 and n = 100.A) ±1.645 B) ±2.33 C) ±2.575 D) ±1.9614) Find the critical value for a left-tailed test with α = 0.025 and n = 50.A) -1.96 B) -2.33 C) -2.575 D) -1.64515) Find the critical value for a two-tailed test with α = 0.07 and n = 36.A) ±1.81 B) ±2.33 C) ±2.575 D) ±1.964 Test a Claim About a MeanMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.16) You wish to test the claim that μ > 32 at a level of significance of α = 0.05 and are given sample statistics n = 50,x = 32.3, and s = 1.2. Compute the value of the standardized test statistic. Round your answer to two decimalplaces.A) 1.77 B) 2.31 C) 0.98 D) 3.1117) You wish to test the claim that μ ≠ 22 at a level of significance of α = 0.05 and are given sample statistics n = 35,x = 21.1, and s = 2.7. Compute the value of the standardized test statistic. Round your answer to two decimalplaces.A) -1.97 B) -3.12 C) -2.86 D) -1.8318) You wish to test the claim that μ ≤ 38 at a level of significance of α = 0.01 and are given sample statistics n = 40,x = 39.8, and s = 4.3. Compute the value of the standardized test statistic. Round your answer to two decimalplaces.A) 2.65 B) 3.51 C) 2.12 D) 1.96Page 137Copyright © 2012 Pearson Education, Inc.19) You wish to test the claim that μ = 1240 at a level of significance of α = 0.01 and are given sample statisticsn = 35, x = 1210 and s = 82. Compute the value of the standardized test statistic. Round your answer to twodecimal places.A) -2.16 B) -3.82 C) -4.67 D) -5.1820) Suppose you want to test the claim that μ ≠ 3.5. Given a sample size of n = 47 and a level of significance ofα = 0.10, when should you reject H0 ?A) Reject H0 if the standardized test statistic is greater than 1.645 or less than -1.645.B) Reject H0 if the standardized test statistic is greater than 2.575 or less than -2.575.C) Reject H0 if the standardized test statistic is greater than 2.33 or less than -2.33D) Reject H0 if the standardized test statistic is greater than 1.96 or less than -1.9621) Suppose you want to test the claim that μ ≤ 25.6. Given a sample size of n = 53 and a level of significance ofα = 0.01, when should you reject H0?A) Reject H0 if the standardized test statistic is greater than 2.33.B) Reject H0 if the standardized test statistic is greater than 1.96.C) Reject H0 if the standardized test statistic is greater than 1.645.D) Reject H0 if the standardized test statistic is greater than 2.575.22) Suppose you want to test the claim that μ ≥ 65.4. Given a sample size of n = 35 and a level of significance ofα = 0.05, when should you reject H0?A) Reject H0 if the standardized test statistic is less than -1.645.B) Reject H0 if the standardized test is less than -1.96.C) Reject H0 if the standardized test statistic is less than -2.575.D) Reject H0 if the standardized test statistic is less than -1.28.SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.23) Test the claim that μ > 19, given that α = 0.05 and the sample statistics are n = 50, x = 19.3, and s = 1.2.24) Test the claim that μ ≠ 38, given that α = 0.05 and the sample statistics are n = 35, x = 37.1 and s = 2.7.25) Test the claim that μ ≤ 22, given that α = 0.01 and the sample statistics are n = 40, x = 23.8, and s = 4.3.26) Test the claim that μ = 1210, given that α = 0.01 and the sample statistics are n = 35, x = 1180, and s = 82.27) A local brewery distributes beer in bottles labeled 12 ounces. A government agency thinks that the brewery ischeating its customers. The agency selects 50 of these bottles, measures their contents, and obtains a samplemean of 11.7 ounces with a standard deviation of 0.70 ounce. Use a 0.01 significance level to test the agencyʹsclaim that the brewery is cheating its customers.28) A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1000 hours. A homeowner selects 40bulbs and finds the mean lifetime to be 980 hours with a standard deviation of 80 hours. Test themanufacturerʹs claim. Use α = 0.05.29) A trucking firm suspects that the mean lifetime of a certain tire it uses is less than 34,000 miles. To check theclaim, the firm randomly selects and tests 54 of these tires and gets a mean lifetime of 33,390 miles with astandard deviation of 1200 miles. At α = 0.05, test the trucking firmʹs claim.Page 138Copyright © 2012 Pearson