Chapter 10.3, Problem 89E

(a)

To determine

To construct and interpret a $99\%$ confidence interval for the true mean difference.

Answer to Problem 89E

We are $99\%$ confident that the true mean difference in the number of words recalled is between $0.2114\text{ and }2.9286$ .

Explanation of Solution

It is given that:

  $\begin{array}{l}\alpha =0.01\\ n=30\\ {\overline{x}}_D=1.57\\ s_D=2.70\end{array}$

Given claim is that: Difference.

The claim is either null hypothesis or an alternative hypothesis.

  $\begin{array}{l}{H}_0:{\mu }_D=0\\ H_a:{\mu }_D\ne 0\end{array}$

And the degree of freedom will be:

  $\begin{array}{l}df=n-1\\ =30-1\\ =29\end{array}$

The t -value will be as:

  $t_{\alpha /2}=2.756$

The confidence interval will be:

  $\begin{array}{l}\overline{x}-t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =1.57-2.756\times \frac{2.70}{\sqrt{30}}\\ =0.2114\\ \overline{x}+t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =1.57+2.756\times \frac{2.70}{\sqrt{30}}\\ =2.9286\end{array}$

Thus, we conclude that we are $99\%$ confident that the true mean difference in the number of words recalled is between $0.2114\text{ and }2.9286$ .

(b)

To determine

To explain how the confidence interval provide more information than the test.

Explanation of Solution

The confidence interval gives us more information than a significance test because the confidence interval gives us a range of possible values for the difference between the true means and the significance test only checks one possible value for the difference between the means, that is if the means are equal.