Stats Modeling the World 3rd Edition By David E. Bock - Test Bank

Stats Modeling the World 3rd Edition By David E. Bock – Test Bank

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ExamName___________________________________MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.1)A college alumni fund appeals for donations by phoning or emailing recent graduates. A randomsample of 300 alumni shows that 40% of the 150 who were contacted by telephone actually madecontributions compared to only 30% of the 150 who received email requests. Which formulacalculates the 98% confidence interval for the difference in the proportions of alumni who maymake donations if contacted by phone or by email?1)A)(0.40-0.30)±2.33(0.35)(0.65)300B)(0.40-0.30)±2.33(0.40)(0.60)150+(0.30)(0.70)150C)(0.40-0.30)±2.33(0.40)(0.60)300+(0.30)(0.70)300D)(0.40-0.30)±2.33(0.35)(0.65)150E)(0.40-0.30)±2.33(0.35)(0.65)150+(0.35)(0.65)150Answer:BExplanation:A)B)C)D)E)1
2)An online catalog company wants on-time delivery for at least 90% of the orders they ship. Theyhave been shipping orders via UPS and FedEx but will switch to a more expensive service(ShipFast) if there is evidence that this service can exceed the 90% on-time goal. As a test thecompany sends a random sample of orders via ShipFast, and then makes follow-up phone calls tosee if these orders arrived on time. Which hypotheses should they test?2)A)H0:p=0.90HA:p<0.90B)H0:p>0.90HA:p=0.90C)H0:p=0.90HA:pJ0.90D)H0:p<0.90HA:p=0.90E)H0:p=0.90HA:p>0.90Answer:EExplanation:A)B)C)D)E)3)A researcher investigating whether joggers are less likely to get colds than people who do not jogfound aP-value of 3%. This means that:3)A)There’s a 3% chance that joggers get fewer colds.B)There’s a 3% chance that joggers don’t get fewer colds.C)None of these.D)Joggers get 3% fewer colds than non-joggers.E)3% of joggers get colds.Answer:CExplanation:A)B)C)D)E)2
4)To plan the course offerings for the next year a university department dean needs to estimate whatimpact the “No Child Left Behind” legislation might have on the teacher credentialing program.Historically, 40% of this university’s pre-service teachers have qualified for paid internshippositions each year. The Dean of Education looks at a random sample of internship applications tosee what proportion indicate the applicant has achieved the content-mastery that is required forthe internship. Based on these data he creates a 90% confidence interval of (33%, 41%). Could thisconfidence interval be used to test the hypothesisH0:p=0.40 versusHA:p<0.40 at the ΅=0.05level of significance?4)A)No, because the dean only reviewed a sample of the applicants instead of all of them.B)Yes, since 40% is not the center of the confidence interval he rejects the null hypothesis,concluding that the percentage of qualified applicants will decrease.C)Yes, since 40% is in the confidence interval he accepts the null hypothesis, concluding that thepercentage of applicants qualified for paid internship positions will stay the same.D)Yes, since 40% is in the confidence interval he fails to reject the null hypothesis, concludingthat there is not strong enough evidence of any change in the percent of qualified applicants.E)No, because the dean should have used a 95% confidence interval.Answer:DExplanation:A)B)C)D)E)5)To plan the budget for next year a college needs to estimate what impact the current economicdownturn might have on student requests for financial aid. Historically, this college has providedaid to 35% of its students. Officials look at a random sample of this year’s applications to see whatproportion indicate a need for financial aid. Based on these data they create a 90% confidenceinterval of (32%, 40%). Could this interval be used to test the hypothesisH0:p=0.35 versusHA:pJ0.35 at the ΅=0.10 level of significance?5)A)Yes; since 35% is in the confidence interval they fail to reject the null hypothesis, concludingthat there is not strong evidence of any change in financial aid requests.B)No, because financial aid amounts may not be normally distributed.C)Yes; since 35% is in the confidence interval they accept the null hypothesis, concluding thatthe percentage of students requiring financial aid will stay the same.D)No, because they only used a sample of the applicants instead of all of them.E)Yes; since 35% is not at the center of the confidence interval they reject the null hypothesis,concluding that the percentage of students requiring aid will increase.Answer:AExplanation:A)B)C)D)E)3
6)A pharmaceutical company investigating whether drug stores are less likely than food stores toremove over-the-counter drugs from the shelves when the drugs are past the expiration datefound aP-value of 2.8%. This means that:6)A)There is a 2.8% chance the drug stores remove more expired over-the-counter drugs.B)97.2% more drug stores remove over-the-counter drugs from the shelves when the drugs arepast the expiration date than food stores.C)There is a 97.2% chance the drug stores remove more expired over-the-counter drugs.D)2.8% more drug stores remove over-the-counter drugs from the shelves when the drugs arepast the expiration date.E)None of these.Answer:EExplanation:A)B)C)D)E)7)A statistics professor wants to see if more than 80% of her students enjoyed taking her class. At theend of the term, she takes a random sample of students from her large class and asks, in ananonymous survey, if the students enjoyed taking her class. Which set of hypotheses should shetest?7)A)H0:p>0.80HA:p=0.80B)H0:p<0.80HA:pJ0.80C)H0:p<0.80HA:p>0.80D)H0:p=0.80HA:p<0.80E)H0:p=0.80HA:p>0.80Answer:EExplanation:A)B)C)D)E)4
8)Suppose that a conveyor used to sort packages by size does not work properly. We test theconveyor on several packages (withH0: incorrect sort) and our data results in aP-value of 0.016.What probably happens as a result of our testing?8)A)We rejectH0, making a Type II error.B)We fail to rejectH0, committing a Type II error.C)We correctly rejectH0.D)We correctly fail to rejectH0.E)We rejectH0, making a Type I error.Answer:EExplanation:A)B)C)D)E)9)A relief fund is set up to collect donations for the families affected by recent storms. A randomsample of 400 people shows that 28% of those 200 who were contacted by telephone actually madecontributions compared to only 18% of the 200 who received first class mail requests. Whichformula calculates the 95% confidence interval for the difference in the proportions of people whomake donations if contacted by telephone or first class mail?9)A)(0.28-0.18)±1.96(0.28)(0.72)200+(0.18)(0.82)200B)(0.28-0.18)±1.96(0.23)(0.77)400C)(0.28-0.18)±1.96(0.23)(0.77)200D)(0.28-0.18)±1.96(0.23)(0.77)200+(0.23)(0.77)200E)(0.28-0.18)±1.96(0.28)(0.72)400+(0.18)(0.82)400Answer:AExplanation:A)B)C)D)E)5
10)Suppose that a manufacturer is testing one of its machines to make sure that the machine isproducing more than 97% good parts (H0:p=0.97 andHA:p>0.97) . The test results in aP-value of 0.122. Unknown to the manufacturer, the machine is actually producing 99% goodparts. What probably happens as a result of the testing?10)A)They fail to rejectH0, making a Type I error.B)They correctly fail to rejectH0.C)They correctly rejectH0.D)They fail to rejectH0, making a Type II error.E)They rejectH0, making a Type I error.Answer:DExplanation:A)B)C)D)E)11)We test the hypothesis thatp=35% versusp<35%. We don’t know it but actuallyp=26%. Withwhich sample size and significance level will our test have the greatest power?11)A)΅=0.03,n=250B)΅=0.01,n=400C)The power will be the same as long as the true proportionpremains 26%D)΅=0.01,n=250E)΅=0.03,n=400Answer:EExplanation:A)B)C)D)E)12)We will test the hypothesis thatp=60% versusp>60%. We don’t know it, but actuallypis 70%.With which sample size and significance level will our test have the greatest power?12)A)΅=0.05,n=200B)΅=0.01,n=500C)The power will be the same so long as the true proportionpremains 70%.D)΅=0.05,n=500E)΅=0.01,n=200Answer:DExplanation:A)B)C)D)E)6
13)Suppose that a device advertised to increase a car’s gas mileage really does not work. We test it on asmall fleet of cars (withH0: not effective), and our data results in aP-value of 0.004. Whatprobably happens as a result of our experiment?13)A)We rejectH0, making a Type II error.B)We correctly rejectH0.C)We rejectH0, making a Type I error.D)We fail to rejectH0, committing aType II error.E)We correctly fail to rejectH0.Answer:CExplanation:A)B)C)D)E)14)AP-value indicates14)A)the probability of the observed statistic given that the null hypothesis is true.B)the probability the null is true given the observed statistic.C)the probability of the observed statistic given that the alternative hypothesis is true.D)the probability that the null hypothesis is true.E)the probability that the alternative hypothesis is true.Answer:AExplanation:A)B)C)D)E)7
15)A truck company wants on-time delivery for 98% of the parts they order from a metalmanufacturing plant. They have been ordering from Hudson Manufacturing but will switch to anew, cheaper manufacturer (Steel-R-Us) unless there is evidence that this new manufacturercannot meet the 98% on-time goal. As a test the truck company purchases a random sample ofmetal parts from Steel-R-Us, and then determines if these parts were delivered on-time. Whichhypothesis should they test?15)A)H0:p=0.98HA:p>0.98B)H0:p>0.98HA:p=0.98C)H0:p=0.98HA:p<0.98D)H0:p<0.98HA:p>0.98E)H0:p=0.98HA:pJ0.98Answer:CExplanation:A)B)C)D)E)SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.16)Approval ratingA newspaper article reported that a poll based on a sample of 1150residents of a state showed that the state’s Governor’s job approval rating stood at 58%.They claimed a margin of error of±3%. What level of confidence were the pollsters using?16)Answer:^ ^SinceME=z*pqn, we have 0.03=z*(0.58)(0.42)1150or z*Y2.06. Confidence level is96%.Explanation:The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. Amarketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latestjeans…). He loans her several different “people” to scatter around the store for a two-week trial period. The owner carefullycounts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’llpurchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy somethingincreases.17)Over the trial month the rate of in-store sales rose to 30% of shoppers. The store’s ownerdecided this increase was statistically significant. Now that she’s convinced themannequins work, why might she still choose not to purchase them?17)Answer:Although statistically significant, the small increase in sales from 28% to 30% ofshoppers might not be enough to justify the expense of purchasing the mannequins.Explanation:8
A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concernedthat employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defectivechips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentiveprogram is instituted for one month. If successful, the company will continue with the incentive program.18)Based on the data they collected during the trial program, management found that a 95%confidence interval for the percentage of defective chips was (5.0%, 7.0%). What conclusionshould management reach about the new incentive program? Explain.18)Answer:The confidence interval contains values that are all below the hypothesized value of8%, so the data provide convincing evidence that the incentive program lowers thedefect rate of the computer chips.Explanation:19)BirthsA city has two hospitals, with many more births recorded at the larger hospital thanat the smaller one. Records indicate that in general babies are about equally likely to beboys or girls, but the actual gender ratio varies from week to week. Which hospital is morelikely to report a week when over two-thirds of the babies born were girls? Explain.19)Answer:The smaller hospital should experience greater variability in the weekly percentageof male babies. We’d expect the larger hospital to stay closer to the expected 50-50ratio.Explanation:A state’s Department of Education reports that 12% of the high school students in that state attend private high schools. TheState University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a randomsample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attendprivate schools.20)Interpret the confidence interval in this context.20)Answer:We are 90% confident that between 7.9% and 12.6% of the applicants attend privatehigh schools.Explanation:9
21)Pumpkin pieA can of pumpkin pie mix contains a mean of 30 ounces and a standarddeviation of 2 ounces. The contents of the cans are normally distributed. What is theprobability that four randomly selected cans of pumpkin pie mix contain a total of morethan 126 ounces?21)Answer:Two methods are shown below to solve this problem:Method 1:Let P=one can of pumpkin pie mix and T=four cans of pumpkin pie mix.We are told that the contents of the cans are normally distributed, and can assumethat the content amounts are independent from can to can.E(T)=E(P1+P2+P3+P4)=E(P1)+E(P2)+E(P3)+E(P4)=120 ouncesSince the content amounts are independent,Var(T)=Var(P1+P2+P3+P4)=Var(P1)+Var(P2)+Var(P3)+Var(P4)=16SD(T)=Var(T)=16=4 ouncesWe model T withN(120, 4)z=126-1204=1.5P=P(T>126)=P(z>1.5)=0.067There is a 6.7% chance that four randomly selected cans of pumpkin pie mix containmore than 126 ounces.Method 2:Using the Central Limit Theorem approach, lety=average content of cans insampleSince the contents are Normally distributed,yis modeled byN30,24.Py>1264=P(y>31.5)=Pz>31.5-301=P(z>1.5)=0.067There is about a 6.7% chance that 4 randomly selected cans will contain a total ofover 126 ounces.Explanation:10