Education, Inc.30) A local politician, running for reelection, claims that the mean prison time for car thieves is less than therequired 4 years. A sample of 80 convicted car thieves was randomly selected, and the mean length of prisontime was found to be 3 years and 6 months, with a standard deviation of 1 year and 3 months. At α = 0.05, testthe politicianʹs claim.31) A local group claims that the police issue at least 60 speeding tickets a day in their area. To prove their point,they randomly select one month. Their research yields the number of tickets issued for each day. The data arelisted below. At α = 0.01, test the groupʹs claim.70 48 41 68 69 55 70 57 60 8332 60 72 58 88 48 59 60 56 6566 60 68 42 57 59 49 70 75 634432) A fast food outlet claims that the mean waiting time in line is less than 3.5 minutes. A random sample of 60customers has a mean of 3.6 minutes with a standard deviation of 0.6 minute. If α = 0.05, test the fast foodoutletʹs claim using critical values and rejection regions.33) A fast food outlet claims that the mean waiting time in line is less than 3.5 minutes. A random sample of 60customers has a mean of 3.6 minutes with a standard deviation of 0.6 minute. If α = 0.05, test the fast foodoutletʹs claim using confidence intervals.34) A local school district claims that the number of school days missed by teachers due to illness is below thenational average of 5 days. A random sample of 40 teachers provided the data below. At α = 0.05, test thedistrictʹs claim using critical values and rejection regions.0 3 6 3 3 5 4 1 3 57 3 1 2 3 3 2 4 1 62 5 2 8 3 1 2 5 4 11 1 2 1 5 7 5 4 9 335) A local school district claims that the number of school days missed by teachers due to illness is below thenational average of 5. A random sample of 40 teachers provided the data below. At α = 0.05, test the districtʹsclaim using confidence intervals.0 3 6 3 3 5 4 1 3 57 3 1 2 3 3 2 4 1 62 5 2 8 3 1 2 5 4 11 1 2 1 5 7 5 4 9 37.3 Hypothesis Testing for the Mean (Small Samples)1 Find Critical Values in a t-distributionMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.1) Find the critical values for a sample with n = 15 and α = 0.05 if H0: μ ≤ 20.A) 1.761 B) 2.977 C) 2.625 D) 1.345Page 139Copyright © 2012 Pearson Education, Inc.2 Test a Claim About a MeanMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.2) Find the standardized test statistic t for a sample with n = 12, x = 22.2, s = 2.2, and α = 0.01 if H0: μ = 21. Roundyour answer to three decimal places.A) 1.890 B) 1.991 C) 2.132 D) 2.0013) Find the standardized test statistic t for a sample with n = 10, x = 9.7, s = 1.3, and α = 0.05 if H0: μ ≥ 10.6. Roundyour answer to three decimal places.A) -2.189 B) -3.186 C) -3.010 D) -2.6174) Find the standardized test statistic t for a sample with n = 15, x = 7.4, s = 0.8, and α = 0.05 if H0: μ ≤ 7.1. Roundyour answer to three decimal places.A) 1.452 B) 1.728 C) 1.631 D) 1.3125) Find the standardized test statistic t for a sample with n = 20, x = 13, s = 2.0, and α = 0.05 if Ha: μ < 13.4. Roundyour answer to three decimal places.A) -0.894 B) -0.872 C) -1.265 D) -1.2336) Find the standardized test statistic t for a sample with n = 25, x = 36, s = 3, and α = 0.005 if Ha: μ > 35. Roundyour answer to three decimal places.A) 1.667 B) 1.997 C) 1.452 D) 1.2397) Find the standardized test statistic t for a sample with n = 12, x = 18.8, s = 2.1, and α = 0.01 if Ha: μ ≠ 19.3.Round your answer to three decimal places.A) -0.825 B) -0.008 C) -0.037 D) -0.381SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.8) Use a t-test to test the claim μ = 16 at α = 0.01, given the sample statistics n = 12, x = 17.2, and s = 2.2.9) Use a t-test to test the claim μ ≥ 10.2 at α = 0.05, given the sample statistics n = 10, x = 9.3, and s = 1.3.10) Use a t-test to test the claim μ ≤ 8.8 at α = 0.05, given the sample statistics n = 15, x = 9.1, and s = 0.8.11) Use a t-test to test the claim μ < 11 at α = 0.10, given the sample statistics n = 20, x = 10.6, and s = 2.0.12) Use a t-test to test the claim μ > 39 at α = 0.005, given the sample statistics n = 25, x = 40, and s = 3.13) Use a t-test to test the claim μ = 23.3 at α = 0.01, given the sample statistics n = 12, x = 22.8, and s = 2.1.14) The Metropolitan Bus