STAT303: Secs 102 and 103Summer I 2000Exam #3Form AInstructor: Julie Hagen Carroll1. Don’t EVEN open this until you are told to do so.2. Be sure to mark your section number and your test form (A, B, C or D) on the scantron!3. Sign your name where indicated on your scantron and write your section number, seat number andcomputer number beside it. You will get your scantrons back tomorrow in class. You may keep thisexam.4. There are 20 multiple-choice questions on this exam, each worth 5 points. There is partial credit. Pleasemark your answers clearly on the scantron. Multiple marks will be counted wrong.5. You will have 60 minutes to finish this exam.6. If you are caught cheating or helping someone to cheat on this exam, you both will receive a grade ofzero on the exam. You must work alone.7. This exam is worth 100 points, and will constitute 20% of your final grade.8. Good luck!1STAT303: 102 and 103 Exam #3, Form A Summer 20001. Oprah wants to increase revenue. She has a lotof female targeted advertising, so her only outletis to increase the advertising targeted at males.You need at least a 20% viewing audience to at-tract advertisers, so she needs to find out if heraudience is at least 20% male. What hypothesesshould she test? Let F indicate females and Mmales.A. H0: µF= µMvs. HA: µF6= µMB. H0: πF= πMvs. HA: πF> πMC. H0: µF= µMvs. HA: µF> µMD. H0: πM= 0.20 vs. HA: πM< 0.20E. H0: πM= 0.20 vs. HA: πM> 0.202. Same scenario: Advertisers are tight with theirmoney, so they want strong proof that there isat least a 20% viewing audience. What α-levelwould the advertising company want Oprahto use?A. α = 0.05 =5% because neither Type I norType II error is critical.B. α = 0.10 =10% because Type I is morecritical.C. α = 0.01 =1% because Type I is more crit-ical.D. α = 0.10 =10% because Type II is morecritical.E. α = 0.01 =1% because Type II is more crit-ical.3. Which of the following is true?A. A p-value is how often we would get dataas contradictory as we got even though H0is true.B. A p-value is a measure of the strength ofthe evidence against the null hypothesis.C. A p-value can be used to perform a test ofhypotheses at any significance level.D. All of the above are true.E. Exactly two of the above are true.4. Which of the following is NOT a property of thet distribution?A. It is always centered at zero.B. It is always wider than the z distribution.C. It is always symmetric.D. It’s height, at center, increases as the de-grees of freedom increase.E. All of the above are properties of the t.5. What is the correct conclusion that can be madefrom the output above which is testing H0: µ =10 vs. HA: µ > 10?A. Reject the null at the 10, 5 and 1% levelsand conclude that the true mean is greaterthan 10.B. Reject the null at the 10, 5 and 1% levelsand conclude that the true mean is 10.C. Reject the null at the 10, 5 and 1% levelsand conclude that the true mean is NOT10.D. Reject the null at the 10% level ONLY andconclude that the true mean is greater than10.E. Reject the null at the 10% level ONLY andconclude that the true mean is NOT 10.6. In the test above, the sample mean,x = 12.6.What would change if we had gotten x = 11instead?A. The conclusion would be the same, sincethe hypotheses are the same.B. The p-value would be larger since the newsample mean is closer to the hypothesizedmean, µ = 10.C. The p-value would be smaller since the newsample mean is closer to the hypothesizedmean, µ = 10.D. The sign of the alternative, HA, wouldchange from > to <.E. It’s impossible to tell without rerunning thetest.2STAT303: 102 and 103 Exam #3, Form A Summer 20007. What do we mean by the term confidence in ref-erence to confidence intervals?A. We are confident that our data is random.B. We are confident that our method producesintervals that contain the parameter (1 −α)100% of the time.C. We are confident that our method pro-duces intervals that contain the parameterα ∗ 100% of the time.D. We are confident that our interval containsthe parameter.E. We are confident that our method pro-duces intervals that contain the statistic(1 − α)100% of the time.90% |Lower Limit = 6.1404199|Upper Limit = 6.859580195% |Lower Limit = 6.0689941|Upper Limit = 6.931005999% |Lower Limit = 5.9252143|Upper Limit = 7.07478578. Using the information above, what is the correctrange of the p-value if I wanted to test H0: µ = 5vs. HA: µ 6= 5?A. p-value > 0.10B. 0.10 > p-value > 0.05C. 0.05 > p-value > 0.01D. p-value< 0.01E. You need a test statistic value to determinethe p-value9. Which of the following would define a Type Ierror for the test above, H0: µ = 5 vs. HA: µ 6=5?A. We claim the true mean is 5 when it actu-ally is NOT 5.B. We claim the true mean is NOT 5 when itactually IS 5.C. We claim the true mean is NOT 5 when itactually is greater than 5.D. We conclude there is insufficient evidence toclaim the true mean is NOT 5 even thoughit actually is NOT 5.E. We conclude there is insufficient evidence toclaim the true mean is NOT 5 even thoughit actually IS 5.10. Which of the following is a TRUE statementabout hypothesis testing in general?A. If we reject H0, we conclude that the alter-native hypothesis is true.B. If we fail to reject H0, we conclude that thenull hypothesis is true.C. If we fail to reject H0, we conclude that thealternative hypothesis is false.D. We initially assume the alternative is trueand then gather data to support it.E. Two of the above are true.11. Texas A&M wants to find out whether most stu-dents are opposed to the proposed Rec Center feeincrease from $50 to $60. Which of the followingis the correct statement of the hypotheses thatA&M should test?A. H0: µ = $50 vs. HA: µ = $60B. H0: π = 0.5 vs. HA: π > 0.5C. H0: µ = 0.5 vs. HA: µ > 0.5D. H0: µ = $10 vs. HA: µ > $10E. H0: µ1− µ2= $10 vs. HA: µ1− µ2> $1012. Suppose A&M concluded that the studentsthought the increase was acceptable (less thanhalf were opposed) when in reality they onlyasked students who don’t use the Rec Center.Students using the Rec Center are strongly op-posed to the increase. What happened?A. A&M took a biased sample which causedthem to make a Type I error.B. A&M took a biased sample which causedthem to make a Type II error.C. A&M made a correct decision.D. A&M tested the wrong hypotheses.E. A&M will always go for the increase.13. Suppose you were testing H0: µ = 5 vs HA: µ >5 and ended up with a p-value = 0.03. Whichof the following is the best interpretation of
View Full Document
Unlocking...