22)Employment programA city council must decide whether to fund a new”welfare-to-work” program to assist long-time unemployed people in finding jobs. Thisprogram would help clients fill out job applications and give them advice about dealingwith job interviews. A six-month trial has just ended. At the start of this trial a number ofunemployed residents were randomly divided into two groups; one group went throughthe help program and the other group did not. Data about employment at the end of thistrial are shown in the table. Should the city council fund this program? Test an appropriatehypothesis and state your conclusion.22)Answer:^ ^ ^
^ ^H0:p1-p2=0HA:p1-p2>0People were randomly assigned to groups, we assume the groups are independent,and 20, 34, 13, 33 are allL10. OK to do a 2-proportion z-test.p1=0.370,p2=0.283,p=20+1354+46=0.33z=(0.370-0.283)-0(0.33)(0.67)54-(0.33)(0.67)46=0.93P=P(p1-p2>0.087)=P(z>0.93)=0.176We fail to reject the null hypothesis becausePis very large. We do not haveevidence that the help program is beneficial, so it should not be funded.Explanation:A state’s Department of Education reports that 12% of the high school students in that state attend private high schools. TheState University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a randomsample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attendprivate schools.23)The admissions officers want to estimate the true percentage of private school applicants towithin±4%, with 90% confidence. How many applications should they sample?23)Answer:^ ^
^ME=z*pqn0.04=1.645(0.12)(0.88)nn=1.645(0.12)(0.88)0.04n=178.60Y179They should sample at least 179 applicants. (423 ifp=0.5 is used)Explanation:11
24)DepressionA recent psychiatric study from the University of Southampton observed ahigher incidence of depression among women whose birth weight was less than 6.6pounds than in women whose birth weight was over 6.6 pounds. Based on aP-value of0.0248 the researchers concluded there was evidence that low birth weights may be a riskfactor for susceptibility to depression. Explain in context what the reported P-value means.24)Answer:If birth weight was not a risk factor for susceptibility to depression, an observeddifference in incidence of depression this large (or larger) would occur in only 2.48%of such samples.Explanation:The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. Amarketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latestjeans…). He loans her several different “people” to scatter around the store for a two-week trial period. The owner carefullycounts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’llpurchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy somethingincreases.25)In this context describe a Type I error and the impact such an error would have on thestore.25)Answer:A Type I error would be deciding the percentage of customers who’ll makepurchases will go up when in fact it won’t. The store’s owner would waste moneybuying useless mannequins.Explanation:The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of femalesin the labor force is the same in the United States. Representatives from the United States Department of Labor plan to checka random sample of over 10,000 employment records on file to estimate a percentage of females in the United States laborforce.26)Interpret the confidence interval in this context.26)Answer:We are 90% confident that between 40.0% and 47.2% of the employment recordsfrom the United States labor force are for females.Explanation:12
Great Britain has a great literary tradition that spans centuries. One might assume, then, that Britons read more than citizensof other countries. Some Canadians, however, feel that a higher percentage of Canadians than Britons read. A recent GallupPoll reported that 86% of 1004 randomly sampled Canadians read at least one book in the past year, compared to 81% of 1009randomly sampled Britons. Do these results confirm a higher reading rate in Canada?27)Find a 99% confidence interval for the difference in the proportion of Britons andCanadians who read at least one book in the last year. Interpret your interval.27)Answer:^ ^
^ ^
^ ^
^ ^ ^ ^
^ ^
^ ^With the conditions satisfied (from Problem 1), the sampling distribution of thedifference inproportions is approximately Normal with a mean ofpB-pC, the true differencebetween thepopulation proportions. We can find a two-proportion z-interval.We know:nB=1009,pB=0.81,nC=1004,pC=0.86We estimateSD(pB-pC) asSE(pB-pC)=pBqBnB+pCqCnC=(0.81)(0.19)1009+(0.86)(0.14)1004=0.0165ME=z*×SE(pB-pC)=2.576(0.0165)=0.0425The observed difference in sample proportions=pB-pC=0.81-0.186=-0.05, sothe 99%confidence interval is-0.05±0.0425, or-9.3% to-0.8%.We are 99% confident that the proportion of Britons who read at least one book inthe past year isbetween 0.8-percentage points and 9.3-percentage points lower than the proportionof Canadians who read at least one book in the past year.Explanation:The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Cresttoothpaste. A marketing director suggests that the company invest in a new marketing campaign which will includeadvertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for onemonth to determine if the market share increases with new labels.28)Write the company’s null and alternative hypotheses.28)Answer:The null hypothesis is that 19.5% of all people who use toothpaste buy Crest. Thealternative hypothesis is that the percentage of all people who use toothpaste whouse Crest is greater than 19.5%. In symbols:H0:p=0.195 andHA:p>0.195Explanation:13
A report on health care in the US said that 28% of Americans have experienced times when they haven’t been able to affordmedical care. A news organization randomly sampled 801 black Americans, of whom 38% reported that there had beentimes in the last year when they had not been able to afford medical care. Does this indicate that this problem is more severeamong black Americans?29)Test an appropriate hypothesis and state your conclusion. (Make sure to check anynecessary conditions and to state a conclusion in the context of the problem.)29)Answer:^
^Hypotheses:H0:p=0.28. The proportion of all black Americans that were unableto afford medical care in the last year is 28%.HA:p>0.28. The proportion of all black Americans that were unableto afford medical care in the last year is greater than 28%.Model: Okay to use the Normal model because the sample is random, these 801black Americans are less than 10% of all black Americans, andnp=(801)(0.28)=224.28L10 andnq=(801)(0.72)=576.72L10.We will do a one-proportionz-test.Mechanics:n=801,p=0.38z=0.38-0.28(0.28)(0.72)801P(p>0.38)=Pz>6.29Y0Conclusion: With a P-value so small (just about zero), I reject the null hypothesis.There is enough evidence to suggest that the proportion of black Americans whowere not able to afford medical care in the past year is more than 28%.Explanation:14
30)It is generally believed that nearsightedness affects about 12% of children. A school districtgives vision tests to 133 incoming kindergarten children.a. Describe the sampling distribution model for the sample proportion by naming themodel and telling its mean and standard deviation. Justify your answer.b. Sketch and clearly label the model.c. What is the probability that in this group over 15% of the children will be found to benearsighted?30)Answer:^a. We can assume these kids are a random sample of all children, andcertainly less than 10% of them. We expectnp=(133)(0.12)=15.96successes and 117.04 failures so the sample size is large enough to usethe sampling modelN(0.12, 0.028).b.c.z=0.15-0.120.028=1.07P(p>0.15)=P(z>1.07)=0.142Explanation:The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of femalesin the labor force is the same in the United States. Representatives from the United States Department of Labor plan to checka random sample of over 10,000 employment records on file to estimate a percentage of females in the United States laborforce.31)Explain what 90% confidence means in this context.31)Answer:If many random samples were taken, 90% of the confidence intervals producedwould contain the actual percentage of all female employment records in the UnitedStates labor force.Explanation:The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. Amarketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latestjeans…). He loans her several different “people” to scatter around the store for a two-week trial period. The owner carefullycounts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’llpurchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy somethingincreases.32)What alpha level did the store’s owner use?32)Answer:΅=1-0.982=0.01=1%Explanation:15
33)In this context describe a Type II error and the impact such an error would have on thestore.33)Answer:A Type II error would be deciding the percentage of customers who’ll makepurchases won’t go up when in fact it would have. The store’s owner would miss anopportunity to increase sales.Explanation:34)SleepDo more than 50% of U.S. adults feel they get enough sleep? According to Gallup’sDecember 2004 Lifestyle poll, 55% of U.S. adults said that that they get enough sleep. Thepoll was based on a random sample of 1003 U.S. adults. Test an appropriate hypothesisand state your conclusion in the context of the problem.34)Answer:^
^Hypothesis:H0:p=0.50HA:p>0.50Plan: Okay to use the Normal model because the trials are independent (randomsample of U.S. adults), these 1003 U.S. adults are less than 10% of all U.S. adults, andnp0=(1003)(0.50)=501.5L10andnq0=(1003)(0.50)=501.5L10.We will do a one-proportion z-test.Mechanics:SD(p0)=p0q0n=(0.50)(0.50)1003=0.0158; sample proportion:p=0.55P(p>0.55)=P(z>0.55-0.500.0158)=P(z>3.16)=0.0008With a P-value of 0.0008, I reject the null hypothesis. There is strong evidence thatthe proportion of U.S. adults who feel they get enough sleep is more than 50%.Explanation:The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Cresttoothpaste. A marketing director suggests that the company invest in a new marketing campaign which will includeadvertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for onemonth to determine if the market share increases with new labels.35)The board of directors asked the research department to extend the trial period so that thedecision can be made on two months worth of data. Will the power increase, decrease, orremain the same?35)Answer:The power would increase because of the larger sample size.Explanation:A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concernedthat employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defectivechips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentiveprogram is instituted for one month. If successful, the company will continue with the incentive program.36)In this context describe a Type I error and the impact such an error would have on thecompany.36)Answer:A Type I error would be deciding the percentage of defective chips has decreased,when in fact it has not. The company would waste money on a new incentiveprogram that does not decrease the defect rate of the chips.Explanation:16
A report on health care in the US said that 28% of Americans have experienced times when they haven’t been able to affordmedical care. A news organization randomly sampled 801 black Americans, of whom 38% reported that there had beentimes in the last year when they had not been able to afford medical care. Does this indicate that this problem is more severeamong black Americans?37)Was your test one-tail upper tail, one-tail lower tail, or two-tail? Explain why you chosethat kind of test in this situation.37)Answer:One-tail, upper tail test. We are concerned that the proportion of people who are notable to afford medical care is higher among black Americans.Explanation:38)CerealA box of Raspberry Crunch cereal contains a mean of 13 ounces with a standarddeviation of 0.5 ounce. The distribution of the contents of cereal boxes is approximatelyNormal. What is the probability that a case of 12 cereal boxes contains a total of more than160 ounces?38)Answer:Two methods can be used to solve this problem:Method 1:Let B=weight of one box of cereal and T=weight of 12 boxes of cereal. We are toldthat the contents of the boxes are approximately Normal, and we can assume thatthe content amounts are independent from box to box.E(T)=E(B1+B2+…+B12)=E(B1)+E(B2)+…+E(B12)=156 ouncesSince the content amounts are independent,Var(T)=Var(B1+B2+…+B12)=Var(B1)+Var(B2)+…+Var(B12)=3SD(T)=Var(T)=3=1.73 oz.We model T withN(156, 1.73).z=160-1561.73=2.31 andP(T>160)=P(z>2.31)=0.0104There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160ounces.Method 2:Using the Central Limit Theorem approach, lety=average content of boxes in thecase. Since the contents are Normally distributed,yis modeled byN13,0.512.Py>16012=P(y>13.33)=Pz>13.33-130.512=P(z>2.31)=0.0104.There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160ounces.Explanation:17
