Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal.
-A researcher was interested in comparing the resting pulse rates of people who exercise regularly and the pulse rates of people who do not exercise regularly. She obtained independent simple random samples of 16 people who do not exercise regularly and 12 people who do exercise regularly. The resting pulse rates (in beats per minute) were recorded and the summary statistics
Are as follows.
$\begin{array}{c|l}
\text { Do not exercise regularly } & \text { Exercise regularly } \
\hline \overline{\mathrm{x}}_{1}=72.3 \text { beats } / \mathrm{min} & \overline{\mathrm{x}}_{2}=68.2 \text { beats } / \mathrm{min} \
\mathrm{s}_{1}=10.9 \text { beats } / \mathrm{min} & \mathrm{s}_{2}=8.2 \text { beats } / \mathrm{min} \
\mathrm{n}_{1}=16 & \mathrm{n}_{2}=12
\end{array}$
Construct a 95\% confidence interval for $\mu _ { 1 } - \mu _ { 2 }$, the difference between the mean pulse rate of people who do not exercise regularly and the mean pulse rate of people who exercise regularly.
A) $- 3.32$ beats $/ \mathrm { min }

Question

Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal.
-A researcher was interested in comparing the resting pulse rates of people who exercise regularly and the pulse rates of people who do not exercise regularly. She obtained independent simple random samples of 16 people who do not exercise regularly and 12 people who do exercise regularly. The resting pulse rates (in beats per minute) were recorded and the summary statistics
Are as follows.
 $\begin{array}{c|l}
\text { Do not exercise regularly } & \text { Exercise regularly } \
\hline \overline{\mathrm{x}}_{1}=72.3 \text { beats } / \mathrm{min} & \overline{\mathrm{x}}_{2}=68.2 \text { beats } / \mathrm{min} \
\mathrm{s}_{1}=10.9 \text { beats } / \mathrm{min} & \mathrm{s}_{2}=8.2 \text { beats } / \mathrm{min} \
\mathrm{n}_{1}=16 & \mathrm{n}_{2}=12
\end{array}$
Construct a 95\% confidence interval for $\mu _ { 1 } - \mu _ { 2 }$, the difference between the mean pulse rate of people who do not exercise regularly and the mean pulse rate of people who exercise regularly.
A) $- 3.32$ beats $/ \mathrm { min } < \mu _ { 1 } - \mu _ { 2 } < 11.52$ beats $/ \mathrm { min }$
B) $- 3.34$ beats $/ \mathrm { min } < \mu _ { 1 } - \mu _ { 2 } < 11.54$ beats $/ \mathrm { min }$
C) $- 3.63$ beats $/ \mathrm { min } < \mu _ { 1 } - \mu _ { 2 } < 11.83$ beats $/ \mathrm { min }$
D) $- 3.65$ beats $/ \mathrm { min } < \mu _ { 1 } - \mu _ { 2 } < 11.85$ beats $/ \mathrm { min }$

Quizplus · Accepted Answer

The correct answer is A because the confidence interval is calculated as:$\left(\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}_{2}\right) \pm t_{\alpha/2} \sqrt{\frac{\mathrm{s}_{1}^{2}}{\mathrm{n}_{1}}+\frac{\mathrm{s}_{2}^{2}}{\mathrm{n}_{2}}}$where $t_{\alpha/2}$ is the critical value from the t-distribution with $\nu$ degrees of freedom, where $\nu$ is the smaller of $\mathrm{n}_{1}-1$ and $\mathrm{n}_{2}-1$.In this case, $\nu$ is the smaller of 15 and 11, which is 11. The critical value from the t-distribution with 11 degrees of freedom and a significance level of 0.05 is 2.201.Plugging in the values, we get:$(72.3-68.2) \pm 2.201 \sqrt{\frac{10.9^{2}}{16}+\frac{8.2^{2}}{12}}$which simplifies to:$-4.1 \pm 7.4$or$-3.32$ beats $/ \mathrm { min } < \mu _ { 1 } - \mu _ { 2 } < 11.52$ beats $/ \mathrm { min }$

Construct the Indicated Confidence Interval for the Difference Between the Two