STAT 302 Section 503Fall 2017Exam #2Form AInstructor: Raanju SundararajanName:1. DO NOT open this until you are told to do so.2. You need a 8.5 × 11” GRAY scantron, pencil, calculator and you may have one page as a tip sheet(formula sheet).3. There are 20 multiple-choice questions on this exam, each worth 5 points. There is NO partial credit. Please markyour answers clearly. Multiple marks will be counted wrong.4. You will have 60 minutes to finish this exam.5. If you are caught cheating or helping someone to cheat on this exam, you both will receive a grade of zero on theexam. You must work alone.6. When you are finished please make sure you have filled in your name and marked your FORM (A or B) and 20answers, then turn in JUST your scantron.7. All the very best!ANSWERS: 1-B, 2-B, 3-C, 4-C, 5-E, 6-B, 7-B, 8-A, 9-E, 10-B, 11-A, 12-D, 13-A, 14-A, 15-C, 16-D, 17-A,18-A, 19-A, 20-D1STAT 302 sec 503 Exam #2, Form A Fall 20171. Researchers are testing the alternative hypothesis: µ >166.3 lb. Suppose the actual true value of the mean is170 lb. For the researcher’s experimental design (sam-ple size, etc.) the power of their test is 0.2203. Whichof the following is TRUE?A. A power of 0.2203 can be interpreted to meanthere is a 22.03% chance of selecting a samplewhose sample mean is larger than 166.3 lb.B. A power of 0.2203 can be interpreted to meanthere is a 22.03% chance of selecting a samplewhich leads to the decision to reject Ho when thetrue mean is 170 lb.C. A power of 0.2203 can be interpreted to meanthere is a 22.03% chance of selecting a samplewhich leads to the decision to fail to reject Howhen the true mean is 170 lb.D. Power equals one minus the probability of a TypeI error.E. Two of the above are true.2. While testing for a difference in mean SAT scores ofstudents in Texas and Florida, H0: µ1− µ2= 0 vsHa: µ1− µ26= 0 at 5% level, a p-value of 0.15 impliesA. The mean SAT scores of students in Texas andFlorida are significantly different.B. A 95% confidence interval for the difference in themean scores would contain 0.C. The mean SAT scores of students in Texas ishigher than Florida.D. All of the above are true.E. None of the above are true.3. Suppose that you are a writer for a university newspa-per and due to time constraints you conduct a surveyof only 50 randomly selected students. You ask them,”Will you vote during the elections this year?” and 20of them say they plan to vote. Create a 95% confi-dence interval for the proportion of students that planon voting.A. (0.29, 0.51)B. (0.74, 1)C. (0.26, 0.54)D. (0.44, 0.68)E. Cannot be determined.4. What type of problem would be used to test the follow-ing alternative hypothesis? Europeans on averagewalk less miles per day than Americans.A. Comparing two proportions from independentsamples.B. Comparing two proportions from dependent sam-ples.C. Comparing two means from independent samples.D. Comparing two medians from independent sam-plesE. None of the above.5. While testing for a difference in the proportion of stu-dents at A&M (Group 1) and UT (Group 2) who areRepublicans, H0: p1− p2= 0 vs Ha: p1− p26= 0 at5% level, a p-value of 0.068 impliesA. The proportion of students in A&M and UT whoare Republicans are significantly different.B. A 95% confidence interval for the difference in theproportion would not contain 0.C. The A&M proportion is higher than the UT pro-portion.D. Two of the above are true.E. None of the above are true.6. Do more women (w) get greater satisfaction from theirfamily life than men (m)? A survey asked 234 men and270 women if they got a very great deal of satisfactionfrom their family life. What is the correct alternativehypothesis?A. Ha: pm− pw> 0B. Ha: pm− pw< 0C. Ha: pm− pw= 0D. Ha: pm− pw6= 0E. None of the above.7. Suppose that you were interested in determining theproportion of Americans that agreed with the state-ment that The word super is used excessively inAmerica. Suppose that you know roughly half theAmericans will agree with this statement, determinethe sample size needed to estimate the proportion ofpeople that agree for the current year within 0.05 mar-gin of error at 95% confidence.A. 9604B. 384C. 4458D. 490E. 101282STAT 302 sec 503 Exam #2, Form A Fall 20178. A survey was aimed at determining if teenagers spenta different amount of money on recreation every monththan middle aged people. The output below is the re-sult of the study. What would be your p-value for atwo sided test based on the confidence interval below?The difference µ1− µ2has a 95% Confidence Intervalas (20.8, 67.7).A. The p-value would be less than 0.05.B. The p-value would be greater than 0.05.C. The p-value would be equal to 0.05.D. The p-value would be lesser than 0.01.E. None of the above.9. A recent Pew research study reports, “In 2015, 17% ofall U.S. newlyweds had a spouse of a different race orethnicity, marking more than a fivefold increase since1967, according to a new Pew Research Center analysis. Later, they report that the margin of error for thedata from 2015 was 4 percentage points (m = 0.04 forthe proportion CI) with 95% confidence. Construct a95% confidence interval for the proportion of all U.S.newlyweds marrying a spouse of a different race or eth-nicity in 2015.A. (0.07, 0.27)B. (0.09, 0.25)C. (0.10, 0.24)D. (0.11, 0.23)E. (0.13, 0.21)10. Interpret the margin of error for the study above.A. The typical difference between the sample andpopulation proportions is approximately 0.04.B. The estimated maximum distance (with 95% con-fidence) between the sample and population pro-portions is 0.04.C. The confidence interval will fail to capture thepopulation proportion 4% of the time, in repeatedsampling.D. There is a 4% chance that the population propor-tion is not in this interval.E. The confidence interval will fail to capture thesample proportion 4% of the time, in repeatedsampling.11. A study of bulimia among college women studied theconnection between childhood sexual abuse and a mea-sure of family cohesion (the higher the score, the greaterthe cohesion). The sample mean on the family cohesionscale was 1.9 for 130 sexually abused students (s1=2.1)and 4.9 for 170 nonabused students (s2=3.5). The stan-dard error for comparing the difference of means isA. 0.3255B. 2.12/130 + 3.52/170C. 1.92/130 + 4.92/170D. 0.1812E. None of the above.12. Using the data from the previous question, constructa 95% confidence interval for the difference of means.Use a z∗critical value.A.