The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Cresttoothpaste. A marketing director suggests that the company invest in a new marketing campaign which will includeadvertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for onemonth to determine if the market share increases with new labels.39)Based on the data they collected during the trial the research department found that a 98%confidence interval for the proportion of all consumers who might buy Crest toothpastewas (16%, 28%). What conclusion should the company reach about the new marketingcampaign? Explain.39)Answer:The confidence interval contains the current value of 19.5%, so the product trials donot provide convincing evidence that the new marketing campaign would increasesales.Explanation:40)Dolphin birthsA state has two aquariums that have dolphins, with more births recordedat the larger aquarium than at the smaller one. Records indicate that in general babies areequally likely to be male or female, but the gender ratio varies from season to season.Which aquarium is more likely to report a season when over two-thirds of the dolphinsborn were males? Explain.40)Answer:The smaller aquarium would experience more variability in the season percentageof male births. We would expect the larger aquarium to stay more consistent andcloser to the 50-50 ratio for gender births.Explanation:A state’s Department of Education reports that 12% of the high school students in that state attend private high schools. TheState University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a randomsample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attendprivate schools.41)They actually select a random sample of 450 applications, and find that 46 of thosestudents attend private schools. Create the confidence interval.41)Answer:We have a random sample of less than 10% of the applicants, with 46 successes and404 failures, so a Normal model applies. The confidence interval is (0.079, 0.126).Explanation:The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. Amarketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latestjeans…). He loans her several different “people” to scatter around the store for a two-week trial period. The owner carefullycounts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’llpurchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy somethingincreases.42)Describe to the owner an advantage and a disadvantage of using an alpha level of 5%instead.42)Answer:Advantage: the test would have greater power to detect a positive effect of themannequins. Disadvantage: she’d be more likely to think the mannequins wereeffective even if they were not.Explanation:18
A statistics professor asked her students whether or not they were registered to vote. In a sample of 50 of her students(randomly sampled from her 700 students), 35 said they were registered to vote.43)What is the probability that the true proportion of the professor’s students who wereregistered to vote is in your confidence interval?43)Answer:There is no probability involved-once the interval is constructed, the trueproportion of the professor’s students who were registered to vote is in the intervalor it is not.Explanation:44)ExerciseA random sample of 150 men found that 88 of the men exercise regularly, while arandom sample of 200 women found that 130 of the women exercise regularly.a. Based on the results, construct and interpret a 95% confidence interval for the differencein the proportions of women and men who exercise regularly.b. A friend says that she believes that a higher proportion of women than men exerciseregularly. Does your confidence interval support this conclusion? Explain.44)Answer:^ ^
^ ^
^ ^
^ ^ ^ ^
^ ^
^ ^a. Conditions:* Randomization Condition: We are told that we have random samples.* 10% Condition: We have less than 10% of all men and less than 10% of all women.* Independent samples condition: The two groups are clearly independent of eachother.* Success/Failure Condition: Of the men, 88 exercise regularly and 62 do not; of thewomen, 130 exercise regularly and 70 do not. The observed number of bothsuccesses and failures in both groups is at least 10.With the conditions satisfied, the sampling distribution of the difference inproportions is approximately Normal with a mean ofpM-pW, the true differencebetween the population proportions. We can find a two-proportion z-interval.We know:nM=150,pM=88150=0.587,nW=200,pW=130200=0.650We estimateSD(pM-pW) asSE(pM-pW)=pMqMnM+pWqWnW=(0.587)(0.413)150+(0.65)(0.35)200=0.0525ME=z*×SE(pM-pW)=1.96(0.0525)=0.1029The observed difference in sample proportions=pM-pW=0.587-0.650=-0.063,so the 95%confidence interval is-0.063±0.1029, or-16.6% to 4.0%.We are 95% confident that the proportion of women who exercise regularly isbetween 4.0% lower and 16.6% higher than the proportion of men who exerciseregularly.b. Since zero is contained in my confidence interval, I cannot say that a higherproportion of women than men exercise regularly. My confidence interval does notsupport my friend’s claim.Explanation:19
In 2000, the United Nations claimed that there was a higher rate of illiteracy in men than in women from the country ofQatar. A humanitarian organization went to Qatar to conduct a random sample. The results revealed that 45 out of 234 menand 42 out of 251 women were classified as illiterate on the same measurement test. Do these results indicate that the UnitedNations findings were correct?45)Test an appropriate hypothesis and state your conclusion.45)Answer:^ ^ ^
^ ^
^ ^ ^ ^
^ ^
^ ^
^ ^H0:pm-pw=0,HA:pm-pw>0, wherem=men andw=women* Randomization Condition: The men and women in the sample were randomlyselected by thehumanitarian organization.* 10% Condition: The number of men and women in Qatar is greater than 2340 (10×234) and2510 (10×251), respectively.* Independent samples condition: The two groups are independent of each otherbecause thesamples were selected at random.* Success/Failure Condition: In men, 45 were illiterate and 189 were not. In women,42 wereilliterate and 209 were not. The observed number of both successes and failures inboth groups is larger than 10.Because the conditions are satisfied, we can model the sampling distribution of thedifference in proportions with a Normal model. We can perform a two-proportionz-test.We know:nm=234,pm=0.192,nw=251,pw=0.167, andppooled=45+42234+251=0.179.SEpooled(pm-pw)=ppooledqpoolednm+ppooledqpoolednw=(0.179)(0.821)234+(0.179)(0.821)251=0.0349The observed difference in sample proportions=pm-pw=0.192-0.167=0.025z=(pm-pw)-0SEpooled(pm-pw)=0.025-00.0349=0.716P=P(z>0.716)=0.24The P-value of 0.24 is high, so we fail to reject the null hypothesis. There isinsufficient evidence to conclude that illiteracy rate in men is higher than for womenin Qatar.Explanation:20
46)According to Gallup, about 33% of Americans polled said they frequently experience stressin their daily lives. Suppose you are in a class of 45 students.a. What is the probability that no more than 12 students in the class will say that theyfrequently experience stress in their daily lives? (Make sure to identify the samplingdistribution you use and check all necessary conditions.)b. If 20 students in the class said they frequently experience stress in their daily lives,would you be surprised? Explain, and use statistics to support your answer.46)Answer:^a. We want to find the probability that no more than 12 students in the class will saythat they frequently experience stress. This is the same as asking the probability offinding less than 26.7% of “stressed” students in a class of 45 students.Check the conditions:1. 10% condition: 45 students is less than 10% of all students who could take theclass2. Success/failure cond.:np=45(0.33)=14.85,nq=45(0.67)=30.15, which bothexceed 10We can use theN(0.33(0.33)(0.67)45=0.070) to model the sampling distribution.We need to standardize the 26.7% and then find the probability of getting az-scoreless than or equal to the one we find:z=0.267-0.330.070=-0.90P(p<0.267)=P(z<-0.90)=0.1841, so the probability is about 18.4% that no morethan 12 students will say that they frequently experience stress in their daily lives.b. From part a, we can use N(0.33, 0.070) to model the sampling distribution. Twentystudents is about 44.4% of the class. This is about 1.63 standard deviations abovewhat we would expect, which is not a surprising result.Explanation:A statistics professor asked her students whether or not they were registered to vote. In a sample of 50 of her students(randomly sampled from her 700 students), 35 said they were registered to vote.47)Explain what 95% confidence means in this context.47)Answer:If many random samples were taken, 95% of the confidence intervals producedwould contain the actual percentage of the professor’s students who are registered tovote.Explanation:48)Wildlife scientists studying a certain species of frogs know that past records indicate theadults should weigh an average of 118 grams with a standard deviation of 14 grams. Theresearchers collect a random sample of 50 adult frogs and weigh them. In their sample themean weight was only 110 grams. One of the scientists is alarmed, fearing thatenvironmental changes may be adversely affecting the frogs. Do you think this sampleresult is unusually low? Explain.48)Answer:We have a random sample of frogs drawn from a much larger population. With asample of size 50 the CLT says that the approximate sampling model for samplemeans will be N(118, 1.98). A sample mean of only 110 grams is about 4 standarddeviations below what we expect, a very unusual result.Explanation:21
A company claims to have invented a hand-held sensor that can detect the presence of explosives inside a closed container.Law enforcement and security agencies are very interested in purchasing several of the devices if they are shown to performeffectively. An independent laboratory arranged a preliminary test. If the device can detect explosives at a rate greater thanchance would predict, a more rigorous test will be performed. They placed four empty boxes in the corners of an otherwiseempty room. For each trial they put a small quantity of an explosive in one of the boxes selected at random. The company’stechnician then entered the room and used the sensor to try to determine which of the four boxes contained the explosive.The experiment consisted of 50 trials, and the technician was successful in finding the explosive 16 times. Does this indicatethat the device is effective in sensing the presence of explosives, and should undergo more rigorous testing?49)Explain what yourP-value means in this context.49)Answer:Even if the device actually performs no better than guessing, we could expect to findthe explosives 16 or more times out of 50 about 13% of the time.Explanation:A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concernedthat employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defectivechips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentiveprogram is instituted for one month. If successful, the company will continue with the incentive program.50)Describe to management an advantage and disadvantage of using a 1% alpha level ofsignificance instead.50)Answer:Advantage: management would be less likely to conclude the incentive programwas effective if it really were not. Disadvantage: the test would have less power todetect a positive effect of the new incentive program.Explanation:51)Tax adviceEach year people who have income file income tax reports with thegovernment. In some instances people seek advice from accountants and financial advisorsregarding their income tax situations. This advice is meant to lower the percentage of taxespaid to the government each year. A random sample of people who filed tax reportsresulted in the data in the table below. Does this data indicate that people should seek taxadvice from an accountant or financial advisor? Test an appropriate hypothesis and stateyour conclusion.51)Answer:We want to know whether the percentage of people who paid lower taxes wasdifferent based on whether they sought tax advice or not.H0:padvice-pno advice=0,HA:padvice-pno advice>0Conditions:* Independence: The people who filed tax reports were randomly selected and donot influence each other.* Random Condition: The people who filed tax reports were randomly selected.* 10% Condition: 72 is less than 10% of people who didn’t get tax advice, and 105 isless than 10% of people who did get tax advice.* Success/Failure: All observed counts (48, 19, 24, and 86) are at least 10.Because all conditions have been satisfied, we can model the sampling distributionof the difference in proportions with a Normal model. We can perform atwo-proportionz-test.Let ‘Advice’ group be the people who sought tax advice and the ‘No advice’ group22
Answer:^ ^
^
^ ^
^ ^
^ ^
^ ^be the people who did not seek tax advice.We know:nadvice=105,nno advice=72,padvice=0.819,pno advice=0.333.ppooled=24+8672+105=0.621SEpooled(padvice-pno advice)=(0.62)(0.38)72+(0.62)(0.38)105=0.074The observed difference in sample proportions ispadvice-pno advice=0.819-0.333=0.486z=(padvice-pno advice)-0SEpooled(padvice-pno advice)=0.49-00.074=6.62P=P(z>6.62)<0.0001The P-value is small, so we reject the null hypothesis. There is strong evidence of adifference in the tax percentages paid between the group who has tax advice and thegroup who did not have tax advice. It appears that if you have tax advice you aremore likely to pay a lower percentage of taxes to the government.Explanation:The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of femalesin the labor force is the same in the United States. Representatives from the United States Department of Labor plan to checka random sample of over 10,000 employment records on file to estimate a percentage of females in the United States laborforce.52)The representatives from the Department of Labor want to estimate a percentage offemales in the United States labor force to within±5%, with 90% confidence. How manyemployment records should they sample?52)Answer:^ ^ME=z*pqn0.05=1.645(0.46)(0.54)nn=1.645(0.46)(0.54)0.05n=268.87Y269They should sample at least 269 employment records.Explanation:23
