lab4 - STAT 350 – Spring 2009 Lab#4 SOLUTION(corrected...

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Unformatted text preview: STAT 350 – Spring 2009 Lab #4 SOLUTION (corrected) t-tests All the following analyses should be conducted using SAS. For all tests, use a significance level of 0.05. Attach the SAS input (editor window content) as an appendix at the end of your assignment. For this assignment, sample means, sample standard deviations, test statistics, and p-values must be taken from the SAS output and without any additional rounding. 1. The following in a contrived example to make sure you understand how to obtain the appropriate p-value from SAS. I have arbitrarily labeled the two groups “dogs” and “cats”. Assume you wish to test the null hypothesis: µd - µc = 0, where µd denotes the true mean [whatever] for dogs and µc denotes the true mean [whatever] for cats. The data from a random sample of dogs and cats follow. dogs: 47.0 28.2 37.2 47.7 52.4 61.0 62.8 39.1 62.0 43.1 67.3 33.1 cats: 65.7 88.1 79.3 73.2 43.6 76.9 74.6 56.4 51.2 51.2 a. Paste the SAS output from the t-test as your answer to part (a). The TTEST Procedure Statistics Variable species x x x c d Diff (1-2) N 10 12 Lower CL Mean Mean Upper CL Mean Lower CL Std Dev Std Dev Upper CL Std Dev Std Err 55.512 40.261 5.3802 66.02 48.408 17.612 76.528 56.556 29.843 10.104 9.0842 10.477 14.689 12.824 13.695 26.817 21.773 19.776 4.6452 3.7019 5.8637 T-Tests Variable Method Variances x x Pooled Satterthwaite Equal Unequal DF t Value Pr > |t| 20 18.1 3.00 2.97 0.0070 0.0083 Equality of Variances Variable x Lab #4 - SOLUTION Method Folded F Num DF 9 Den DF 11 F Value 1.31 Pr > F 0.6603 Page 1 of 1 b. Complete the table below by filling in the appropriate p-value for each condition. Alternative Hypothesis µd - µc ≠ 0 same as µc - µd ≠ 0 µd - µc < 0 same as µc - µd > 0 µd - µc > 0 same as µc - µd < 0 do not assume equal variance 0.0083 assume equal variance 0.0070 0.00415 0.0035 0.99585 0.9965 Note: SAS made “cat” group #1 and “dog” group #2, because cat comes before dog alphabetically. So the test statistics (t = 3.00 and t = 2.97) given were for the null hypothesis: µc - µd = 0. The test statistics for the null hypothesis: µd - µc = 0 would have been -3.00 and -2.97. Also note that there is absolutely no evidence that µd - µc > 0 (µd > µc), because the sample mean for dogs was actually LESS THAN the sample mean for cats. If you had a small p-value (0.0035 or 0.00415) for that alternative hypothesis, you would conclude that µd - µc > 0 (µd > µc), which, logically, must be wrong! Do not disconnect your calculations from logic and reason! c. Is the assumption of equal variance appropriate? Give both the appropriate p-value and your conclusion. Use α = 0.05. Yes, the assumption of equal variance is appropriate (p-value = 0.6603) 2. The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students' study habits and attitude toward school. Scores range from 0 to 200. Higher scores indicate better attitudes. An educational psychologist believes that older students have a better attitude toward school than younger students. To test this theory, she gives the SSHA test to a group of students with ages between 18 and 21 and to another group with ages between 30 and 40. She brings the results of their tests (below) to you for help with the analysis. young: 115 119 124 124 90 86 118 79 116 161 157 108 127 108 80 130 old: 116 122 127 131 122 119 122 121 a. State the null and alternative hypotheses. Use µy to denote the true mean SSHA score for the population of “young” students and use µo to denote the true mean SSHA score for the population of “old” students. H0: µo - µy = 0 Ha: µo - µy > 0 Ha: µy - µo < 0 Another acceptable set of hypotheses: H0: µy - µo = 0 Lab #4 - SOLUTION Page 2 of 2 b. Conduct a t-test, and copy and paste the output of the analysis onto your assignment as your answer to part (b). The TTEST Procedure Statistics Variable age N score score score old young Diff (1-2) 8 16 Lower CL Mean Mean Upper CL Mean Lower CL Std Dev Std Dev Upper CL Std Dev Std Err 118.63 102.42 -10.45 122.5 115.13 7.375 126.37 127.83 25.204 3.0606 17.607 15.355 4.6291 23.835 19.854 9.4215 36.89 28.1 1.6366 5.9588 8.5969 T-Tests Variable Method Variances score score Pooled Satterthwaite Equal Unequal DF t Value Pr > |t| 22 17.1 0.86 1.19 0.4002 0.2489 Equality of Variances Variable Method score Folded F Num DF Den DF F Value Pr > F 15 7 26.51 0.0002 c. Is the assumption of equal variances appropriate? Give both the appropriate p-value and your conclusion. No, the assumption of equal variance is NOT appropriate (p-value = 0.0002) d. What was the standard deviation of SSHA scores for the "young" students? What was the standard deviation of SSHA scores for the "old" students? sy = 23.835 so = 4.6291 e. What was the mean SSHA score for the "young" students? What was the mean