Company claims that the mean waiting time for a bus during rush hour is less than 10minutes. A random sample of 20 waiting times has a mean of 8.6 minutes with a standard deviation of 2.1minutes. At α = 0.01, test the bus companyʹs claim. Assume the distribution is normally distributed.15) A local brewery distributes beer in bottles labeled 12 ounces. A government agency thinks that the brewery ischeating its customers. The agency selects 20 of these bottles, measures their contents, and obtains a samplemean of 11.7 ounces with a standard deviation of 0.7 ounce. Use a 0.01 significance level to test the agencyʹsclaim that the brewery is cheating its customers.Page 140Copyright © 2012 Pearson Education, Inc.16) A local group claims that the police issue more than 60 speeding tickets a day in their area. To prove theirpoint, they randomly select two weeks. Their research yields the number of tickets issued for each day. Thedata are listed below. At α = 0.01, test the groupʹs claim.70 48 41 68 69 55 7057 60 83 32 60 72 5817) A local group claims that the police issue more than 60 speeding tickets a day in their area. To prove theirpoint, they randomly select two weeks. Their research yields the number of tickets issued for each day. Thedata are listed below. At α = 0.02, test the groupʹs claim using confidence intervals.70 48 41 68 69 55 7057 60 83 32 60 72 5818) A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1100 hours. A homeowner selects 25bulbs and finds the mean lifetime to be 1070 hours with a standard deviation of 80 hours. Test themanufacturerʹs claim. Use α = 0.05.19) A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1000 hours. A homeowner selects 25bulbs and finds the mean lifetime to be 980 hours with a standard deviation of 80 hours. If α = 0.05, test themanufacturerʹs claim using confidence intervals.20) A trucking firm suspects that the mean life of a certain tire it uses is less than 35,000 miles. To check the claim,the firm randomly selects and tests 18 of these tires and gets a mean lifetime of 34,400 miles with a standarddeviation of 1200 miles. At α = 0.05, test the trucking firmʹs claim.3 Test a Claim About a Mean Using a P-valueSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.Provide an appropriate response.21) A local group claims that the police issue more than 60 speeding tickets a day in their area. To prove theirpoint, they randomly select two weeks. Their research yields the number of tickets issued for each day. Thedata are listed below. At α = 0.01, test the groupʹs claim using P-values.70 48 41 68 69 55 7057 60 83 32 60 72 5822) A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1000 hours. A homeowner selects 25bulbs and finds the mean lifetime to be 980 hours with a standard deviation of 80 hours. If α = 0.05, test themanufacturerʹs claim using P-values.23) A fast food outlet claims that the mean waiting time in line is less than 3.3 minutes. A random sample of 20customers has a mean of 3.1 minutes with a standard deviation of 0.8 minute. If α = 0.05, test the fast foodoutletʹs claim using P-values.Page 141Copyright © 2012 Pearson Education, Inc.24) A local school district claims that the number of school days missed by its teachers due to illness is below thenational average of μ = 5. A random sample of 28 teachers provided the data below. At α = 0.05, test thedistrictʹs claim using P-values.0 3 6 3 3 5 4 1 3 57 3 1 2 3 3 2 4 1 62 5 2 8 3 1 2 57.4 Hypothesis Testing for Proportions1 Test a Claim About a ProportionMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.1) Determine whether the normal sampling distribution can be used. The claim is p < 0.25 and the sample size isn = 18.A) Do not use the normal distribution. B) Use the normal distribution.2) Determine whether the normal sampling distribution can be used. The claim is p ≥ 0.325 and the sample size isn = 42.A) Use the normal distribution. B) Do not use the normal distribution.3) Determine the critical value, z0, to test the claim about the population proportion p = 0.250 given n = 48and p ^= 0.231. Use α = 0.01.A) ±2.575 B) ±1.96 C) ±1.645 D) ±2.334) Determine the standardized test statistic, z, to test the claim about the population proportion p = 0.250 givenn = 48 and p ^= 0.231. Use