Great Britain has a great literary tradition that spans centuries. One might assume, then, that Britons read more than citizensof other countries. Some Canadians, however, feel that a higher percentage of Canadians than Britons read. A recent GallupPoll reported that 86% of 1004 randomly sampled Canadians read at least one book in the past year, compared to 81% of 1009randomly sampled Britons. Do these results confirm a higher reading rate in Canada?53)Test an appropriate hypothesis and state your conclusions.53)Answer:^ ^
^
^ ^
^ ^ ^ ^
^ ^
^ ^
^ ^H0:pB-pC=0,HA:pB-pC<0, where B=Britons and C=CanadiansConditions:* Randomization Condition: The Britons and Canadians were randomly sampled byGallup.* 10% Condition: The number of Britons and Canadians is greater than 10,090 (10×1009) and10,040 (10×1004), respectively.* Independent samples condition: The two groups are clearly independent of eachother.* Success/Failure Condition: Of the Britons, approximately 817 read at least onebook and 192 did not; of the Canadians, approximately 863 read at least one bookand 141 did not. The observed number of both successes and failures in both groupsis larger than 10.Because the conditions are satisfied, we can model the sampling distribution of thedifferencein proportions with a Normal model. We can perform a two-proportion z-test.We know:nB=1009,pB=0.81,nC=1004,pC=0.86, andppooled=817+8631009+1004=16802013=0.835SEpooled(pB-pC)=ppooledqpoolednB+ppooledqpoolednC=(0.835)(0.165)1009+(0.835)(0.165)1004=0.0165The observed difference in sample proportions=pB-pC=0.81-0.86=-0.05z=(pB-pC)-0SEpooled(pB-pC)=-0.05-00.0165=-3.03, so theP-value=P(z<-3.03)=0.0012The P-value of 0.0012 is low, so we reject the null hypothesis.There is strongevidence that the percentage of Britons who read at least one book in the past year isless than the percentage of Canadians who read at least one book in the past year.Explanation:24
A company claims to have invented a hand-held sensor that can detect the presence of explosives inside a closed container.Law enforcement and security agencies are very interested in purchasing several of the devices if they are shown to performeffectively. An independent laboratory arranged a preliminary test. If the device can detect explosives at a rate greater thanchance would predict, a more rigorous test will be performed. They placed four empty boxes in the corners of an otherwiseempty room. For each trial they put a small quantity of an explosive in one of the boxes selected at random. The company’stechnician then entered the room and used the sensor to try to determine which of the four boxes contained the explosive.The experiment consisted of 50 trials, and the technician was successful in finding the explosive 16 times. Does this indicatethat the device is effective in sensing the presence of explosives, and should undergo more rigorous testing?54)Was your test one-tail upper tail, lower tail, or two-tail? Explain why you chose that kindof test in this situation.54)Answer:One-tail, upper tail. The device is effective only if it can detect explosives at a ratehigher than chance (25%).Explanation:The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. Amarketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latestjeans…). He loans her several different “people” to scatter around the store for a two-week trial period. The owner carefullycounts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’llpurchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy somethingincreases.55)The owner talked the salesman into extending the trial period so that she can base herdecision on data for a full month. Will the power of the test increase, decrease, or remainthe same?55)Answer:The power of the test will increase.Explanation:The International Olympic Committee states that the female participation in the 2004 Summer Olympic Games was 42%,even with new sports such as weight lifting, hammer throw, and modern pentathlon being added to the Games. Broadcastingand clothing companies want to change their advertising and marketing strategies if the female participation increases at thenext games. An independent sports expert arranged for a random sample of pre-Olympic exhibitions. The sports expertreported that 202 of 454 athletes in the random sample were women. Is this strong evidence that the participation rate mayincrease?56)Was your test one-tail upper tail, lower tail, or two-tail? Explain why you choose that kindof test in this situation.56)Answer:One-tail, upper test. The companies will change strategies only if there is strongevidence of an increase in female participation rate from the current rate of 42%.Explanation:The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Cresttoothpaste. A marketing director suggests that the company invest in a new marketing campaign which will includeadvertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for onemonth to determine if the market share increases with new labels.57)What level of significance did the research department use?57)Answer:΅=1-0.982=0.01=1%Explanation:25
58)Describe to the board of directors an advantage and a disadvantage of using a 5% alphalevel of significance instead.58)Answer:Advantage: the test would have greater power to detect a positive effect of the newmarketing campaign. Disadvantage: they would be more likely to think themarketing campaign was effective even if it were not.Explanation:59)TruckersOn many highways state police officers conduct inspections of driving logbooksfrom large trucks to see if the trucker has driven too many hours in a day. At one truckinspection station they issued citations to 49 of 348 truckers that they reviewed.a. Based on the results of this inspection station, construct and interpret a 95% confidenceinterval for the proportion of truck drivers that have driven too many hours in a day.b. Explain the meaning of “95% confidence” in part A.59)Answer:^ ^ ^ ^
^a. Conditions:* Independence: We assume that one trucker’s driving times do not influence othertrucker’s driving times.* Random Condition: We assume that trucks are stopped at random.* 10% Condition: This sample of 348 truckers is less than 10% of all truckers.* Success/Failure: 49 tickets and 299 tickets are both at least 10, so our sample is largeenough.Under these conditions the sampling distribution of the proportion can be modeledby a Normal model. We will find a one-proportionz-interval.We known=348 andp=0.14 , soSE(p)=pqn=(0.14)(0.86)348=0.0186The sampling model is Normal, for a 95% confidence interval the critical value isz*=1.96.The margin of error isME=z*×SE(p)=1.96(0.0186)=0.0365.The 95% confidence interval is 0.14±0.0365 or (0.1035, 0.1765).We are 95% confident that between 10.4% and 17.7% of truck drivers have driventoo many hours in a day.b. If we repeated the sampling and created new confidence intervals many times wewould expect about 95% of those intervals to contain the actual proportion of truckdrivers that have driven too many hours in a day.Explanation:26
60)The average composite ACT score for Ohio students who took the test in 2003 was 21.4.Assume that the standard deviation is 1.05. In a random sample of 25 students who tookthe exam in 2003, what is the probability that the average composite ACT score is 22 ormore? (Make sure to identify the sampling distribution you use and check all necessaryconditions.)60)Answer:Check the conditions:1. Random sampling condition: We have been told that this is a random sample.2. Independence assumption: It’s reasonable to think that the scores of the 25students are mutually independent.3. 10% condition: 25 students is certainly less than 10% of all students who took theexam.We’re assuming that the model for composite ACT scores has mean μ=21.4 andstandard deviationΗ=1.05.Since the sample size is large enough and thedistribution of ACT scores is most likely unimodal and symmetric, CLT allows us todescribe the sampling distribution ofywith a Normal model with mean 21.4 andSD=(y)=1.0525=0.21.An average score of 22 isz=22-21.40.21=2.86 SDs above the mean.P(x>22)=P(Z>2.86)=0.0021, so the probability that the average composite ACTscore for a sample of 25 randomly selected students is 22 or more is 0.0021.Explanation:61)Approval ratingThe President’s job approval rating is always a hot topic. Your localpaper conducts a poll of 100 randomly selected adults to determine the President’s jobapproval rating. A CNN/USA Today/Gallup poll conducts a poll of 1010 randomly selectedadults. Which poll is more likely to report that the President’s approval rating is below50%, assuming that his actual approval rating is 54%? Explain.61)Answer:The smaller poll would have more variability and would thus be more likely to varyfrom the actual approval rating of 54%. We would expect the larger poll to be moreconsistent with the 54% rating. So, it is more likely that the smaller poll wouldreport that the President’s approval rating is below 50%.Explanation:A statistics professor asked her students whether or not they were registered to vote. In a sample of 50 of her students(randomly sampled from her 700 students), 35 said they were registered to vote.62)If the professor only knew the information from the September 2004 Gallup poll andwanted to estimate the percentage of her students who were registered to vote to within±4% with 95% confidence, how many students should she sample?62)Answer:ME=z*pqn0.04=1.96(0.73)(0.27)nn=(1.96)2(0.73)(0.27)(0.04)2=473.24n=474Note: Since there are only 700 students in the professor’s class, she cannot samplethis many students without violating the 10% condition!Explanation:27
The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Cresttoothpaste. A marketing director suggests that the company invest in a new marketing campaign which will includeadvertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for onemonth to determine if the market share increases with new labels.63)In this context describe a Type I error and the impact such an error would have on thecompany.63)Answer:A Type I error would be concluding the proportion of people who will buy Cresttoothpaste will go up when in fact it won’t. The company would waste money on anew marketing campaign that will not increase sales.Explanation:The International Olympic Committee states that the female participation in the 2004 Summer Olympic Games was 42%,even with new sports such as weight lifting, hammer throw, and modern pentathlon being added to the Games. Broadcastingand clothing companies want to change their advertising and marketing strategies if the female participation increases at thenext games. An independent sports expert arranged for a random sample of pre-Olympic exhibitions. The sports expertreported that 202 of 454 athletes in the random sample were women. Is this strong evidence that the participation rate mayincrease?64)Explain what yourP-value means in this context.64)Answer:If the proportion of female athletes has not increased, we could expect to find at least202 females of 454 pre-Olympic athletes about 13.8% of the time.Explanation:The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. Amarketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latestjeans…). He loans her several different “people” to scatter around the store for a two-week trial period. The owner carefullycounts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’llpurchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy somethingincreases.65)Write the owner’s null and alternative hypotheses.65)Answer:The null hypothesis is that the percentage of all customers who buy something is28%. The alternative hypothesis is that the percentage of all customers who buysomething is greater than 28%. In symbols:H0:p=0.28HA:p>0.28Explanation:28
66)RoadblocksFrom time to time police set up roadblocks to check cars to see if the safetyinspection is up to date. At one such roadblock they issued tickets for expired inspectionstickers to 22 of 628 cars they stopped.a. Based on the results at this roadblock, construct and interpret a 95% confidence intervalfor the proportion of autos in that region whose safety inspections have expired.b. Explain the meaning of “95% confidence” in part a.66)Answer:^a. We assume the cars stopped are representative of all cars in the area, and that 628<10% of the cars. The 22 successes (expired stickers) and 608 failures are bothgreater than 10. It’s okay to use a Normal model and find a one-proportionz-interval.p=0.035;0.035±1.96(0.035)(0.965)628=(0.021, 0.049)The interval (0.021,0.049) means that we are 95% confident that between 2% and 5%of local cars have expired safety inspection stickers.b. If we repeated the sampling and created new confidence intervals many timeswe’d expect about 95% of those intervals to contain the actual proportion of carswith expired stickers.Explanation:A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concernedthat employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defectivechips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentiveprogram is instituted for one month. If successful, the company will continue with the incentive program.67)What level of significance did management use?67)Answer:΅=1-0.952=0.0252.5%Explanation:29
