OUTPUT FOR #5 The SAS system The TTEST Procedure Variable: sbp sports N Mean Std
ID: 3231322 • Letter: O
Question
OUTPUT FOR #5 The SAS system The TTEST Procedure Variable: sbp sports N Mean Std Dev Sud Em Minimum Maximum No 869 45.7 0.5469 0.3578 0,0 180.0 Yes 231 147.3 2198 0.7382 119.2 180.7 Diff (1-2) 5857 10.6914 0.7914 Mean 95% CL Mean Std Dev 95% CL Std Dev sports Method 145.7 145.0 146.4 10.5469 10 0733 11.0676 Yes 147.3 145.9 148,8 i 11.2198 10.2815 12 3481 l Diff (1-2 Pooled 5657 3.1386 -0.0328 10.6914 10.2624 1.158 Diff (1-2) Satterthwaite 5857 -3.1992 0.0277 Variances OF t Value Pr Equal Pooled 1098 2.00 0.0454 Satterthwaite Unequal 345.69 93 0.0540 Equally Vanances Method Num DF Den DF F Value Pr F Folded F 230 868 113 02251 Distribution of sbp Q-Q Plots of sbp Mao vesExplanation / Answer
A) From the t-test procedure results we can see that the p value associated with the t statistic of -2.0 is 0.0454.
As p value is less than 0.05, we can reject the null hypothesis. So at 5% significance level we can conclude that the systolic blood pressure is different for subjects who participated in sports and for those who did not.
B) In independent t test, the dependent variable should be continuous, which is fine here.
The dependent variable should be normally distributed within each group. From the histograms and qq plots, we can see that this assumption is also met.
Also the independent t test assumes homogeneity of variance assumption. For this we can see the p value associated with the F test. As p value is 0.2251, so we fail to reject the null hypothesis. That means the group variances can be treated as equal.
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