α = 0.01.A) -0.304 B) -1.18 C) -0.23 D) -2.87SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.5) Test the claim about the population proportion p ≠ 0.325 given n = 42 and p ^= 0.247. Use α = 0.05.6) Fifty-five percent of registered voters in a congressional district are registered Democrats. The Republicancandidate takes a poll to assess his chances in a two-candidate race. He polls 1200 potential voters and findsthat 621 plan to vote for the Republican candidate. Does the Republican candidate have a chance to win? Useα = 0.05.7) An airline claims that the no-show rate for passengers is less than 5%. In a sample of 420 randomly selectedreservations, 19 were no-shows. At α = 0.01, test the airlineʹs claim.8) A recent study claimed that at least 15% of junior high students are overweight. In a sample of 160 students, 18were found to be overweight. At α = 0.05, test the claim.9) A recent study claimed that at least 15% of junior high students are overweight. In a sample of 160 students, 18were found to be overweight. If α = 0.05, test the claim using P-values.10) A recent study claimed that at least 15% of junior high students are overweight. In a sample of 160 students, 18were found to be overweight. If α = 0.05, test the claim using confidence intervals.11) The engineering school at a major university claims that 20% of its graduates are women. In a graduating classof 210 students, 58 were women. Does this suggest that the school is believable? Use α = 0.05.Page 142Copyright © 2012 Pearson Education, Inc.12) A coin is tossed 1000 times and 570 heads appear. At α = 0.05, test the claim that this is not a biased coin. Doesthis suggest the coin is fair?13) A telephone company claims that 20% of its customers have at least two telephone lines. The company selects arandom sample of 500 customers and finds that 88 have two or more telephone lines. At α = 0.05, does the datasupport the claim? Use a P-value.14) A telephone company claims that 20% of its customers have at least two telephone lines. The company selectsa random sample of 500 customers and finds that 88 have two or more telephone lines. If α = 0.05, test thecompanyʹs claim using critical values and rejection regions.15) A telephone company claims that 20% of its customers have at least two telephone lines. The company selectsa random sample of 500 customers and finds that 88 have two or more telephone lines. If α = 0.05, test thecompanyʹs claim using confidence intervals.16) A coin is tossed 1000 times and 530 heads appear. At α = 0.05, test the claim that this is not a biased coin. Use aP-value. Does this suggest the coin is fair?7.5 Hypothesis Testing for Variance and Standard Deviation1 Find Critical ValuesMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.1) Find the critical X2 -values to test the claim σ2 = 4.3 if n = 12 and α = 0.05.A) 3.816, 21.920 B) 2.603, 26.757 C) 3.053, 24.725 D) 4.575, 19.6752) Find the critical X2 -value to test the claim σ2 ≥ 1.8 if n = 15 and α = 0.05.A) 6.571 B) 4.075 C) 4.660 D) 5.6293) Find the critical X2 -value to test the claim σ2 ≤ 3.2 if n = 20 and α = 0.01.A) 36.191 B) 27.204 C) 30.144 D) 32.8524) Find the critical X2 -value to test the claim σ2 > 1.9 if n = 18 and α = 0.01.A) 33.409 B) 27.587 C) 30.181 D) 35.7185) Find the critical X2 -value to test the claim σ2 < 5.6 if n = 28 and α = 0.10.A) 18.114 B) 14.573 C) 16.151 D) 36.7416) Find the critical X2 -values to test the claim σ2 ≠ 6.8 if n = 10 and α = 0.01.A) 1.735, 23.589 B) 2.088, 21.666 C) 2.700, 19.023 D) 3.325, 16.9192 Test Claims About Variances and Standard DeviationsMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Provide an appropriate response.7) Compute the standardized test statistic, X2, to test the claim σ2 = 21.5 if n = 12, s2 = 18, and α = 0.05.A) 9.209 B) 12.961 C) 18.490 D) 0.4928) Compute the standardized test statistic, X2, to test the claim σ2 ≥ 14.4 if n = 15, s2 = 12, and α = 0.05.A) 11.667 B) 8.713 C) 12.823 D) 23.891Page 143Copyright © 2012 Pearson Education, Inc.9) Compute the standardized test statistic, X2, to test the claim σ2 ≤ 28.8 if n = 20, s2 = 55.8, and α = 0.01.A) 36.813 B) 9.322 C) 12.82 D) 33.4110) Compute the standardized test statistic, X2, to test the claim σ2 > 1.9 if n = 18, s2 = 