Researchers conduct a study to test a potential side effect of a new allergy medication. A random sample of 160 subjects withallergies was selected for the study. The new “improved” Brand I medication was randomly assigned to 80 subjects, and thecurrent Brand C medication was randomly assigned to the other 80 subjects. 14 of the 80 patients with Brand I reporteddrowsiness, and 22 of the 80 patients with Brand C reported drowsiness.68)Compute a 95% confidence interval for the difference in proportions of subjects reportingdrowsiness. Show all steps.68)Answer:^ ^
^ ^
^ ^
^ ^ ^ ^
^ ^
^ ^Conditions:* Randomization Condition: The treatments were randomly assigned to subjects.* 10% Condition: The subjects were randomly selected. We assume it is from a largepopulation ofallergy sufferers.* Independent samples condition: The two groups are independent of each otherbecause the treatments were assigned at random.* Success/Failure Condition: For Brand C, 22 were drowsy and 58 were not. ForBrand I, 14 weredrowsy and 66 were not. The observed number of both successes and failures inboth groups is larger than 10.Because the conditions are satisfied, we can model the sampling distribution of thedifference inproportions with a Normal model.We know:nI=80,pI=0.175,nC=80,pC=0.275We estimateSD(pC-pI) asSE(pC-pI)=pCqCnC+pIqInI=(0.275)(0.725)80+(0.175)(0.825)80=0.066ME=z*×SE(pC-pI)=1.96(0.066)=0.128The observed difference in sample proportions=pC-pI=0.275-0.175=0.10, sothe 95% confidence interval is 0.10±0.128, or (-0.028, 0.228)We are 95% confident that the difference between the population proportions ofpatients that reported drowsiness for Brand C and Brand I is between-2.8% and22.8%.Explanation:The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Cresttoothpaste. A marketing director suggests that the company invest in a new marketing campaign which will includeadvertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for onemonth to determine if the market share increases with new labels.69)In this context describe a Type II error and the impact such an error would have on thecompany.69)Answer:A Type II error would be deciding the proportion of people who will buy Cresttoothpaste won’t go up when in fact it would have. The company would miss anopportunity to increase sales.Explanation:30
70)Herpetologists (snake specialists) found that a certain species of reticulated python has anaverage length of 20.5 feet with a standard deviation of 2.3 feet. The scientists collect arandom sample of 30 adult pythons and measure their lengths. In their sample the meanlength was 19.5 feet long. One of the herpetologists fears that pollution might be affectingthe natural growth of the pythons. Do you think this sample result is unusually small?Explain.70)Answer:We have a random sample of adult pythons drawn from a much larger population.With a sample size of 30, the CLT says that the approximate sampling model forsample means will beN(20.5, 0.42).A sample mean of only 19.5 feet is about 2.38standard deviations below what we expect. The sample mean of 19.5 feet isunusually small.Explanation:A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concernedthat employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defectivechips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentiveprogram is instituted for one month. If successful, the company will continue with the incentive program.71)In this context describe a Type II error and the impact such an error would have on thecompany.71)Answer:A Type II error would be deciding the percentage of defective chips has notdecreased, when in fact it has. The company would miss an opportunity to decreasethe defect rate of the chips.Explanation:Researchers conduct a study to test a potential side effect of a new allergy medication. A random sample of 160 subjects withallergies was selected for the study. The new “improved” Brand I medication was randomly assigned to 80 subjects, and thecurrent Brand C medication was randomly assigned to the other 80 subjects. 14 of the 80 patients with Brand I reporteddrowsiness, and 22 of the 80 patients with Brand C reported drowsiness.72)Would you make the same conclusion as question 2 if you conducted a hypothesis test?Explain.72)Answer:A hypothesis test at a 0.05 level should reach the same conclusion if a two-sided testis used. If thealternate hypothesis were “less drowsy”, the significance level would need to bechanged, or adifferent conclusion could occur.Explanation:31
A company claims to have invented a hand-held sensor that can detect the presence of explosives inside a closed container.Law enforcement and security agencies are very interested in purchasing several of the devices if they are shown to performeffectively. An independent laboratory arranged a preliminary test. If the device can detect explosives at a rate greater thanchance would predict, a more rigorous test will be performed. They placed four empty boxes in the corners of an otherwiseempty room. For each trial they put a small quantity of an explosive in one of the boxes selected at random. The company’stechnician then entered the room and used the sensor to try to determine which of the four boxes contained the explosive.The experiment consisted of 50 trials, and the technician was successful in finding the explosive 16 times. Does this indicatethat the device is effective in sensing the presence of explosives, and should undergo more rigorous testing?73)Test an appropriate hypothesis and state your conclusion.73)Answer:^Hypotheses:H0:p=0.25. The device can detect explosives at the same level asguessing.HA:p>0.25. The device can detect explosives at a level greaterthan chance.Model: OK to use a Normal model because trials are independent (box is randomlychosen each time), andnp=12.5,nq=37.5. Do a 1-proportion z-testMechanics:p=0.32,z=1.14,P=0.13Conclusion: With aP-value so high I fail to reject the null hypothesis. This test doesnot provide convincing evidence that the sensor can detect the presence ofexplosives inside a box.Explanation:The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of femalesin the labor force is the same in the United States. Representatives from the United States Department of Labor plan to checka random sample of over 10,000 employment records on file to estimate a percentage of females in the United States laborforce.74)They actually select a random sample of 525 employment records, and find that 229 of thepeople are females. Create the confidence interval.74)Answer:^ ^ ^ ^ ^
^
^We have a random sample of less than 10% of the employment records, with 229successes (females) and 296 failures (males), so a Normal model applies.n=525,p=0.436 andq=0.564, soSE(p)=pqn=(0.436)(0.564)525=0.022margin of error:ME=z*×SE(p)=(1.645)(0.022)=0.0362.Confidence interval:p±ME=0.436±0.0362 or (0.3998, 0.4722)Explanation:32
In 2000, the United Nations claimed that there was a higher rate of illiteracy in men than in women from the country ofQatar. A humanitarian organization went to Qatar to conduct a random sample. The results revealed that 45 out of 234 menand 42 out of 251 women were classified as illiterate on the same measurement test. Do these results indicate that the UnitedNations findings were correct?75)Find a 95% confidence interval for the difference in the proportions of illiteracy in men andwomen from Qatar. Interpret your interval.75)Answer:^ ^
^ ^
^ ^
^ ^ ^ ^
^ ^
^ ^With the conditions satisfied (from Problem 1), the sampling distribution of thedifference inproportions is approximately Normal with a mean ofpm-pw, the true differencebetween thepopulation proportions.We can find a two-proportion z-interval.We know:nm=234,pm=0.192,nw=251,pw=0.167We estimateSD(pm-pw) asSE(pm-pw)=pmqmnm+pwqwnw=(0.192)(0.808)234+(0.167)(0.833)251=0.0349ME=z*×SE(pm-pw)=1.96(0.0349)=0.0684The observed difference in sample proportions=pm-pw=0.192-0.167=0.025, sothe 95%confidence interval is 0.025±0.0684, or-4% to 9%.We 95% confident that the proportion of illiterate men in Qatar is between4-percentage pointslower and 9-percentage points higher than the proportion of illiterate women.Explanation:A state’s Department of Education reports that 12% of the high school students in that state attend private high schools. TheState University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a randomsample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attendprivate schools.76)Explain what 90% confidence means in this context.76)Answer:If many random samples were taken, 90% of the confidence intervals producedwould contain the actual percentage of all applicants who attend private schools.Explanation:33
The owner of a small clothing store is concerned that only 28% of people who enter her store actually buy something. Amarketing salesman suggests that she invest in a new line of celebrity mannequins (think Seth Rogan modeling the latestjeans…). He loans her several different “people” to scatter around the store for a two-week trial period. The owner carefullycounts how many shoppers enter the store and how many buy something so that at the end of the trial she can decide if she’llpurchase the mannequins. She’ll buy the mannequins if there is evidence that the percentage of people that buy somethingincreases.77)Based on data that she collected during the trial period the store’s owner found that a 98%confidence interval for the proportion of all shoppers who might buy something was (27%,35%). What conclusion should she reach about the mannequins? Explain.77)Answer:Because the hypothesized value of 28% is in the confidence interval the trial resultsdo not provide convincing evidence that the mannequins would help increase sales.Explanation:A report on health care in the US said that 28% of Americans have experienced times when they haven’t been able to affordmedical care. A news organization randomly sampled 801 black Americans, of whom 38% reported that there had beentimes in the last year when they had not been able to afford medical care. Does this indicate that this problem is more severeamong black Americans?78)Explain what yourP-value means in this context.78)Answer:If the proportion of black Americans was 28%, we would almost never expect to findat least 38% of 801 randomly selected black Americans responding “yes”.Explanation:A state’s Department of Education reports that 12% of the high school students in that state attend private high schools. TheState University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a randomsample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attendprivate schools.79)Should the admissions officers conclude that the percentage of private school students intheir applicant pool is lower than the statewide enrollment rate of 12%? Explain.79)Answer:No. Since 12% lies in the confidence interval it’s possible that the percentage ofprivate school students in the applicant pool matches the statewide enrollment rate.Explanation:80)Internet accessA recent Gallup poll found that 28% of U.S. teens aged 13-17 have acomputer with Internet access in their rooms. The poll was based on a random sample of1028 teens and reported a margin of error of±3%. What level of confidence did Gallup usefor this poll?80)Answer:^ ^SinceME=z*pqn, we have 0.03=z*(0.28)(0.72)1028orz*=0.03(0.28)(0.72)1028Y2.14.Our confidence level is approximatelyP(-2.14<z<2.14)=0.9676, or 97%.Explanation:81)Baldness and heart attacksA recent medical study observed a higher frequency of heartattacks among a group of bald men than among another group of men who were not bald.Based on aP-value of 0.062 the researchers concluded there was some evidence that malebaldness may be a risk factor for predicting heart attacks. Explain in context what theirP-value means.81)Answer:If baldness is not a risk factor an observed level of heart attacks this much higher (ormore) would occur in only 6% of such samples.Explanation:34
