Question 7.8: Refer to Example 7.7. Can you conclude that the mean number ...
Refer to Example 7.7. Can you conclude that the mean number of items identified with the new structured design exceeds that of the conventional design by more than 2?
The null and alternate hypotheses are
H_0: \mu_X-\mu_Y \leq 2 \quad \text{versus}\quad H_1: \mu_X-\mu_Y>2
We observe \bar{X}=44.1, \bar{Y}=32.3, s_X=10.09, s_Y=8.56, n_X=10 and n_Y = 10. Under H_0 , we take \mu_X-\mu_Y=2, The test statistic given by expression (7.13) is
t=\frac{(\bar{X}-\bar{Y})-\Delta_0}{\sqrt{s_X^2 / n_X+s_Y^2 / n_Y}} (7.13)
t=\frac{(\bar{X}-\bar{Y})-2}{\sqrt{s_X^2 / n_X+s_Y^2 / n_Y}}
The number of degrees of freedom is calculated in the same way as in Example 7.7, so there are 17 degrees of freedom. The value of the test statistic is t = 2.342. This is a one-tailed test. The P-value is between 0.01 and 0.025. We conclude that the mean number of items identified with the new structured design exceeds that of the conventional design by more than 2.