Chapter 10, Problem 10.83SE

(a)

To determine

To find: whether the given data is an evidence for the difference in mean times to absorption for the two drugs.

Answer to Problem 10.83SE

There is sufficient evidence to indicate a difference in mean times to absorption for the two drugs.

Explanation of Solution

Given:

Drug A	Drug B
$\overline{x_1}=27.2$	$\overline{x_2}=33.5$
$s_1{}^2=16.36$	$s_2{}^2=18.92$

Calculation:

The null and alternative hypotheses are,

  $H_0:({\mu }_1-{\mu }_2)=0$ versus $H_a:({\mu }_1-{\mu }_2)\ne 0$

The alternative hypothesis $H_a:({\mu }_1-{\mu }_2)\ne 0$ implies that you should use two-tailed test of the t distribution with $df=n_1+n_2-2=8$ . You can find the approximate critical value for a rejection region with $\alpha =0.05$ in Table $4$ of Appendix I, and $H_0$ will be rejected if

  $t<-t_{0.05/2}=-2.306$ or $t>t_{0.05/2}=2.306$ .

For Drug A: $\overline{x_1}=27.2$ , $s_1{}^2=16.36$ , and $n_1=5$

For Drug B: $\overline{x_2}=33.5$ , $s_2{}^2=18.92$ , and $n_2=5$

The pooled estimator for ${\sigma }^2$ is,

  $\begin{array}{l}{s}^2=\frac{(n_1-1)s_1{}^2+(n_2-1)s_2{}^2}{n_1+n_2-2}\\ =\frac{(5-1)16.36+(5-1)18.92}{5+5-2}\\ =\frac{141.12}{8}\\ =17.64\end{array}$

The test static is,

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{s^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{(27.2-33.5)}{\sqrt{17.64\left(\frac{1}{5}+\frac{1}{5}\right)}}\\ =\frac{-6.3}{2.6563}\\ =-2.372\end{array}$

Comparing the observed value of the test statistic $t=-2.372$ with critical value $-2.306$ , you can observe that it falls in the rejected region.

Therefore, you can reject the null hypothesis.

Therefore, there is sufficient evidence to indicate a difference in mean times to absorption for the two drugs.

Conclusion: Therefore, there is sufficient evidence to indicate a difference in mean times to absorption for the two drugs.

(b)

To determine

To find: the approximate p-value and explain whether the result confirms the conclusion.

Answer to Problem 10.83SE

There is sufficient evidence to indicate a difference in mean times to absorption for the two drugs. Yes. It confirms our conclusion in part a.

Explanation of Solution

Given:

Drug A	Drug B
$\overline{x_1}=27.2$	$\overline{x_2}=33.5$
$s_1{}^2=16.36$	$s_2{}^2=18.92$

Calculation:

Since the value, $t=2.372$ lies between $t_{0.025}=2.306$ and $t_{0.01}=2.896$ , the tail area to the right of $2.372$ is between $0.01$ and $0.025$ . Since this area represents only half of the p-value, you can write

  $0.01<\frac{1}{2}(p-value)<0.025$ or $0.02<p-value<0.05$

Since, the p-value is less than $\alpha$ , you can reject the null hypothesis.

Thus, there is sufficient evidence to indicate a difference in mean times to absorption for the two drugs.

Yes. It confirms our conclusion in part a.

Conclusion: Thus, there is sufficient evidence to indicate a difference in mean times to absorption for the two drugs. Yes. It confirms our conclusion in part a.

(c)

To determine

To find: $95\%$ confidence intervals for the difference in mean times to absorption and explain whether the interval confirms part a conclusions.

Answer to Problem 10.83SE

A $95\%$ confidence interval for the difference in mean times to absorptionis $(-12.425,-0.175)$ .

Yes. It confirms our conclusion in part a.

Explanation of Solution

Given:

Drug A	Drug B
$\overline{x_1}=27.2$	$\overline{x_2}=33.5$
$s_1{}^2=16.36$	$s_2{}^2=18.92$

Calculation:

If you choose a $5\%$ level of significance ( $\alpha =0.05$ ), the critical value can be found from the values of t from Table $4$ of Appendix I. With $df=n_1+n_2-2=8$ the critical value is $t_{\alpha /2}=t_{0.05/2}=2.306$ .

A $95\%$ confidence interval gives the upper and lower limits for ${\mu }_1-{\mu }_2$ as,

  $\begin{array}{l}(\overline{x_1}-\overline{x_2})\pm t_{\alpha /2}\sqrt{s^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ =>(27.2-33.5)\pm 2.306\sqrt{17.64\left(\frac{1}{5}+\frac{1}{5}\right)}\\ =>-6.3\pm 2.306(2.6563)\\ =>-6.3\pm 6.125\\ =>(-12.425,-0.175)\end{array}$

Since, the interval do not contain zero, you can reject the null hypothesis.

Thus, there is sufficient evidence to indicate a difference in mean times to absorption for the two drugs.

Note that the confidence interval does not contain 0, which indicates that the population means are difference and thus this confirms our conclusion in part a.

Conclusion: A $95\%$ confidence interval for the difference in mean times to absorptionis $(-12.425,-0.175)$ . Yes. It confirms our conclusion in part a.