A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concernedthat employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defectivechips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentiveprogram is instituted for one month. If successful, the company will continue with the incentive program.82)Management decided to extend the incentive program so that the decision can be made onthree months of data instead. Will the power increase, decrease, or remain the same?82)Answer:The power would increase because of the larger sample size.Explanation:83)Write the company’s null and alternative hypotheses.83)Answer:The null hypothesis is that the defect rate is 8%. The alternative hypothesis is thatthe defect rate is lower than 8%. In symbols:H0:p=0.08 andHA:p<0.08Explanation:The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of femalesin the labor force is the same in the United States. Representatives from the United States Department of Labor plan to checka random sample of over 10,000 employment records on file to estimate a percentage of females in the United States laborforce.84)Should the representatives from the Department of Labor conclude that the percentage offemales in their labor force is lower than Europe’s rate of 46%? Explain.84)Answer:No. Since 46% lies in the confidence interval, (0.3998, 0.4722), it is possible that thepercentage of females in the labor force matches Europe’s rate of 46% females in thelabor force.Explanation:85)Egg weightsThe weights of hens’ eggs are normally distributed with a mean of 56 gramsand a standard deviation of 4.8 grams. What is the probability that a dozen randomlyselected eggs weighs over 690 grams?85)Answer:Py>69012=Pz>57.5-564812=P(z>1.08)=14%Explanation:86)Approval ratingA newspaper article reported that a poll based on a sample of 800 votersshowed the President’s job approval rating stood at 62%. They claimed a margin of error of±3%. What level of confidence were the pollsters using?86)Answer:0.03=z*(0.62)(0.38)800; z*=1.75;P(-1.75<z<1.75)=92% confidenceExplanation:Researchers conduct a study to test a potential side effect of a new allergy medication. A random sample of 160 subjects withallergies was selected for the study. The new “improved” Brand I medication was randomly assigned to 80 subjects, and thecurrent Brand C medication was randomly assigned to the other 80 subjects. 14 of the 80 patients with Brand I reporteddrowsiness, and 22 of the 80 patients with Brand C reported drowsiness.87)Does the interval in question 1 provide evidence that the side effect of drowsiness isdifferent with the new medication?87)Answer:There is not sufficient evidence because 0 is contained in the interval. There may beno difference in the proportion of drowsiness reported.Explanation:35
A company manufacturing computer chips finds that 8% of all chips manufactured are defective. Management is concernedthat employee inattention is partially responsible for the high defect rate. In an effort to decrease the percentage of defectivechips, management decides to offer incentives to employees who have lower defect rates on their shifts. The incentiveprogram is instituted for one month. If successful, the company will continue with the incentive program.88)Over the trial month, 6% of the computer chips manufactured were defective. Managementdecided that this decrease was significant. Why might management might choose not topermanently institute the employee incentive program?88)Answer:Although statistically significant, the practical significance (cost of the incentiveprogram compared to the savings due to the decrease in defect rate of the chips)might not be great enough to warrant instituting the program permanently.Explanation:The board of directors for Procter and Gamble is concerned that only 19.5% of the people who use toothpaste buy Cresttoothpaste. A marketing director suggests that the company invest in a new marketing campaign which will includeadvertisements and new labeling for the toothpaste. The research department conducts product trials in test markets for onemonth to determine if the market share increases with new labels.89)Over the trial month the market share in the sample rose to 22% of shoppers. Thecompany’s board of directors decided this increase was significant. Now that they haveconcluded the new marketing campaign works, why might they still choose not to invest inthe campaign?89)Answer:Although statistically significant, the small increase in sales from 19.5% to 22%might not be enough to justify the expense of the new marketing campaignnationwide.Explanation:A statistics professor asked her students whether or not they were registered to vote. In a sample of 50 of her students(randomly sampled from her 700 students), 35 said they were registered to vote.90)According to a September 2004 Gallup poll, about 73% of 18-to 29-year-olds said thatthey were registered to vote. Does the 73% figure from Gallup seem reasonable for theprofessor’s class? Explain.90)Answer:The 73% figure from Gallup seems reasonable since 73% lies in our confidenceinterval.Explanation:36
The International Olympic Committee states that the female participation in the 2004 Summer Olympic Games was 42%,even with new sports such as weight lifting, hammer throw, and modern pentathlon being added to the Games. Broadcastingand clothing companies want to change their advertising and marketing strategies if the female participation increases at thenext games. An independent sports expert arranged for a random sample of pre-Olympic exhibitions. The sports expertreported that 202 of 454 athletes in the random sample were women. Is this strong evidence that the participation rate mayincrease?91)Test an appropriate hypothesis and state your conclusion.91)Answer:^
^Hypotheses:H0:p=0.42. The female participation rate in the Olympics is 42%.HA: p>0.42 . The female participation rate in the Olympics is greaterthan 42%.Model: Okay to use the Normal model because the sample is random, these 454athletes are less than 10% of all athletes at exhibitions, andnp=(454)(0.42)=190.68L10 andnq=(454)(0.58)=263.32L10.Use aN(0.42, 0.023) model, do a 1-proportionz-test.Mechanics:n=454, x=202,p=202454=0.445z=0.445-0.42(0.42)(0.58)454=1.09P=P(p>0.445)=Pz>1.09=0.138Conclusion: With a p-value (0.138) so large, I fail to reject the null hypothesis thatthe proportion of female athletes is 0.42. There is not enough evidence to suggestthat the proportion of female athletes will increase.Explanation:37
A statistics professor asked her students whether or not they were registered to vote. In a sample of 50 of her students(randomly sampled from her 700 students), 35 said they were registered to vote.92)Find a 95% confidence interval for the true proportion of the professor’s students who wereregistered to vote. (Make sure to check any necessary conditions and to state a conclusionin the context of the problem.)92)Answer:^ ^ ^ ^ ^ ^
^ ^We have a random sample of less than 10% of the professor’s students, with 35expected successes (registered) and 15 expected failures (not registered), so aNormal model applies.n=50,p=3550=0.70,q=1-p=0.30, soSE(p)=pqn=(0.70)(0.30)50=0.065Our 95% confidence interval is:p±z*SE(p)=0.70±1.96(0.065)=0.70±0.127=0.573 to 0.827We are 95% confident that between 57.3% and 82.7% of the professor’s students areregistered to vote.Explanation:93)It is generally believed that electrical problems affect about 14% of new cars. Anautomobile mechanic conducts diagnostic tests on 128 new cars on the lot.a. Describe the sampling distribution for the sample proportion by naming the model andtelling its mean and standard deviation. Justify your answer.b. Sketch and clearly label the model.c. What is the probability that in this group over 18% of the new cars will be found to haveelectrical problems?93)Answer:^
^
^
^a. We can assume these cars are a representative sample of all new cars, andcertainly less than 10% of them. We expectnp=(128)(0.14)=17.92 successes(electrical problems) andnq=(128)(0.86)=110.08failures (no problems) so thesample is large enough to use the sampling modelN(0.14, 0.031).SD(p)=pqn=(0.14)(0.86)128=0.031b.c.z=p-pSD(p)P(p>0.18)=P(z>0.18-0.140.031)=P(z>1.30)=0.096, about 10%Explanation:38
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.94)We are about to test a hypothesis using data from a well-designed study. Which is true?I. A smallP-value would be strong evidence against the null hypothesis.II. We can set a higher standard of proof by choosing ΅=10% instead of 5%.III. If we reduce the alpha level, we reduce the power of the test.94)A)NoneB)I and III onlyC)II onlyD)III onlyE)I onlyAnswer:BExplanation:A)B)C)D)E)95)We have calculated a confidence interval based on a sample of sizen=100. Now we want to get abetter estimate with a margin of error that is only one-fourth as large. How large does our newsample need to be?95)A)50B)1600C)25D)400E)200Answer:BExplanation:A)B)C)D)E)96)We are about to test a hypothesis using data from a well-designed study. Which is true?I. A large P-value would be strong evidence against the null hypothesis.II. We can set a higher standard of proof by choosing ΅=10% instead of 5%.III. If we reduce the risk of committing a Type I error, then the risk of a Type II error will alsodecrease.96)A)I and II onlyB)II onlyC)III onlyD)noneE)I onlyAnswer:DExplanation:A)B)C)D)E)39
97)We have calculated a 95% confidence interval and would prefer for our next confidence interval tohave a smaller margin of error without losing any confidence. In order to do this, we canI. change thez* value to a smaller number.II. take a larger sample.III. take a smaller sample.97)A)I onlyB)I and IIC)II onlyD)I and IIIE)III onlyAnswer:CExplanation:A)B)C)D)E)98)Which is true about a 99% confidence interval based on a given sample?I. The interval contains 99% of the population.II. Results from 99% of all samples will lie in this interval.III. The interval is wider than a 95% confidence interval would be.98)A)III onlyB)I onlyC)II onlyD)noneE)II and III onlyAnswer:AExplanation:A)B)C)D)E)99)A certain population is strongly skewed to the right. We want to estimate its mean, so we willcollect a sample. Which should be true if we use a large sample rather than a small one?I. The distribution of our sample data will be closer to normal.II. The sampling model of the sample means will be closer to normal.III. The variability of the sample means will be greater.99)A)II onlyB)II and III onlyC)I and III onlyD)I onlyE)III onlyAnswer:AExplanation:A)B)C)D)E)40
100)Which of the following is true about Type I and Type II errors?I. Type I errors are always worse than Type II errors.II. The severity of Type I and Type II errors depends on the situation being tested.III. In any given situation, the higher the risk of Type I error, the lower the risk of Type II error.100)A)I onlyB)II and IIIC)II onlyD)I and IIIE)III onlyAnswer:BExplanation:A)B)C)D)E)101)Which is true about a 98% confidence interval for a population proportion based on a givensample?I. We are 98% confident that other sample proportions will be in our interval.II. There is a 98% chance that our interval contains the population proportion.III. The interval is wider than a 95% confidence interval would be.101)A)III onlyB)I and IIC)NoneD)II onlyE)I onlyAnswer:AExplanation:A)B)C)D)E)102)Not wanting to risk poor sales for a new soda flavor, a company decides to run one more taste teston potential customers, this time requiring a higher approval rating than they had for earlier tests.This higher standard of proof will increaseI. the risk of Type I errorII. the risk of Type II errorIII. power102)A)I and IIB)I and IIIC)I onlyD)II onlyE)III onlyAnswer:DExplanation:A)B)C)D)E)103)The manager of an orchard expects about 70% of his apples to exceed the weight requirement for”Grade A” designation. At least how many apples must he sample to be 90% confident ofestimating the true proportion within±4%?103)A)19B)356C)505D)89E)23Answer:BExplanation:A)B)C)D)E)41
104)A certain population is bimodal. We want to estimate its mean, so we will collect a sample. Whichshould be true if we use a large sample rather than a small one?I. The distribution of our sample data will be clearly bimodal.II. The sampling distribution of the sample means will be approximately normal.III. The variability of the sample means will be smaller.104)A)II onlyB)I onlyC)II and IIID)I, II, and IIIE)III onlyAnswer:DExplanation:A)B)C)D)E)105)We have calculated a confidence interval based upon a sample ofn=200. Now we want to get abetter estimate with a margin of error only one fifth as large. We need a new sample withnat least…105)A)450B)40C)5000D)240E)1000Answer:CExplanation:A)B)C)D)E)106)A certain population is strongly skewed to the left. We want to estimate its mean, so we collect asample. Which should be true if we use a large sample rather than a small one?I. The distribution of our sample data will be more clearly skewed to the left.II. The sampling model of the sample means will be more skewed to the left.III. The variability of the sample means will greater.106)A)III onlyB)I onlyC)I and III onlyD)II and III onlyE)II onlyAnswer:BExplanation:A)B)C)D)E)42
107)We have calculated a confidence interval based on a sample ofn=180. Now we want to get a betterestimate with a margin of error only one third as large. We need a new sample withnat least…107)A)60B)1620C)20D)540E)312Answer:BExplanation:A)B)C)D)E)108)Which is true about a 95% confidence interval based on a given sample?I. The interval contains 95% of the population.II. Results from 95% of all samples will lie in the interval.III. The interval is narrower than a 98% confidence interval would be.108)A)II onlyB)III onlyC)II and III onlyD)NoneE)I onlyAnswer:BExplanation:A)B)C)D)E)43
Answer KeyTestname: PART51)B2)E3)C4)D5)A6)E7)E8)E9)A10)D11)E12)D13)C14)A15)C16)^ ^SinceME=z*pqn, we have 0.03=z*(0.58)(0.42)1150or z*Y2.06. Confidence level is 96%.17)Although statistically significant, the small increase in sales from 28% to 30% of shoppers might not be enough tojustify the expense of purchasing the mannequins.18)The confidence interval contains values that are all below the hypothesized value of 8%, so the data provideconvincing evidence that the incentive program lowers the defect rate of the computer chips.19)The smaller hospital should experience greater variability in the weekly percentage of male babies. We’d expect thelarger hospital to stay closer to the expected 50-50 ratio.20)We are 90% confident that between 7.9% and 12.6% of the applicants attend private high schools.21)Two methods are shown below to solve this problem:Method 1:Let P=one can of pumpkin pie mix and T=four cans of pumpkin pie mix.We are told that the contents of the cans are normally distributed, and can assume that the content amounts areindependent from can to can.E(T)=E(P1+P2+P3+P4)=E(P1)+E(P2)+E(P3)+E(P4)=120 ouncesSince the content amounts are independent,Var(T)=Var(P1+P2+P3+P4)=Var(P1)+Var(P2)+Var(P3)+Var(P4)=16SD(T)=Var(T)=16=4 ouncesWe model T withN(120, 4)z=126-1204=1.5P=P(T>126)=P(z>1.5)=0.067There is a 6.7% chance that four randomly selected cans of pumpkin pie mix contain more than 126 ounces.Method 2:Using the Central Limit Theorem approach, lety=average content of cans in sampleSince the contents are Normally distributed,yis modeled byN30,24.Py>1264=P(y>31.5)=Pz>31.5-301=P(z>1.5)=0.067There is about a 6.7% chance that 4 randomly selected cans will contain a total of over 126 ounces.44
