Chapter 11, Problem 11.56SE

(a)

To determine

To conduct: an analysis of variance and test for a difference in the mean reaction times due to the five stimuli.

Answer to Problem 11.56SE

The data provide sufficient evidence to indicate a difference in mean reaction times due to the five stimuli.

Explanation of Solution

Given:

Stimulus	Reaction Time (sec)							Total	Mean
A	$.8$	$.6$	$.6$	$.5$				$2.5$	$.625$
B	$.7$	$.8$	$.5$	$.5$	$.6$	$.9$	$.7$	$4.7$	$.671$
C	$1.2$	$1.0$	$.9$	$1.2$	$\begin{array}{l}1.3\hfill \end{array}$	$.8$		$6.4$	$1.067$
D	$1.0$	$.9$	$.9$	$\begin{array}{l}1.1\hfill \end{array}$	$.7$			$4.6$	$.920$
E	$.6$	$.4$	$.4$	$.7$	$\begin{array}{l}.3\hfill \end{array}$			$2.4$	$.480$

  

Calculation:

Now, we want to test there is any difference in mean reaction times.

To test the null hypothesis,

  $H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5$ .

Versus the alternative hypothesis,

  $H_a:$ At least one of the means is different from the others.

We have,

  $\begin{array}{l}MSE=0.3030\\ MST=0.026\end{array}$

The test statistic is given by,

  $\begin{array}{l}F=\frac{MST}{MSE}\\ =\frac{0.3030}{0.026}\\ =11.67\end{array}$

Interpretation: The P-value from the above printout is $0.000$ . Since, the $P-value=0.000<\alpha =.05$ , we reject $H_0$ .

Hence, we can conclude that the data provide sufficient evidence to indicate a difference in mean reaction times due to the five stimuli.

Conclusion: Therefore, the data provide sufficient evidence to indicate a difference in mean reaction times due to the five stimuli.

(b)

To determine

To compare: stimuli A and D to check whether there is a difference in mean reaction times.

Answer to Problem 11.56SE

The data provide sufficient evidence to indicate a difference between ${\mu }_A$ and ${\mu }_D$ .

Explanation of Solution

Calculation:

We can test the null hypothesis,

  $H_0:{\mu }_A={\mu }_D=0$ .

Versus the alternative hypothesis,

  $H_a:{\mu }_A={\mu }_D\ne 0$ .

Furthermore, the standard deviations of the two samples, calculated as

  $s_A=0.1258$ , $s_D=0.1483$ .

The pooled estimate of the common variance as,

  $\begin{array}{l}{s}^2=\frac{(n_A-1)s_A{}^2+(n_D-1)s_D{}^2}{n_A+n_D-2}\\ =\frac{(4-1)(0.015833)+(5-1)(0.022)}{4+5-2}\\ =\frac{0.0475+0.088}{7}\\ =\frac{0.1355}{7}\\ s^2=0.019357\end{array}$

We have,

  $\begin{array}{l}{n}_A=4\\ n_D=5\\ {\overline{x}}_A=0.625\\ {\overline{x}}_D=0.92\end{array}$

The test statistic is given by,

  $\begin{array}{l}t=\frac{({\overline{x}}_A-{\overline{x}}_D)}{\sqrt{s^2\left(\frac{1}{n_A}+\frac{1}{n_D}\right)}}\\ =\frac{(0.625-0.92)}{\sqrt{0.019357\left(0.2+0.25\right)}}\\ =\frac{-0.295}{\sqrt{0.008710714}}\\ =\frac{-0.295}{0.093331207}\\ =-3.16078628\\ \approx -3.16\end{array}$

We can find the appropriate critical value for a rejection region with $\alpha =.05$ in Table $4$ of Appendix I, and $H_0$ will be rejected if $t>2.365$ . Comparing the observed value of the test statistic $t=-3.16$ with the critical value $t_{0.025}=2.365$ . And the P-value from the Minitab is $P-value=0.016$ . We can reject the null hypothesis. Hence we conclude that the data provide sufficient evidence to indicate a difference between ${\mu }_A$ and ${\mu }_D$ .

Conclusion: Hence we conclude that the data provide sufficient evidence to indicate a difference between ${\mu }_A$ and ${\mu }_D$ .