2.7, and α = 0.01.A) 24.158 B) 28.175 C) 33.233 D) 43.15611) Compute the standardized test statistic, X2, to test the claim σ2 < 50.4 if n = 28, s2 = 31.5, and α = 0.10.A) 16.875 B) 14.324 C) 18.132 D) 21.47812) Compute the standardized test statistic, X2 to test the claim σ2 ≠ 61.2 if n = 10, s2 = 67.5, and α = 0.01.A) 9.926 B) 3.276 C) 4.919 D) 12.008SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.13) Test the claim that σ2 = 4.3 if n = 12, s2 = 3.6 and α = 0.05. Assume that the population is normally distributed.14) Test the claim that σ2 ≥ 1.8 if n = 15, s2 = 1.5, and α = 0.05. Assume that the population is normally distributed.15) Test the claim that σ2 ≤ 3.2 if n = 20, s2 = 6.2, and α = 0.01. Assume that the population is normally distributed.16) Test the claim that σ2 > 17.1 if n = 18, s2 = 24.3, and α = 0.01. Assume that the population is normallydistributed.17) Test the claim that σ2 < 39.2 if n = 28, s2 = 24.5, and α = 0.10. Assume that the population is normallydistributed.18) Test the claim that σ2 ≠ 6.8 if n = 10, s2 = 7.5, and α = 0.01. Assume that the population is normally distributed.19) Test the claim that σ = 10.35 if n = 12, s = 9.5, and α = 0.05. Assume that the population is normally distributed.20) Test the claim that σ ≥ 12.06 if n = 15, s = 10.98, and α = 0.05. Assume that the population is normallydistributed.21) Test the claim that σ ≤ 1.79 if n = 20, s = 2.49, and α = 0.01. Assume that the population is normally distributed.22) Test the claim that σ > 5.52 if n = 18, s = 6.56, and α = 0.01. Assume that the population is normally distributed.23) Test the claim that σ < 4.74 if n = 28, s = 3.74 and α = 0.10. Assume that the population is normally distributed.24) Test the claim that σ ≠ 10.44 if n = 10, s = 10.96, and α = 0.01. Assume that the population is normallydistributed.25) Listed below is the number of tickets issued by a local police department. Assuming that the data is normallydistributed, test the claim that the standard deviation for the data is 15 tickets. Use α = 0.01.70 48 41 68 69 55 7057 60 83 32 60 72 58Page 144Copyright © 2012 Pearson Education, Inc.26) The heights (in inches) of 20 randomly selected adult males are listed below. Test the claim that the variance isless than 6.25. Use α = 0.05. Assume the population is normally distributed.70 72 71 70 69 73 69 68 70 7167 71 70 74 69 68 71 71 71 7227) The heights (in inches) of 20 randomly selected adult males are listed below. Test the claim that the variance isless than 6.25. Assume the population is normally distributed. Use α = 0.05 and confidence intervals.70 72 71 70 69 73 69 68 70 7167 71 70 74 69 68 71 71 71 7228) A trucking firm suspects that the variance for a certain tire is greater than 1,000,000. To check the claim, thefirm puts 101 of these tires on its trucks and gets a standard deviation of 1200 miles. At α = 0.05, test thetrucking firmʹs claim.29) A trucking firm suspects that the variance for a certain tire is greater than 1,000,000. To check the claim, thefirm puts 101 of these tires on its trucks and gets a standard deviation of 1200 miles. If α = 0.05, test thetrucking firmʹs claim using confidence intervals.30) A local bank needs information concerning the standard deviation of the checking account balances of itscustomers. From previous information it was assumed to be $250. A random sample of 61 accounts waschecked. The standard deviation was $286.20. At α = 0.01, test the bankʹs assumption. Assume that the accountbalances are normally distributed.31) In one area, monthly incomes of college graduates have a standard deviation of $650. It is believed that thestandard deviation of monthly incomes of non-college graduates is higher. A sample of 71 non-collegegraduates are randomly selected and found to have a standard deviation of $950. Test the claim thatnon-college graduates have a higher standard deviation. Use α = 0.05.32) A statistics professor at an all-womenʹs college determined that the standard deviation of womenʹs heights is2.5 inches. The professor then randomly selected 41 male students from a nearby all-male college and foundthe standard deviation to be 2.9 inches. Test the professorʹs claim that the standard deviation of male heights isgreater than 2.5 