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 paint manufacturer wanted to compare the drying times of two different types of paint. Independent simple random samples of 11 cans of type A and 9 cans of type B were selected and applied to similar surfaces. The drying times, in hours, were recorded. The summary statistics are as follows.
$\begin{array}{|l|l|}
\hline {\text { Type A }} & {\text { Type B }} \
\hline \bar{x}_{1}=71.5 \mathrm{hrs} & \bar{x}_{2}=68.5 \mathrm{hrs} \
\hline s_{1}=3.4 \mathrm{hrs} & s_{2}=3.6 \mathrm{hrs} \
\hline n_{1}=11 & n_{2}=9 \
\hline
\end{array}$

Construct a $99 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$, the difference between the mean drying time for paint type A and the mean drying time for paint type B.
A) $- 1.51 \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 paint manufacturer wanted to compare the drying times of two different types of paint. Independent simple random samples of 11 cans of type A and 9 cans of type B were selected and applied to similar surfaces. The drying times, in hours, were recorded. The summary statistics are as follows.
$\begin{array}{|l|l|}
\hline {\text { Type A }} & {\text { Type B }} \
\hline \bar{x}_{1}=71.5 \mathrm{hrs} & \bar{x}_{2}=68.5 \mathrm{hrs} \
\hline s_{1}=3.4 \mathrm{hrs} & s_{2}=3.6 \mathrm{hrs} \
\hline n_{1}=11 & n_{2}=9 \
\hline
\end{array}$

Construct a $99 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$, the difference between the mean drying time for paint type A and the mean drying time for paint type B.
A) $- 1.51 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 7.51 \mathrm { hrs }$
B) $- 2.24 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 8.24 \mathrm { hrs }$
C) $- 1.00 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 7.00 \mathrm { hrs }$
D) $- 0.14 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 6.14 \mathrm { hrs }$

Quizplus · Accepted Answer

The correct answer is A because the confidence interval is calculated as:$$\bar{x}_1 - \bar{x}_2 \pm t^* \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$$where $t^*$ is the critical value from the t-distribution with $n_1 + n_2 - 2$ degrees of freedom and a significance level of 0.01. Plugging in the values, we get:$$71.5 - 68.5 \pm 3.250 \sqrt{\frac{3.4^2}{11} + \frac{3.6^2}{9}}$$$$= 3 \pm 3.250 \sqrt{0.986 + 1.44}$$$$= 3 \pm 3.250 \sqrt{2.426}$$$$= 3 \pm 3.250 (1.557)$$$$= 3 \pm 5.046$$$$= (-2.046, 8.046)$$Therefore, the 99% confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ is $- 2.046 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 8.046 \mathrm { hrs }$.

Construct the Indicated Confidence Interval for the Difference Between the Two