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 amount of time spent watching television by women and by men. Independent simple random samples of 14 women and 17 men were selected and each person was asked how many hours he or she had watched television during the previous week.
The summary statistics are as follows.

$\begin{array}{r|r}
\text { Women } & \text { Men } \
\hline \overline{\mathrm{x}}_{1}=12.5 \mathrm{hr} & \mathrm{x}_{2}=17.6 \mathrm{hr} \
\mathrm{s}_{1}=4.0 \mathrm{hr} & \mathrm{s}_{2}=4.4 \mathrm{hr} \
\mathrm{n}_{1}=14 & \mathrm{n}_{2}=17
\end{array}$
Construct a $95 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$, the difference between the mean amount of time spent watching television for women and the mean amount of time spent watching television for men.
A) $- 8.22 \mathrm { hrs }

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 amount of time spent watching television by women and by men. Independent simple random samples of 14 women and 17 men were selected and each person was asked how many hours he or she had watched television during the previous week.
The summary statistics are as follows.

$\begin{array}{r|r}
\text { Women } & \text { Men } \
\hline \overline{\mathrm{x}}_{1}=12.5 \mathrm{hr} & \mathrm{x}_{2}=17.6 \mathrm{hr} \
\mathrm{s}_{1}=4.0 \mathrm{hr} & \mathrm{s}_{2}=4.4 \mathrm{hr} \
\mathrm{n}_{1}=14 & \mathrm{n}_{2}=17
\end{array}$
Construct a $95 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$, the difference between the mean amount of time spent watching television for women and the mean amount of time spent watching television for men.
A) $- 8.22 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < - 1.98 \mathrm { hrs }$
B) $- 8.48 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < - 1.72 \mathrm { hrs }$
C) $- 7.69 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < - 2.51 \mathrm { hrs }$
D) $- 8.33$ hrs $< \mu _ { 1 } - \mu _ { 2 } < - 1.87 \mathrm { hrs }$

Quizplus · Accepted Answer

The correct answer is A because the confidence interval is calculated as:$\left( \overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}_{2}\right) \pm \mathrm{t}_{(\alpha/2, n_1+n_2-2)} \sqrt{\frac{\mathrm{s}_1^2}{\mathrm{n}_1}+\frac{\mathrm{s}_2^2}{\mathrm{n}_2}}$where $\mathrm{t}_{(\alpha/2, n_1+n_2-2)}$ is the critical value from the t-distribution with $n_1+n_2-2$ degrees of freedom and $\alpha$ is the level of significance.Plugging in the values from the problem, we get:$(12.5-17.6) \pm \mathrm{t}_{(0.025, 29)} \sqrt{\frac{4.0^2}{14}+\frac{4.4^2}{17}}$$\approx -5.1 \pm 3.12$$\approx (-8.22, -1.98)$

Construct the Indicated Confidence Interval for the Difference Between the Two