Chapter 9.1, Problem 14E

To determine

Construct and interpret a confidence interval for the true difference between the two population means.

Answer to Problem 14E

Solution:

The value of the margin of error is 0.59352 and 95% confidence interval for the true difference between the mean amounts of time spent exercising each week by people who work out in the morning and those who work out in the afternoon or evening at the three health centers is $\left(-0.39352,0.79352\right)$

Explanation of Solution

Given Information:

Mean hours of exercise who work out in the morning.

i.e. ${\overline{x}}_1=4.1hoursperweek$.

Mean hours of exercise who work out in the afternoon or evening.

i.e. ${\overline{x}}_2=3.7hoursperweek$.

The Standard deviation for people who exercise in the morning.

i.e. ${\sigma }_1=0.7hoursperweek$.

Standard deviation for people who exercise in the afternoon or evening.

i.e. ${\sigma }_2=0.5hoursperweek$.

Number of people who exercise in the morning i.e. $n_1=49$.

Number of people who exercise in the afternoon or evening i.e. $n_1=54$.

Formula used:

The confidence interval for the difference between two population means for independent data sets is given by,

$\left({\overline{x}}_1-{\overline{x}}_2\right)-E<{\mu }_1-{\mu }_2<\left({\overline{x}}_1-{\overline{x}}_2\right)+E$

Or

$\left(\left({\overline{x}}_1-{\overline{x}}_2\right)-E,\left({\overline{x}}_1-{\overline{x}}_2\right)+E\right)$

Where ${\overline{x}}_1\text{ and }{\overline{x}}_2$ are the two sample means,

$\left({\overline{x}}_1-{\overline{x}}_2\right)$ is the point estimate for the difference between the population means, ${\mu }_1-{\mu }_2$, and E is the margin of error and it is given as

$E=z_{\alpha /2}\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$

Where $z_{\alpha /2}$ is the critical value for the level of confidence, $c=1-\alpha$, such that the area under the standard normal distribution to the right of $z_{\alpha /2}$ is equal to $\frac{\alpha }{2}$,

${\sigma }_1\text{ and }{\sigma }_2\text{ are the two population standard deviations, and}$

$n_1\text{ and }{n}_2\text{ are the two sample sizes}\text{.}$

Calculation:

The margin of error and it is given as,

$E=z_{\frac{\alpha }{2}}\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$…$\left(1\right)$

Since the level of confidence is 95%, then the level of significance is given as,

$\begin{array}{rll}\alpha & =& 1-0.95\\ & =& 0.05\end{array}$

The value of $z$ from table at $\alpha =0.05$ is $1.96$.

Substitute the values $\text{4}\text{.1}$ for ${\overline{x}}_1$, $3.7$ for ${\overline{x}}_2$, $0.7$ for ${\sigma }_1$, $49$ for $n_1$, $0.5$ for ${\sigma }_2$, $54$ for $n_2$ and 1.96 for $z_{\frac{\alpha }{2}}$ in the above equation $\left(1\right)$.

$\begin{array}{rll}E& =& 1.96\times \sqrt{\frac{{\left(0.7\right)}^2}{49}+\frac{{\left(0.5\right)}^2}{54}}\\ & =& 0.23706\end{array}$

The confidence interval is $\left(\left({\overline{x}}_1-{\overline{x}}_2\right)-E,\left({\overline{x}}_1-{\overline{x}}_2\right)+E\right)$, that is,

$\begin{array}{rll}\text{Confidence interval}& =& \left(\left(4.1-3.7\right)-0.23706,\left(4.1-3.7\right)+0.23706\right)\\ & =& \left(0.16294,0.63706\right)\end{array}$

Interpretation:

The 95% confidence interval for the true difference between the mean amounts of time spent exercising each week by people who work out in the morning and those who work out in the afternoon or evening at the three health centers, ranges from 0.16294 to 0.63706.