Answer KeyTestname: PART522)^ ^ ^
^ ^H0:p1-p2=0HA:p1-p2>0People were randomly assigned to groups, we assume the groups are independent, and 20, 34, 13, 33 are allL10. OK todo a 2-proportion z-test.p1=0.370,p2=0.283,p=20+1354+46=0.33z=(0.370-0.283)-0(0.33)(0.67)54-(0.33)(0.67)46=0.93P=P(p1-p2>0.087)=P(z>0.93)=0.176We fail to reject the null hypothesis becausePis very large. We do not have evidence that the help program isbeneficial, so it should not be funded.23)^ ^
^ME=z*pqn0.04=1.645(0.12)(0.88)nn=1.645(0.12)(0.88)0.04n=178.60Y179They should sample at least 179 applicants. (423 ifp=0.5 is used)24)If birth weight was not a risk factor for susceptibility to depression, an observed difference in incidence of depressionthis large (or larger) would occur in only 2.48% of such samples.25)A Type I error would be deciding the percentage of customers who’ll make purchases will go up when in fact it won’t.The store’s owner would waste money buying useless mannequins.26)We are 90% confident that between 40.0% and 47.2% of the employment records from the United States labor force arefor females.27)^ ^
^ ^
^ ^
^ ^ ^ ^
^ ^
^ ^With the conditions satisfied (from Problem 1), the sampling distribution of the difference inproportions is approximately Normal with a mean ofpB-pC, the true difference between thepopulation proportions. We can find a two-proportion z-interval.We know:nB=1009,pB=0.81,nC=1004,pC=0.86We estimateSD(pB-pC) asSE(pB-pC)=pBqBnB+pCqCnC=(0.81)(0.19)1009+(0.86)(0.14)1004=0.0165ME=z*×SE(pB-pC)=2.576(0.0165)=0.0425The observed difference in sample proportions=pB-pC=0.81-0.186=-0.05, so the 99%confidence interval is-0.05±0.0425, or-9.3% to-0.8%.We are 99% confident that the proportion of Britons who read at least one book in the past year isbetween 0.8-percentage points and 9.3-percentage points lower than the proportion of Canadians who read at leastone book in the past year.28)The null hypothesis is that 19.5% of all people who use toothpaste buy Crest. The alternative hypothesis is that thepercentage of all people who use toothpaste who use Crest is greater than 19.5%. In symbols:H0:p=0.195 andHA:p>0.19545
Answer KeyTestname: PART529)^
^Hypotheses:H0:p=0.28. The proportion of all black Americans that were unable to afford medical care in the lastyear is 28%.HA:p>0.28. The proportion of all black Americans that were unable to afford medical care in the lastyear is greater than 28%.Model: Okay to use the Normal model because the sample is random, these 801 black Americans are less than 10% ofall black Americans, andnp=(801)(0.28)=224.28L10 andnq=(801)(0.72)=576.72L10.We will do a one-proportionz-test.Mechanics:n=801,p=0.38z=0.38-0.28(0.28)(0.72)801P(p>0.38)=Pz>6.29Y0Conclusion: With a P-value so small (just about zero), I reject the null hypothesis. There is enough evidence to suggestthat the proportion of black Americans who were not able to afford medical care in the past year is more than 28%.30)^a. We can assume these kids are a random sample of all children, and certainly less than 10% of them. Weexpectnp=(133)(0.12)=15.96 successes and 117.04 failures so the sample size is large enough to use thesampling modelN(0.12, 0.028).b.c.z=0.15-0.120.028=1.07P(p>0.15)=P(z>1.07)=0.14231)If many random samples were taken, 90% of the confidence intervals produced would contain the actual percentage ofall female employment records in the United States labor force.32)΅=1-0.982=0.01=1%33)A Type II error would be deciding the percentage of customers who’ll make purchases won’t go up when in fact itwould have. The store’s owner would miss an opportunity to increase sales.46
Answer KeyTestname: PART534)^
^Hypothesis:H0:p=0.50HA:p>0.50Plan: Okay to use the Normal model because the trials are independent (random sample of U.S. adults), these 1003 U.S.adults are less than 10% of all U.S. adults, andnp0=(1003)(0.50)=501.5L10andnq0=(1003)(0.50)=501.5L10.We will do a one-proportion z-test.Mechanics:SD(p0)=p0q0n=(0.50)(0.50)1003=0.0158; sample proportion:p=0.55P(p>0.55)=P(z>0.55-0.500.0158)=P(z>3.16)=0.0008With a P-value of 0.0008, I reject the null hypothesis. There is strong evidence that the proportion of U.S. adults whofeel they get enough sleep is more than 50%.35)The power would increase because of the larger sample size.36)A Type I error would be deciding the percentage of defective chips has decreased, when in fact it has not. Thecompany would waste money on a new incentive program that does not decrease the defect rate of the chips.37)One-tail, upper tail test. We are concerned that the proportion of people who are not able to afford medical care ishigher among black Americans.38)Two methods can be used to solve this problem:Method 1:Let B=weight of one box of cereal and T=weight of 12 boxes of cereal. We are told that the contents of the boxes areapproximately Normal, and we can assume that the content amounts are independent from box to box.E(T)=E(B1+B2+…+B12)=E(B1)+E(B2)+…+E(B12)=156 ouncesSince the content amounts are independent,Var(T)=Var(B1+B2+…+B12)=Var(B1)+Var(B2)+…+Var(B12)=3SD(T)=Var(T)=3=1.73 oz.We model T withN(156, 1.73).z=160-1561.73=2.31 andP(T>160)=P(z>2.31)=0.0104There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160 ounces.Method 2:Using the Central Limit Theorem approach, lety=average content of boxes in the case. Since the contents areNormally distributed,yis modeled byN13,0.512.Py>16012=P(y>13.33)=Pz>13.33-130.512=P(z>2.31)=0.0104.There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160 ounces.39)The confidence interval contains the current value of 19.5%, so the product trials do not provide convincing evidencethat the new marketing campaign would increase sales.40)The smaller aquarium would experience more variability in the season percentage of male births. We would expect thelarger aquarium to stay more consistent and closer to the 50-50 ratio for gender births.41)We have a random sample of less than 10% of the applicants, with 46 successes and 404 failures, so a Normal modelapplies. The confidence interval is (0.079, 0.126).42)Advantage: the test would have greater power to detect a positive effect of the mannequins. Disadvantage: she’d bemore likely to think the mannequins were effective even if they were not.47
Answer KeyTestname: PART543)There is no probability involved-once the interval is constructed, the true proportion of the professor’s students whowere registered to vote is in the interval or it is not.44)^ ^
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^ ^a. Conditions:* Randomization Condition: We are told that we have random samples.* 10% Condition: We have less than 10% of all men and less than 10% of all women.* Independent samples condition: The two groups are clearly independent of each other.* Success/Failure Condition: Of the men, 88 exercise regularly and 62 do not; of the women, 130 exercise regularly and70 do not. The observed number of both successes and failures in both groups is at least 10.With the conditions satisfied, the sampling distribution of the difference in proportions is approximately Normal witha mean ofpM-pW, the true difference between the population proportions. We can find a two-proportionz-interval.We know:nM=150,pM=88150=0.587,nW=200,pW=130200=0.650We estimateSD(pM-pW) asSE(pM-pW)=pMqMnM+pWqWnW=(0.587)(0.413)150+(0.65)(0.35)200=0.0525ME=z*×SE(pM-pW)=1.96(0.0525)=0.1029The observed difference in sample proportions=pM-pW=0.587-0.650=-0.063, so the 95%confidence interval is-0.063±0.1029, or-16.6% to 4.0%.We are 95% confident that the proportion of women who exercise regularly is between 4.0% lower and 16.6% higherthan the proportion of men who exercise regularly.b. Since zero is contained in my confidence interval, I cannot say that a higher proportion of women than men exerciseregularly. My confidence interval does not support my friend’s claim.48
Answer KeyTestname: PART545)^ ^ ^
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^ ^H0:pm-pw=0,HA:pm-pw>0, wherem=men andw=women* Randomization Condition: The men and women in the sample were randomly selected by thehumanitarian organization.* 10% Condition: The number of men and women in Qatar is greater than 2340 (10×234) and2510 (10×251), respectively.* Independent samples condition: The two groups are independent of each other because thesamples were selected at random.* Success/Failure Condition: In men, 45 were illiterate and 189 were not. In women, 42 wereilliterate and 209 were not. The observed number of both successes and failures in both groups is larger than 10.Because the conditions are satisfied, we can model the sampling distribution of the difference in proportions with aNormal model. We can perform a two-proportion z-test.We know:nm=234,pm=0.192,nw=251,pw=0.167, andppooled=45+42234+251=0.179.SEpooled(pm-pw)=ppooledqpoolednm+ppooledqpoolednw=(0.179)(0.821)234+(0.179)(0.821)251=0.0349The observed difference in sample proportions=pm-pw=0.192-0.167=0.025z=(pm-pw)-0SEpooled(pm-pw)=0.025-00.0349=0.716P=P(z>0.716)=0.24The P-value of 0.24 is high, so we fail to reject the null hypothesis. There is insufficient evidence to conclude thatilliteracy rate in men is higher than for women in Qatar.46)^a. We want to find the probability that no more than 12 students in the class will say that they frequently experiencestress. This is the same as asking the probability of finding less than 26.7% of “stressed” students in a class of 45students.Check the conditions:1. 10% condition: 45 students is less than 10% of all students who could take the class2. Success/failure cond.:np=45(0.33)=14.85,nq=45(0.67)=30.15, which both exceed 10We can use theN(0.33(0.33)(0.67)45=0.070) to model the sampling distribution.We need to standardize the 26.7% and then find the probability of getting az-score less than or equal to the one wefind:z=0.267-0.330.070=-0.90P(p<0.267)=P(z<-0.90)=0.1841, so the probability is about 18.4% that no more than 12 students will say that theyfrequently experience stress in their daily lives.b. From part a, we can use N(0.33, 0.070) to model the sampling distribution. Twenty students is about 44.4% of theclass. This is about 1.63 standard deviations above what we would expect, which is not a surprising result.49
Answer KeyTestname: PART547)If many random samples were taken, 95% of the confidence intervals produced would contain the actual percentage ofthe professor’s students who are registered to vote.48)We have a random sample of frogs drawn from a much larger population. With a sample of size 50 the CLT says thatthe approximate sampling model for sample means will be N(118, 1.98). A sample mean of only 110 grams is about 4standard deviations below what we expect, a very unusual result.49)Even if the device actually performs no better than guessing, we could expect to find the explosives 16 or more timesout of 50 about 13% of the time.50)Advantage: management would be less likely to conclude the incentive program was effective if it really were not.Disadvantage: the test would have less power to detect a positive effect of the new incentive program.51)^ ^
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^ ^We want to know whether the percentage of people who paid lower taxes was different based on whether they soughttax advice or not.H0:padvice-pno advice=0,HA:padvice-pno advice>0Conditions:* Independence: The people who filed tax reports were randomly selected and do not influence each other.* Random Condition: The people who filed tax reports were randomly selected.* 10% Condition: 72 is less than 10% of people who didn’t get tax advice, and 105 is less than 10% of people who didget tax advice.* Success/Failure: All observed counts (48, 19, 24, and 86) are at least 10.Because all conditions have been satisfied, we can model the sampling distribution of the difference in proportionswith a Normal model. We can perform a two-proportionz-test.Let ‘Advice’ group be the people who sought tax advice and the ‘No advice’ group be the people who did not seek taxadvice.We know:nadvice=105,nno advice=72,padvice=0.819,pno advice=0.333.ppooled=24+8672+105=0.621SEpooled(padvice-pno advice)=(0.62)(0.38)72+(0.62)(0.38)105=0.074The observed difference in sample proportions ispadvice-pno advice=0.819-0.333=0.486z=(padvice-pno advice)-0SEpooled(padvice-pno advice)=0.49-00.074=6.62P=P(z>6.62)<0.0001The P-value is small, so we reject the null hypothesis. There is strong evidence of a difference in the tax percentagespaid between the group who has tax advice and the group who did not have tax advice. It appears that if you have taxadvice you are more likely to pay a lower percentage of taxes to the government.52)^ ^ME=z*pqn0.05=1.645(0.46)(0.54)nn=1.645(0.46)(0.54)0.05n=268.87Y269They should sample at least 269 employment records.50
