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. $$\begin{array} { | c | c | }
\hline \hline \text { Type A } & \text { Type B } \
\hline \bar { x } _ { 1 } = 71.5 \mathrm { hrs } & \bar { x } _ { 2 } = 68.5 \mathrm { hrs } \
\hline \mathrm { s } _ { 1 } = 3.4 \mathrm { hrs } & \mathrm { s } _ { 2 } = 3.6 \mathrm { hrs } \
\hline n _ { 1 } = 11 & n _ { 2 } = 9 \
\hline \hline
\end{array}$$ Construct a $99 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$, the difference between the mean drying time for paint type A ar the mean drying time for paint type B.
A) $- 2.24$ 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. $$\begin{array} { | c | c | } 
\hline \hline \text { Type A } & \text { Type B } \
\hline \bar { x } _ { 1 } = 71.5 \mathrm { hrs } & \bar { x } _ { 2 } = 68.5 \mathrm { hrs } \
\hline \mathrm { s } _ { 1 } = 3.4 \mathrm { hrs } & \mathrm { s } _ { 2 } = 3.6 \mathrm { hrs } \
\hline n _ { 1 } = 11 & n _ { 2 } = 9 \
\hline \hline
\end{array}$$ Construct a $99 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$, the difference between the mean drying time for paint type A ar the mean drying time for paint type B.
A) $- 2.24$ hrs $< \mu _ { 1 } - \mu _ { 2 } < 8.24 \mathrm { hrs }$
B) $- 1.00$ hrs $< \mu _ { 1 } - \mu _ { 2 } < 7.00$ hrs
C) $- 1.51$ hrs $< \mu _ { 1 } - \mu _ { 2 } < 7.51$ hrs
D) $- 0.14 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 6.14 \mathrm { hrs }$

Quizplus · Accepted Answer

First, we need to calculate the pooled standard deviation $\mathrm{s_p}$ using the formula:
$$s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}=\sqrt{\frac{(11-1)(3.4)^2+(9-1)(3.6)^2}{11+9-2}}=3.497$$
Next, we need to calculate the standard error of the difference between means, using the formula:
$$s_{\bar{x_1}-\bar{x_2}}=\sqrt{\frac{s_{p}^2}{n_1}+\frac{s_{p}^2}{n_2}}=\sqrt{\frac{(3.497)^2}{11}+\frac{(3.497)^2}{9}}=2.112$$
Using a t-distribution with degrees of freedom $(n_1+n_2-2)=18$ and a confidence level of 99%, we find the t-value to be $t_{0.005,18}=2.878$. 
Finally, we can calculate the confidence interval using the formula:
$$(\bar{x_1}-\bar{x_2})\pm t_{\alpha/2,n_1+n_2-2}s_{\bar{x_1}-\bar{x_2}}=(71.5-68.5)\pm 2.878 \times 2.112= 3\pm6.08=(-3.08,9.08)$$
Therefore, the 99% confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ is $-1.51$ hrs $< \mu _ { 1 } - \mu _ { 2 } < 7.51$ hrs.

Construct the Indicated Confidence Interval for the Difference Between the Two