Question 8.11: Refer to Example 8.10. Let 𝜇0 represent the mean permeabilit......
Refer to Example 8.10. Let \mu_{0} represent the mean permeability of skin whose resistance is 15 kΩ. Test H_{0} : \mu _{0} ≤ 9 versus H_{1} : \mu _{0} > 9.
Since \mu_{0} is the mean permeability of skin whose resistance is 15 kΩ, \mu_{0} = \beta_{0} + \beta_{1}(15). Now let ŷ =\beta_{0} + \beta_{1}(15). The quantity
\frac{\hat{y}\ −\ [\beta_{0}\ +\ \beta_{1}(15)]}{s_{\hat{y}}} = \frac{\hat{y}\ −\ \mu_{0}}{s_{\hat{y}}}
has a Student’s t distribution with n − 2 = 48 degrees of freedom. Under H_{0}, we take \mu_{0} = 9. The test statistic is therefore
\frac{\hat{y}\ -\ 9 }{s_{\hat{y}}}
We compute ŷ and s_{\hat{y}} :
ŷ = \hat{\beta}_{0} + \hat{\beta}_{1}(15) = 14.3775\ −\ 0.300094(15) = 9.8761
s_{\hat{y}} = 1.80337\sqrt{\frac{1}{50} + \frac{(15\ −\ 21.7548)^{2}}{1886.48}} = 0.37908
The value of the test statistic is
\frac{9.8761\ −\ 9}{0.37908} = 2.31
There are n − 2 = 50 − 2 = 48 degrees of freedom. This number of degrees of freedom is not found in Table A.3; however, the P-value can be determined with a calculator or computer software to be 0.0126. Alternatively, since the number of degrees of freedom is greater than 30, one can use the z table (Table A.2) to approximate the P-value as 0.0104. It is reasonable to conclude that the mean permeability is greater than 9 𝜇m/h.