SSHA score for the "old" students? μy = 115.13 μo = 122.5 Lab #4 - SOLUTION Page 3 of 3 f. Give the value of the appropriate test statistic and the appropriate p-value for the test of your hypotheses in (a). The value of the appropriate test statistic, t = 1.19 p-value: 0.2489/2 = 0.12445 Note: to be technically correct, if your hypotheses were: H0: µo - µy = 0 & Ha: µo - µy > 0, then the test statistic is t = +1.19 if your hypotheses were: H0: µy - µo = 0 & Ha: µy - µo < 0, then the test statistic is t = -1.19 either way, the p-value would be 0.12455 g. What can the educational psychologist conclude based on this test? Be sure to be technically correct! Based on this test, there is no the mean SSHA scores for older students are not statistically significantly higher than the mean SSHA scores for younger students. 3. Does a pretty nurse raise blood pressure? A cardiologist suspects that measurements of his male patients' blood pressure have been artificially elevated since he hired a young, attractive nurse. To test this theory, on a particular day, he randomly selects some male patients to have their blood pressure taken by the new female nurse and the others to have their blood pressure taken by a male nurse. The resulting systolic blood pressure values are given in the table below. female nurse 130 134 146 151 152 155 142 127 139 140 139 male nurse 134 133 145 145 130 137 123 131 138 134 136 135 145 a. State the null and alternative hypotheses. Use µf to denote the true mean blood pressure when read by the female nurse and use µm to denote the true mean blood pressure when read by the male nurse H0: µf - µm = 0 Ha: µf - µm > 0 Another acceptable set of hypotheses: H0: µm - µf = 0 Lab #4 - SOLUTION Ha: µm - µf < 0 Page 4 of 4 b. Conduct a t-test, and copy and paste the output of the analysis onto your assignment as your answer to part (b). The TTEST Procedure Statistics Variable gender N BP BP BP f m Diff (1-2) 11 13 Lower CL Mean Mean Upper CL Mean Lower CL Std Dev Std Dev Upper CL Std Dev Std Err 135.29 131.96 -1.044 141.36 135.85 5.5175 147.43 139.73 12.079 6.3138 4.6088 5.9731 9.0363 6.4271 7.7232 15.858 10.609 10.931 2.7245 1.7826 3.164 T-Tests Variable Method Variances BP BP Pooled Satterthwaite Equal Unequal DF t Value Pr > |t| 22 17.7 1.74 1.69 0.0951 0.1077 Equality of Variances Variable Method BP Folded F Num DF Den DF F Value Pr > F 10 12 1.98 0.2629 c. Is the assumption of equal variances appropriate? Give both the appropriate p-value and your conclusion. Yes, the assumption of equal variances is appropriate (p-value = 0.2629) d. What was the standard deviation of the systolic readings for the patients with the female nurse? What was the standard deviation of the systolic readings for the patients with the male nurse? sf = 9.0363 sm = 6.4271 e. What was the mean systolic reading for the patients with the female nurse? What was the mean systolic reading for the patients with the male nurse? µf = 141.36 µm = 135.85 Lab #4 - SOLUTION Page 5 of 5 f. Give the value of the appropriate test statistic and the appropriate p-value for the test of your hypothesis in (a). Test statistic: t = 1.74 p-value: 0.0951/2 = 0.04755 Note: to be technically correct, if your hypotheses were: H0: µf - µm = 0 & Ha: µf - µm > 0 then the test statistic is t = +1.74 if your hypotheses were: H0: µm - µf = 0 & Ha: µm - µf < 0 then the test statistic is t = -1.74 g. What can the cardiologist conclude based on this test? Be sure to be technically correct! Based on this test, the cardiologist can conclude that the blood pressure readings for his patients are statistically significantly higher when taken by the female nurse compared to when taken by the male nurse. Lab #4 - SOLUTION Page 6 of 6 Appendix Do not forget to include your SAS code (the contents of the Editor window) as an appendix at the end of your assignment. data prob1; input species cards; d 47.0 d d 28.2 d c 65.7 c c 88.1 c ; $ x @@; 37.2 47.7 79.3 73.2 d d c c 52.4 61.0 43.6 76.9 d d c c 62.8 39.1 74.6 56.4 d 62.0 d 43.1 c 51.2 d 67.3 d 33.1 c 51.2 proc ttest data=prob1; class species; var x; run; data ssha; input age $ score cards; young 115 young 116 young 119 young 161 old 116 old 122 old 121 ; @@; young young young young 122 124 157 124 108 old 127 young young young young old 90 127 86 108 131 f f 146 127 f f 151 f 152 139 f 140 m m m 145 131 145 m m 145 138 old young young young young 122 118 80 79 130 old 119 old proc ttest data=ssha; class age; var score; run; data nurse; input gender $ BP @@; cards; f 130 f 134 f 155 f 142 f 139 m 134 m 133 m 137 m 123 m 136 m 135 ; run; m m 130 134 proc ttest data=nurse; class gender; var BP; run; Lab #4 - SOLUTION Page 7 of 7 ...
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This note was uploaded on 02/16/2010 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue.

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