Statistical Analysis: Confidence Intervals and Hypothesis Testing, Exams of Data Analysis & Statistical Methods

Download Statistical Analysis: Confidence Intervals and Hypothesis Testing and more Exams Data Analysis & Statistical Methods in PDF only on Docsity! 1. An interview with 221 randomly chosen investors revealed that 73 believed in “momentum investing,” investing only when the market has been going up regularly. a) Describe the parameter of interest in this situation. b) Set a 95% confidence interval on this parameter. Use the plus-4 method. ANS: .333  .062 c) How large a sample would be necessary in order that the margin of error of your 95% confidence interval be only .03? You may assume you have the results above to aid you. ANS: n = 944 d) In part c), suppose no preliminary estimate was available. Determine a sample size that would be adequate regardless of the true value of p. ANS: n= 1068 2. To estimate the average amount spent per week by students on lattes/coffee at CWU, a random sample of 64 students was taken; the summary data are recorded below.: X =$14.86, S= $3.20 a) What is the parameter of interest in this problem? Be specific! b) Find the best point estimate of this parameter. ANS: X = 14.86 c) Find a 95% confidence interval on this parameter. ANS: 14.86  .78 or the interval is 14.08 <  < 15.64 d) Test, as easily as possible, the hypothesis that average amount spent is $12 against the alternative that it is not $12. Use a 5% level of significance. ANS: reject H0: because $12 is not in the 95% C.I. Conclude  not only greater than 12, but also  > 14.08. e) The 95% margin of error in the above interval was .78. How large a sample would have been needed in order to have the margin of error of a 90% confidence interval be .75? Math 311 Fall 2008 Some Review Problems for Final ANS: Use n = 50 3. To estimate the average beginning salary for actuarial graduates in CY 2008 (who have passed two exams), a random sample of size 6 produced the following summary statistics: 4.8 344S x  (data are in $000). a) Assuming that the beginning salary of actuarial graduates follows a normal distribution, use these data to set a 90% confidence interval on the true unknown mean salary of all beginning actuarial graduates who have two exams by the time they graduate. ANS: 57.333  2.015* 4.8/ 6 =57.333  3.95, or $53383 <  < $61283 b) Test H0:  = 62000 against Ha:  < 62000. Use  = .05. ANS: Use appropriate side of your two-sided 90% C.I. as a one-sided 95% C.I. Here, it is  < $61283. Because ’62,000’ is not in this interval, we reject H0: and conclude  < 62000, but we can “sharpen” this to  < 61283. 4. 80 randomly chosen individuals who are severely overweight were put on two weight-loss diets: A and B---40 for each diet. At the conclusion of a 12-week period, the results below were obtained: Diet A Diet B X = 24.4 lbs Y = 20.8 lbs S = 2.44 S = 2.15 lbs a) Find the best point estimate for  A –  B. Ans: 3.6 lbs. b) Set a 95% confidence interval on  A –  B. ANS: 3.6  1.96 5.954 4.6225 40 40  = 3.6  1.01 or 2.59 <  A –  B < 4.61 c) Test H0:  A –  B =0 versus Ha:  A –  B 0. Use  =.05. ANS: Use your 95% C.I. Check for ‘0’ in the interval. Not there! Conclude  A –  B > 0, but better yet, conclude  A –  B > 2.59. d) Test H0:  A –  B ≤ 2.2 against Ha:  A –  B > 2.2 Use  =.025. ANS: Use your 97.5% C.I. obtained by choosing the bottom half (because it reads in the same direction as Ha: ) of the interval in part b: 2.59 <  A –  B. Because 2.2 is not in this interval, reject H0: and conclude that  A –  B > 2, and you can sharpen this to  A –  B > 2.59