Chapter 11, Problem 11.60SE

(a)

To determine

To find:

Do the data provide sufficient evidence to indicate a difference in mean increase in heart rate among the four age groups? Test by using $\alpha =.05$ .

Answer to Problem 11.60SE

There is no sufficient evidence to indicate a difference in mean increase in heart rate among the four age groups.

Explanation of Solution

Given:

	$10-19$	$20-39$	$40-59$	$60-69$
$1$	$29$	$24$	$37$	$28$
$2$	$33$	$27$	$25$	$29$
$3$	$26$	$33$	$22$	$34$
$4$	$27$	$31$	$33$	$36$
$5$	$39$	$21$	$28$	$21$
$6$	$\begin{array}{l}35\hfill \end{array}$	$28$	$26$	$20$
$7$	$33$	$24$	$30$	$25$
$8$	$29$	$34$	$34$	$24$
$9$	$36$	$21$	$27$	$33$
$10$	$22$	$32$	$33$	$32$
Total	$309$	$275$	$295$	$282$

Calculation:

From the MINITAB output:

One-way ANOVA: $10-19$, $20-39$, $40-59$, $60-69$

Source	$\begin{array}{l}DF\hfill \end{array}$	$SS$	$MS$	$F$	$P$
Factor	$3$	$67.5$	$22.5$	$0.87$	$0.468$
Error	$36$	$935.5$	$26.0$
Total	$39$	$1003.0$

  $S=5.098$$R-Sq=6.73\%$$R-Sq(adj)=0.00\%$

From the above printout the value of $F=0.87$ and the corresponding $P-value=0.468>\alpha =.05$ .

So, we can conclude that there is no sufficient evidence to indicate a difference in mean increase in heart rate among the four age groups.

Conclusion: There is no sufficient evidence to indicate a difference in mean increase in heart rate among the four age groups.

(b)

To determine

To find: a $90\%$ confidence interval for the difference in mean increase in heart rate between age groups $10-19$ and $60-69$ .

Answer to Problem 11.60SE

The $90\%$ confidence interval for the difference in the treatment means is $(6.55,-1.15)$ .

Explanation of Solution

Given:

	$10-19$	$20-39$	$40-59$	$60-69$
$1$	$29$	$24$	$37$	$28$
$2$	$33$	$27$	$25$	$29$
$3$	$26$	$33$	$22$	$34$
$4$	$27$	$31$	$33$	$36$
$5$	$39$	$21$	$28$	$21$
$6$	$\begin{array}{l}35\hfill \end{array}$	$28$	$26$	$20$
$7$	$33$	$24$	$30$	$25$
$8$	$29$	$34$	$34$	$24$
$9$	$36$	$21$	$27$	$33$
$10$	$22$	$32$	$33$	$32$
Total	$309$	$275$	$295$	$282$

Calculation:

Level	N	Mean	$\text{StDev}$
$10-19$	$10$	$30.900$	$5.195$
$20-39$	$10$	$27.500$	$4.882$
$40-59$	$10$	$29.500$	$4.696$
$60-69$	$10$	$28.200$	$5.574$
$\begin{array}{l}\text{Pooled StDev}=5.098\hfill \end{array}$

The $90\%$ confidence interval for the difference $({\mu }_1-{\mu }_3)$ is given by, we have given, $s^2=MSE=26$ so that $s=5.099$ with $df=36$

And we have, ${\overline{x}}_1=30.9$ , ${\overline{x}}_4=28.2$

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_4)\pm t_{\alpha /2}\sqrt{s^2\left(\frac{1}{n_1}+\frac{1}{n_4}\right)}\\ =(30.9-28.2)\pm 1.688\sqrt{26\left(\frac{1}{10}+\frac{1}{10}\right)}\\ =2.7\pm 1.688\sqrt{26(0.2)}\\ =2.7\pm 1.688\sqrt{5.2}\\ =2.7\pm 1.688(2.28035085)\\ =2.7\pm 3.8492\end{array}$

So, the $90\%$ confidence interval for the difference in the treatment means is, $(6.55,-1.15)$ .

Conclusion: Thus, the $90\%$ confidence interval for the difference in the treatment means is, $(6.55,-1.15)$ .

(c)

To determine

To find: a $90\%$ confidence interval for the mean increase in heart rate for the age group $20-39$ .

Answer to Problem 11.60SE

The $90\%$ confidence interval for ${\mu }_1$ is $(30.22,24.78)$ .

Explanation of Solution

Given:

	$10-19$	$20-39$	$40-59$	$60-69$
$1$	$29$	$24$	$37$	$28$
$2$	$33$	$27$	$25$	$29$
$3$	$26$	$33$	$22$	$34$
$4$	$27$	$31$	$33$	$36$
$5$	$39$	$21$	$28$	$21$
$6$	$\begin{array}{l}35\hfill \end{array}$	$28$	$26$	$20$
$7$	$33$	$24$	$30$	$25$
$8$	$29$	$34$	$34$	$24$
$9$	$36$	$21$	$27$	$33$
$10$	$22$	$32$	$33$	$32$
Total	$309$	$275$	$295$	$282$

Calculation:

We have given, $s^2=MSE=26$ so that $s=5.099$ with $df=36$ .

And we have ${\overline{x}}_1=27.5$ .

Now, we want to calculate the $90\%$ confidence interval:

  $\begin{array}{l}{\overline{x}}_1\pm t_{\alpha /2}\left(\frac{s}{\sqrt{n_1}}\right)\\ =27.5\pm 1.688\left(\frac{5.099}{\sqrt{10}}\right)\\ =27.5\pm 2.7218078\end{array}$

The $90\%$ confidence interval for ${\mu }_1$ is $(30.22,24.78)$ .

Conclusion: Thus, the $90\%$ confidence interval for ${\mu }_1$ is $(30.22,24.78)$ .

(d)

To determine

To find: the number ofpeople needed in each group to estimate two beats per minute with probability equal to $.95$ ?

Answer to Problem 11.60SE

The number of people that we need in each group is $25$ .

Explanation of Solution

Given:

	$10-19$	$20-39$	$40-59$	$60-69$
$1$	$29$	$24$	$37$	$28$
$2$	$33$	$27$	$25$	$29$
$3$	$26$	$33$	$22$	$34$
$4$	$27$	$31$	$33$	$36$
$5$	$39$	$21$	$28$	$21$
$6$	$\begin{array}{l}35\hfill \end{array}$	$28$	$26$	$20$
$7$	$33$	$24$	$30$	$25$
$8$	$29$	$34$	$34$	$24$
$9$	$36$	$21$	$27$	$33$
$10$	$22$	$32$	$33$	$32$
Total	$309$	$275$	$295$	$282$

Calculation:

The formula for sample size is given by,

  $\begin{array}{l}n={\left(\frac{z_{\alpha /2}\sigma }{E}\right)}^2\\ ={\left(\frac{1.96(5.099)}{2}\right)}^2\\ $ ={(4.99702)}^2$=24.97\\ =25\end{array}$

The number of people that we need in each group is $25$ .

Conclusion: The number of people that we need in each group is $25$ .