inches. Use α = 0.01.3 Test Claims About Variances and Standard Deviations Using a P -valueSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.Provide an appropriate response.33) The heights (in inches) of 20 randomly selected adult males are listed below. Test the claim that the variance isless than 6.25. Assume the population is normally distributed. Use α = 0.05 and P-values.70 72 71 70 69 73 69 68 70 7167 71 70 74 69 68 71 71 71 7234) A trucking firm suspects that the variance for a certain tire is greater than 1,000,000. To check the claim, thefirm puts 101 of these tires on its trucks and gets a standard deviation of 1200 miles. If α = 0.05, test thetrucking firmʹs claim using P-values.Page 145Copyright © 2012 Pearson Education, Inc.Ch. 7 Hypothesis Testing with One SampleAnswer Key7.1 Introduction to Hypothesis Testing1 State a Null and Alternative Hypothesis1) H0: μ = 48.5, Ha: μ ≠ 48.52) H0: μ ≤ 110, Ha: μ > 1103) H0: μ ≥ 101, Ha: μ < 1014) H0: p ≥ 0.5, Ha: p < 0.55) H0: μ ≤ 4.9, Ha: μ > 4.96) H0: σ ≤ 1, Ha: σ > 17) claim: μ = 53.7; H0: μ = 53.7, Ha: μ ≠ 53.7; claim is H08) claim: μ > 120; H0: μ ≤ 120, Ha: μ > 120; claim is Ha9) claim: μ < 101; H0: μ ≥ 101, Ha: μ < 101; claim is Ha10) claim: p ≥ 0.5; H0: p ≥ 0.5, Ha: p < 0.5; claim is H011) claim: μ ≤ 3.3; H0: μ ≤ 3.3, Ha: μ > 3.3; claim is H012) claim: σ > 1; H0: σ ≤ 1, Ha: σ > 1; claim is Ha2 Identify Whether to Use a One-tailed or Two-tailed Test13) A14) A15) A16) A17) A18) A19) A3 Identify Type I and Type II Errors20) type I: rejecting H0: μ = 52.5 when μ = 52.5type II: failing to reject H0: μ = 52.5 when μ ≠ 52.521) type I: rejecting H0: μ ≤ 120 when μ ≤ 120type II: failing to reject H0: μ ≤ 120 when μ > 12022) type I: rejecting H0: μ ≥ 105 when μ ≥ 105type II: failing to reject H0: μ ≥ 105 when μ < 10523) type I: rejecting H0: p ≥ 0.5 when p ≥ 0.5type II: failing to reject H0: p ≥ 0.5 when p < 0.54 Interpret a Decision Based on the Results of a Statistical Test24) A25) A26) A27) A28) A29) A30) A31) A32) A33) A34) A35) A5 Use Confidence Intervals to Make Decisions36) A37) APage 146Copyright © 2012 Pearson Education, Inc.6 Concepts38) A39) A7.2 Hypothesis Testing for the Mean (Large Samples)1 Find P-values1) A2) A3) A4) A2 Test a Claim About a Mean Using P-values5) A6) A7) A8) Fail to reject H0; There is not enough evidence to support the fast food outletʹs claim that the mean waiting time is lessthan 3.4 minutes.9) P-value = 0.000001, P < α, reject H0; There is sufficient evidence to support the school districtʹs claim.3 Find Critical Values10) A11) A12) A13) A14) A15) A4 Test a Claim About a Mean16) A17) A18) A19) A20) A21) A22) A23) standardized test statistic ≈ 1.77; critical value = 1.645; reject H0; There is enough evidence to support the claim.24) standardized test statistic ≈ -1.97; critical value = ±1.96; reject H0; There is enough evidence to support the claim.25) standardized test statistic ≈ 2.65; critical value = 2.33; reject H0. There is enough evidence to reject the claim.26) standardized test statistic ≈ -2.16, critical value = ±2.575, fail to reject H0; There is not enough evidence to reject theclaim.27) standardized test statistic ≈ -3.03; critical value z0 = -2.33; reject H0; The data support the agencyʹs claim.28) standardized test statistic ≈ -1.58; critical value z0 = ±1.96; fail to reject H0; There is not sufficient evidence to rejectthe manufacturerʹs claim.29) standardized test statistic ≈ -3.74; critical value z0 = -1.645; reject H0; There is sufficient evidence to support thetrucking firmʹs claim.30) standardized test statistic ≈ -3.58; critical value z0 = -1.645; reject H0; There is sufficient evidence to support thepoliticianʹs claim.31) x = 60.4, s = 12.2, standardized test statistic ≈ 0.18; critical value z0 = 2.33; fail to reject H0; There is not sufficientevidence to reject the claim.32) Standardized test statistic ≈ 1.29; critical value z0 = -1.645; fail to reject H0; There is not enough evidence to supportthe fast food outletʹs claim.33) Confidence interval (3.47, 3.73); 3.5 lies in the interval, fail to reject H0; There is not enough evidence to support thefast food outletʹs claim.34) Standardized test statistic ≈ -4.71; critical value z0 = -1.645; reject H0; There is sufficient evidence to support