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. Also assume that the population standard deviations are equal $\left( \sigma _ { 1 } = \sigma _ { 2 } \right)$ , so that the standard error of the difference between means is obtained by pooling the sample variances .
-A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who do not exercise regularly. Independent simple random samples were obtained of 16 people who do not exercise regularly and 12 people who do exercise regularly. The resting pulse
Rate (in beats per minute) of each person was recorded. The summary statistics are as follows. $\begin{array}{r|r}
\text { Do Not Exercise } & \text { Do Exercise } \
\hline \overline{\mathrm{x}}_{1}=73.4 \text { beats } / \mathrm{min} & \overline{\mathrm{x}}_{2}=68.7 \text { beats } / \mathrm{min} \
\mathrm{s}_{1}=10.1 \text { beats } / \mathrm{min} & \mathrm{s}_{2}=8.4 \text { beat } / \mathrm{min} \
\mathrm{n}_{1}=16 & \mathrm{n}_{2}=12
\end{array}$
Construct a $90 \%$ confidence interval for 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) $- 1.44$ 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. Also assume that the population standard deviations are equal  $\left( \sigma _ { 1 } = \sigma _ { 2 } \right)$ , so that the standard error of the difference between means is obtained by pooling the sample variances .
-A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who do not exercise regularly. Independent simple random samples were obtained of 16 people who do not exercise regularly and 12 people who do exercise regularly. The resting pulse
Rate (in beats per minute) of each person was recorded. The summary statistics are as follows. $\begin{array}{r|r}
\text { Do Not Exercise } & \text { Do Exercise } \
\hline \overline{\mathrm{x}}_{1}=73.4 \text { beats } / \mathrm{min} & \overline{\mathrm{x}}_{2}=68.7 \text { beats } / \mathrm{min} \
\mathrm{s}_{1}=10.1 \text { beats } / \mathrm{min} & \mathrm{s}_{2}=8.4 \text { beat } / \mathrm{min} \
\mathrm{n}_{1}=16 & \mathrm{n}_{2}=12
\end{array}$
Construct a $90 \%$ confidence interval for 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) $- 1.44$ beats $/ \mathrm { min } < \mu _ { 1 } - \mu _ { 2 } < 10.84$ beats $/ \mathrm { min }$
B) $- 0.03$ beats $/ \mathrm { min } < \mu _ { 1 } - \mu _ { 2 } < 9.43$ beats $/ \mathrm { min }$
C) $0.74$ beats $/ \mathrm { min } < \mu _ { 1 } - \mu _ { 2 } < 8.66$ beats $/ \mathrm { min }$
D) $- 2.08$ beats $/ \mathrm { min } < \mu _ { 1 } - \mu _ { 2 } < 11.48$ beats $/ \mathrm { min }$

Quizplus · Accepted Answer

The correct answer is A because the confidence interval is calculated as:$\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}_{2} \pm \mathrm{t}_{\alpha/2} \sqrt{\frac{\mathrm{s}_{1}^{2}}{\mathrm{n}_{1}}+\frac{\mathrm{s}_{2}^{2}}{\mathrm{n}_{2}}}$where $\mathrm{t}_{\alpha/2}$ is the critical value from the t-distribution with $\alpha/2$ degrees of freedom.For a 90% confidence interval, $\alpha = 0.1$ and $\alpha/2 = 0.05$. With 26 degrees of freedom (df = n1 + n2 - 2), the critical value is 1.708.Plugging in the values, we get:$73.4 - 68.7 \pm 1.708 \sqrt{\frac{10.1^2}{16}+\frac{8.4^2}{12}}$$= 4.7 \pm 1.708 \times 3.46$$= 4.7 \pm 6.14$$= (-1.44, 10.84)$Therefore, the 90% confidence interval for the difference between the mean pulse rate of people who do not exercise regularly and the mean pulse rate of people who exercise regularly is $-1.44$ beats $/ \mathrm { min } < \mu _ { 1 } - \mu _ { 2 } < 10.84$ beats $/ \mathrm { min }$.

Construct the Indicated Confidence Interval for the Difference Between the Two