Homework 7

8.1  t=2.556>2.093 therefore, reject null hypothesis at 5% level
8.2  since -2.093<1.118<2.093, we accept the null hypothesis at the 5% level
8.3  since 2.907>2.093, we reject the null hypothesis at the 5% level

8.11  Two sample T -test for independent samples with equal variance.
8.12  Since t 23, .975 = 2.069>1.49, we accept the null at the 5% level
8.13  Since t23,.9=1.319<1.49<t23,.95=1.714, it follows that 2 X (1-.95)< 
p < (1-.9) or .10 < p < .20
8.14  (-.85, 5.22)

8.25  3.250<3.37 <4.781 it follows that .001< p < .01 and the results are 
statistically significant
8.26  t=-1.83, therefore results are not statistically significant
8.27  95% CI for meth and topical group = (-2.67, -0.53).  95 CI for 
pressure change is (-1.48, 0.08).
8.28  Accept the null hypothesis and conclude that the change in occular 
pressure is not sig. Different in the two groups.

8.31  null hypothesis:  the mean difference is 0 versus the alternate 
hypothesis that the mean change is not zero
8.32  use a paired t-test to test these hypotheses
8.33  t=3.67; reject the null and conclude that aspirin a has 
significantly higher concentration in the urine.
8.34  3.60 mg%
8.35  (1.38, 5.82)
8.36  If the test result had had been sig. At the 5% level, then the CI 
would have excluded 0; otherwise, it would have included 0.  The former 
possibility is what occurred.

8.81  The appropriate test is the 2-sample t-test assuming equal variances.
8.82  T =-1.92; the results are not statistically significant, but show a 
trend toward women having higher levels of cholesterol than men.
8.83  We accept the null hypothesis given that .95>p>.975.
8.84  Assumption of indep. unlikely to hold.  Why?  You try to figure it 
out, and well give you the answer in section