Two-Tailed Hypothesis Tests and CIs. As we mentioned on page $413$, the following relationship holds between hypothesis tests and confidence intervals: For a two-tailed hypothesis test at the significance level $\alpha$, the null hypothesis $H_0:{\mu }_1={\mu }_2$ will be rejected in favor of the alternative hypothesis $H_2:{\mu }_1\ne {\mu }_2$ if and only if the $(1-\alpha )$-level confidence interval for ${\mu }_1-{\mu }_2$ does not contain 0. In each case, illustrate the preceding relationship by comparing the results of the hypothesis test and confidence interval in the specified exercises.

a. Exercises $10.81$ and $10.87$

b. Excrcises $10.86$ and $10.92$

To discover a comparison of the results of the hypothesis test and the confidence interval.

Consider the upper bound of a confidence interval of 90%. As a result, the 90% interval does not contain zero.

$$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\frac{a}{2}}·\sqrt{\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)} \\ =(25.8-22.1)\pm 1.729·\sqrt{\left(\frac{9.2^2}{32}+\frac{5.7^2}{20}\right)} \end{gathered}$$

$=0.1274\text{to}7.2726$

As a result, it is possible to conclude that there is a difference in the mean age at arrest of East German prisoners with chronic PTSD versus those with remitted PTSD.

To discover a comparison of the results of the hypothesis test and the confidence interval.

Think about the data.

Consider the $99\%$upper bound for a confidence interval.

$$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\frac{a}{2}}·\sqrt{\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)} \\ =(82.1-84.9)\pm 3.012 \end{gathered}$$

$\sqrt{\left(\frac{1.{501}^2}{14}+\frac{1.{698}^2}{14}\right)}$

$=-4.6244\text{to}-0.9756$

As a result, the $99\%$interval does not equal zero. As a result, it is possible to conclude that the mean wing length of the two species differs.