Answer KeyTestname: PART553)^ ^
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^ ^H0:pB-pC=0,HA:pB-pC<0, where B=Britons and C=CanadiansConditions:* Randomization Condition: The Britons and Canadians were randomly sampled by Gallup.* 10% Condition: The number of Britons and Canadians is greater than 10,090 (10×1009) and10,040 (10×1004), respectively.* Independent samples condition: The two groups are clearly independent of each other.* Success/Failure Condition: Of the Britons, approximately 817 read at least one book and 192 did not; of theCanadians, approximately 863 read at least one book and 141 did not. The observed number of both successes andfailures in both groups is larger than 10.Because the conditions are satisfied, we can model the sampling distribution of the differencein proportions with a Normal model. We can perform a two-proportion z-test.We know:nB=1009,pB=0.81,nC=1004,pC=0.86, andppooled=817+8631009+1004=16802013=0.835SEpooled(pB-pC)=ppooledqpoolednB+ppooledqpoolednC=(0.835)(0.165)1009+(0.835)(0.165)1004=0.0165The observed difference in sample proportions=pB-pC=0.81-0.86=-0.05z=(pB-pC)-0SEpooled(pB-pC)=-0.05-00.0165=-3.03, so theP-value=P(z<-3.03)=0.0012The P-value of 0.0012 is low, so we reject the null hypothesis.There is strong evidence that the percentage of Britonswho read at least one book in the past year is less than the percentage of Canadians who read at least one book in thepast year.54)One-tail, upper tail. The device is effective only if it can detect explosives at a rate higher than chance (25%).55)The power of the test will increase.56)One-tail, upper test. The companies will change strategies only if there is strong evidence of an increase in femaleparticipation rate from the current rate of 42%.57)΅=1-0.982=0.01=1%58)Advantage: the test would have greater power to detect a positive effect of the new marketing campaign.Disadvantage: they would be more likely to think the marketing campaign was effective even if it were not.51
Answer KeyTestname: PART559)^ ^ ^ ^
^a. Conditions:* Independence: We assume that one trucker’s driving times do not influence other trucker’s driving times.* Random Condition: We assume that trucks are stopped at random.* 10% Condition: This sample of 348 truckers is less than 10% of all truckers.* Success/Failure: 49 tickets and 299 tickets are both at least 10, so our sample is large enough.Under these conditions the sampling distribution of the proportion can be modeled by a Normal model. We will find aone-proportionz-interval.We known=348 andp=0.14 , soSE(p)=pqn=(0.14)(0.86)348=0.0186The sampling model is Normal, for a 95% confidence interval the critical value isz*=1.96.The margin of error isME=z*×SE(p)=1.96(0.0186)=0.0365.The 95% confidence interval is 0.14±0.0365 or (0.1035, 0.1765).We are 95% confident that between 10.4% and 17.7% of truck drivers have driven too many hours in a day.b. If we repeated the sampling and created new confidence intervals many times we would expect about 95% of thoseintervals to contain the actual proportion of truck drivers that have driven too many hours in a day.60)Check the conditions:1. Random sampling condition: We have been told that this is a random sample.2. Independence assumption: It’s reasonable to think that the scores of the 25 students are mutually independent.3. 10% condition: 25 students is certainly less than 10% of all students who took the exam.We’re assuming that the model for composite ACT scores has mean μ=21.4 and standard deviationΗ=1.05.Since thesample size is large enough and the distribution of ACT scores is most likely unimodal and symmetric, CLT allows usto describe the sampling distribution ofywith a Normal model with mean 21.4 andSD=(y)=1.0525=0.21.An average score of 22 isz=22-21.40.21=2.86 SDs above the mean.P(x>22)=P(Z>2.86)=0.0021, so the probability that the average composite ACT score for a sample of 25 randomlyselected students is 22 or more is 0.0021.61)The smaller poll would have more variability and would thus be more likely to vary from the actual approval rating of54%. We would expect the larger poll to be more consistent with the 54% rating. So, it is more likely that the smallerpoll would report that the President’s approval rating is below 50%.62)ME=z*pqn0.04=1.96(0.73)(0.27)nn=(1.96)2(0.73)(0.27)(0.04)2=473.24n=474Note: Since there are only 700 students in the professor’s class, she cannot sample this many students without violatingthe 10% condition!63)A Type I error would be concluding the proportion of people who will buy Crest toothpaste will go up when in fact itwon’t. The company would waste money on a new marketing campaign that will not increase sales.64)If the proportion of female athletes has not increased, we could expect to find at least 202 females of 454 pre-Olympicathletes about 13.8% of the time.65)The null hypothesis is that the percentage of all customers who buy something is 28%. The alternative hypothesis isthat the percentage of all customers who buy something is greater than 28%. In symbols:H0:p=0.28HA:p>0.2852
Answer KeyTestname: PART566)^a. We assume the cars stopped are representative of all cars in the area, and that 628<10% of the cars. The 22 successes(expired stickers) and 608 failures are both greater than 10. It’s okay to use a Normal model and find a one-proportionz-interval.p=0.035;0.035±1.96(0.035)(0.965)628=(0.021, 0.049)The interval (0.021,0.049) means that we are 95% confident that between 2% and 5% of local cars have expired safetyinspection stickers.b. If we repeated the sampling and created new confidence intervals many times we’d expect about 95% of thoseintervals to contain the actual proportion of cars with expired stickers.67)΅=1-0.952=0.0252.5%68)^ ^
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^ ^Conditions:* Randomization Condition: The treatments were randomly assigned to subjects.* 10% Condition: The subjects were randomly selected. We assume it is from a large population ofallergy sufferers.* Independent samples condition: The two groups are independent of each other because the treatments were assignedat random.* Success/Failure Condition: For Brand C, 22 were drowsy and 58 were not. For Brand I, 14 weredrowsy and 66 were not. The observed number of both successes and failures in both groups is larger than 10.Because the conditions are satisfied, we can model the sampling distribution of the difference inproportions with a Normal model.We know:nI=80,pI=0.175,nC=80,pC=0.275We estimateSD(pC-pI) asSE(pC-pI)=pCqCnC+pIqInI=(0.275)(0.725)80+(0.175)(0.825)80=0.066ME=z*×SE(pC-pI)=1.96(0.066)=0.128The observed difference in sample proportions=pC-pI=0.275-0.175=0.10, so the 95% confidence interval is 0.10±0.128, or (-0.028, 0.228)We are 95% confident that the difference between the population proportions of patients that reported drowsiness forBrand C and Brand I is between-2.8% and 22.8%.69)A Type II error would be deciding the proportion of people who will buy Crest toothpaste won’t go up when in fact itwould have. The company would miss an opportunity to increase sales.70)We have a random sample of adult pythons drawn from a much larger population. With a sample size of 30, the CLTsays that the approximate sampling model for sample means will beN(20.5, 0.42).A sample mean of only 19.5 feet isabout 2.38 standard deviations below what we expect. The sample mean of 19.5 feet is unusually small.71)A Type II error would be deciding the percentage of defective chips has not decreased, when in fact it has. Thecompany would miss an opportunity to decrease the defect rate of the chips.72)A hypothesis test at a 0.05 level should reach the same conclusion if a two-sided test is used. If thealternate hypothesis were “less drowsy”, the significance level would need to be changed, or adifferent conclusion could occur.53
Answer KeyTestname: PART573)^Hypotheses:H0:p=0.25. The device can detect explosives at the same level as guessing.HA:p>0.25. Thedevice can detect explosives at a level greater than chance.Model: OK to use a Normal model because trials are independent (box is randomly chosen each time), andnp=12.5,nq=37.5. Do a 1-proportion z-testMechanics:p=0.32,z=1.14,P=0.13Conclusion: With aP-value so high I fail to reject the null hypothesis. This test does not provide convincing evidencethat the sensor can detect the presence of explosives inside a box.74)^ ^ ^ ^ ^
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^We have a random sample of less than 10% of the employment records, with 229 successes (females) and 296 failures(males), so a Normal model applies.n=525,p=0.436 andq=0.564, soSE(p)=pqn=(0.436)(0.564)525=0.022margin of error:ME=z*×SE(p)=(1.645)(0.022)=0.0362.Confidence interval:p±ME=0.436±0.0362 or (0.3998, 0.4722)75)^ ^
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^ ^With the conditions satisfied (from Problem 1), the sampling distribution of the difference inproportions is approximately Normal with a mean ofpm-pw, the true difference between thepopulation proportions.We can find a two-proportion z-interval.We know:nm=234,pm=0.192,nw=251,pw=0.167We estimateSD(pm-pw) asSE(pm-pw)=pmqmnm+pwqwnw=(0.192)(0.808)234+(0.167)(0.833)251=0.0349ME=z*×SE(pm-pw)=1.96(0.0349)=0.0684The observed difference in sample proportions=pm-pw=0.192-0.167=0.025, so the 95%confidence interval is 0.025±0.0684, or-4% to 9%.We 95% confident that the proportion of illiterate men in Qatar is between 4-percentage pointslower and 9-percentage points higher than the proportion of illiterate women.76)If many random samples were taken, 90% of the confidence intervals produced would contain the actual percentage ofall applicants who attend private schools.54
Answer KeyTestname: PART577)Because the hypothesized value of 28% is in the confidence interval the trial results do not provide convincingevidence that the mannequins would help increase sales.78)If the proportion of black Americans was 28%, we would almost never expect to find at least 38% of 801 randomlyselected black Americans responding “yes”.79)No. Since 12% lies in the confidence interval it’s possible that the percentage of private school students in the applicantpool matches the statewide enrollment rate.80)^ ^SinceME=z*pqn, we have 0.03=z*(0.28)(0.72)1028orz*=0.03(0.28)(0.72)1028Y2.14.Our confidence level is approximatelyP(-2.14<z<2.14)=0.9676, or 97%.81)If baldness is not a risk factor an observed level of heart attacks this much higher (or more) would occur in only 6% ofsuch samples.82)The power would increase because of the larger sample size.83)The null hypothesis is that the defect rate is 8%. The alternative hypothesis is that the defect rate is lower than 8%. Insymbols:H0:p=0.08 andHA:p<0.0884)No. Since 46% lies in the confidence interval, (0.3998, 0.4722), it is possible that the percentage of females in the laborforce matches Europe’s rate of 46% females in the labor force.85)Py>69012=Pz>57.5-564812=P(z>1.08)=14%86)0.03=z*(0.62)(0.38)800; z*=1.75;P(-1.75<z<1.75)=92% confidence87)There is not sufficient evidence because 0 is contained in the interval. There may be no difference in the proportion ofdrowsiness reported.88)Although statistically significant, the practical significance (cost of the incentive program compared to the savings dueto the decrease in defect rate of the chips) might not be great enough to warrant instituting the program permanently.89)Although statistically significant, the small increase in sales from 19.5% to 22% might not be enough to justify theexpense of the new marketing campaign nationwide.90)The 73% figure from Gallup seems reasonable since 73% lies in our confidence interval.55
Answer KeyTestname: PART591)^
^Hypotheses:H0:p=0.42. The female participation rate in the Olympics is 42%.HA: p>0.42 . The female participation rate in the Olympics is greater than 42%.Model: Okay to use the Normal model because the sample is random, these 454 athletes are less than 10% of all athletesat exhibitions, andnp=(454)(0.42)=190.68L10 andnq=(454)(0.58)=263.32L10.Use aN(0.42, 0.023) model, do a1-proportion z-test.Mechanics:n=454, x=202,p=202454=0.445z=0.445-0.42(0.42)(0.58)454=1.09P=P(p>0.445)=Pz>1.09=0.138Conclusion: With a p-value (0.138) so large, I fail to reject the null hypothesis that the proportion of female athletes is0.42. There is not enough evidence to suggest that the proportion of female athletes will increase.92)^ ^ ^ ^ ^ ^
^ ^We have a random sample of less than 10% of the professor’s students, with 35 expected successes (registered) and 15expected failures (not registered), so a Normal model applies.n=50,p=3550=0.70,q=1-p=0.30, soSE(p)=pqn=(0.70)(0.30)50=0.065Our 95% confidence interval is:p±z*SE(p)=0.70±1.96(0.065)=0.70±0.127=0.573 to 0.827We are 95% confident that between 57.3% and 82.7% of the professor’s students are registered to vote.56
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^a. We can assume these cars are a representative sample of all new cars, and certainly less than 10% of them. Weexpectnp=(128)(0.14)=17.92 successes (electrical problems) andnq=(128)(0.86)=110.08failures (no problems) so thesample is large enough to use the sampling modelN(0.14, 0.031).SD(p)=pqn=(0.14)(0.86)128=0.031b.c.z=p-pSD(p)P(p>0.18)=P(z>0.18-0.140.031)=P(z>1.30)=0.096, about 10%94)B95)B96)D97)C98)A99)A100)B101)A102)D103)B104)D105)C106)B107)B108)B57