thedistrictʹs claim.Page 147Copyright © 2012 Pearson Education, Inc.35) Confidence interval (2.84, 3.96); 5 lies outside the interval, reject H0; There is sufficient evidence to support thedistrictʹs claim.7.3 Hypothesis Testing for the Mean (Small Samples)1 Find Critical Values in a t-distribution1) A2 Test a Claim About a Mean2) A3) A4) A5) A6) A7) A8) t0 = ±3.106, standardized test statistic ≈ 1.890, fail to reject H0; There is not sufficient evidence to reject the claim.9) t0 = -1.833, standardized test statistic ≈ -2.189, reject H0; There is sufficient evidence to reject the claim10) t0 = 1.761, standardized test statistic ≈ 1.452, fail to reject H0; There is not sufficient evidence to reject the claim11) t0 = -1.328, standardized test statistic ≈ -0.894, fail to reject H0; There is not sufficient evidence to support the claim12) t0 = 2.797, standardized test statistic ≈ 1.667, fail to reject H0; There is not sufficient evidence to support the claim13) t0 = ±3.106, standardized test statistic ≈ -0.825, fail to reject H0; There is not sufficient evidence to support the claim14) critical value t0 = -2.539; standardized test statistic ≈ -2.981; reject H0; There is sufficient evidence to support theMetropolitan Bus Companyʹs claim.15) critical value t0 = -2.539; standardized test statistic ≈ -1.917; fail to reject H0; There is not sufficient evidence tosupport the government agencyʹs claim.16) x = 60.21, s = 13.43; critical value t0 = 2.650; standardized test statistic ≈ 0.060; fail to reject H0; There is not sufficientevidence to support the claim.17) Confidence interval (50.70, 69.73); 60 lies in the interval, fail to reject H0; There is not sufficient evidence to reject thegroupʹs claim.18) critical value t0 = ±2.064; standardized test statistic ≈ -1.875; fail to reject H0; There is not sufficient evidence to rejectthe manufacturerʹs claim.19) Confidence interval (946.98, 1013.02); 1000 lies in the interval, fail to reject H0; There is not sufficient evidence to rejectthe manufacturerʹs claim.20) critical value t0 = -1.740; standardized test statistic -2.121; reject H0; There is sufficient evidence to support thetrucking firmʹs claim.3 Test a Claim About a Mean Using a P-value21) P-value = 0.4766. Since the P-value is great than α, there is not sufficient evidence to support the the groupʹs claim.22) Standardized test statistic ≈ -1.25; Therefore, at a degree of freedom of 24, P must be between 0.10 and 0.25. P > α, failto reject H0; There is not sufficient evidence to reject the manufacturerʹs claim.23) Standardized test statistic ≈ -1.118; Therefore, at 19 degrees of freedom, P must lie between 0.10 and 0.25. Since P > α,fail to reject H0. There is not sufficient evidence to support the fast food outletʹs claim.24) standardized test statistic ≈ -4.522; Therefore, at a degree of freedom of 27, P must lie between 0.0001 and 0.00003. P <α, reject H0. There is sufficient evidence to support the school districtʹs claim.7.4 Hypothesis Testing for Proportions1 Test a Claim About a Proportion1) A2) A3) A4) A5) critical value z0 = ±1.96; standardized test statistic ≈ -1.08; fail to reject H0; There is not sufficient evidence to supportthe claim.6) critical value z0 = 1.645; standardized test statistic ≈ 1.21; fail to reject H0; There is not sufficient evidence to supportthe claim p > 0.5. The Republican candidate has no chance.Page 148Copyright © 2012 Pearson Education, Inc.7) critical value z0 = -2.33; standardized test statistic ≈ -0.45; fail to reject H0; There is not sufficient evidence to supportthe airlineʹs claim.8) critical value z0 = -1.645; standardized test statistic ≈ -1.33; fail to reject H0; There is not sufficient evidence to rejectthe claim.9) α = 0.05; P-value = 0.0918; P > α, fail to reject H0; There is not sufficient evidence to reject the studyʹs claim.10) Confidence interval (0.071, 0.154); 15% lies in the interval, fail to reject H0; There is not sufficient evidence to reject thestudyʹs claim.11) critical value z0 = ±1.96; standardized test statistic ≈ 2.76; reject H0; There is enough evidence to reject the universityʹsclaim. The school is not believable.12) critical value z0 = ±1.96; standardized test statistic ≈ 4.43; reject H0; There is enough evidence to reject the claim thatthis is not a biased coin. The coin is not fair.13) α = 0.05; P-value = 0.0901; P > α; fail to reject H0; There is not sufficient evidence to reject the telephone companyʹsclaim.14) Standardized test statistic ≈ -1.34; critical value z0 = ±1.96; fail to reject H0; There is not sufficient evidence to rejectthe companyʹs claim.15) Confidence interval (0.143, 0.209); 20% lies in the interval, fail to reject H0; There is not sufficient evidence to reject thecompanyʹs claim.16) α = 0.05; P-value = 0.0574; P > α; fail to reject H0; There is not enough evidence to reject the claim that this is not abiased coin. The coin is fair.7.5 Hypothesis Testing for Variance and Standard Deviation1 Find Critical Values1) A2) A3) A4) A5) A6) A2 Test Claims About Variances and Standard Deviations7) A8) A9) A10) A11) A12) A13) critical values X 2L = 3.816 and X 2R = 21.920; standardized test statistic X2 = 9.209; fail to reject H0; There is notsufficient evidence to reject the claim.14) critical value X 20= 6.571; standardized test statistic X2 ≈ 11.667; fail to reject H0; There is not sufficient evidence toreject the claim.15) critical value X 20= 36.191; standardized test statistic X2 ≈ 36.813; reject H0; There is sufficient evidence to reject theclaim.16) critical value X 20= 33.409; standardized test statistic X2 ≈ 24.158; fail to reject H0; There is not sufficient evidence toreject the claim.17) critical value X 20= 18.114; standardized test statistic X2 ≈ 16.875; reject H0; There is sufficient evidence to support theclaim.Page 149Copyright © 2012 Pearson Education, Inc.18) critical values X 2L = 1.735 and X 2R = 23.589; standardized test statistic X2 ≈ 9.926; fail to reject H0; There is notsufficient evidence to support the claim.19) critical values X 2L = 3.816 and X 2R = 21.920; standardized test statistic X2 ≈ 9.267; fail to reject H0; There is notsufficient evidence to reject the claim.20) critical value X 20= 6.571; standardized test statistic X2 ≈ 11.605; fail to reject H0; There is not sufficient evidence toreject the claim.21) critical value X 20= 36.191; standardized test statistic X2 ≈ 36.766; reject H0; There is sufficient evidence to reject theclaim.22) critical value X 20= 33.409; standardized test statistic X2 ≈ 24.009; fail to reject H0; There is not sufficient evidence tosupport the claim.23) critical value X 20= 18.114; standardized test statistic X2 ≈ 16.809; reject H0; There is sufficient evidence to support theclaim.24) critical values X 2L = 1.735 and X 2R = 23.589; standardized test statistic X2 ≈ 9.919; fail to reject H0; There is notsufficient evidence to support the claim.25) critical values X 2L = 3.565 and X 2R = 29.819; standardized test statistic X2 ≈ 10.42; fail to reject H0; There is notsufficient evidence to reject the claim.26) critical value X 20= 10.117; standardized test statistic X2 ≈ 9.048; reject H0; There is sufficient evidence to support theclaim.27) Confidence interval (1.89, 5.62); 6.25 lies outside the interval, reject H0; There is sufficient evidence to support theclaim.28) critical value X 20= 124.342; standardized test statistic X2 = 144; reject H0; There is sufficient evidence to support theclaim.29) Confidence interval (1,847,835, 1,940,125); 1,000,000 lies outside the interval, reject H0; There is sufficient evidence tosupport the claim.30) critical values X 2L = 35.534 and X 2R = 91.952; standardized test statistic X2 ≈ 78.634; fail to reject H0; There is notsufficient evidence to reject the claim.31) critical value X 20= 90.531; standardized test statistics X2 = 149.527; reject H0; There is sufficient evidence to supportthe claim.32) critical value X 20= 63.691; standardized test statistic X2 = 53.824; fail to reject H0; There is not sufficient evidence tosupport the claim.3 Test Claims About Variances and Standard Deviations Using a P -value33) Standardized test statistic ≈ 9.048; Therefore, at a degree of freedom of 19, P must be between 0.025 and 0.05. P < α,reject H0; There is sufficient evidence to support the claim.34) Standardized test statistic ≈ 144; Therefore, at a degree of freedom of 100, P must be less than 0.005. P < α, reject H0;There is sufficient evidence to support the firmʹs claim.Page 150Copyright © 2012 Pearson Education, Inc.