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 paint manufacturer wished 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. $\begin{array} { | c | c | | }
\hline \hline \text { Type A } & \text { Type B } \
\hline \bar { x } _ { 1 } = 75.7 \mathrm { hrs } & \bar { x } _ { 2 } = 64.3 \mathrm { hrs } \
\hline s _ { 1 } = 4.5 \mathrm { hrs } & s _ { 2 } = 5.1 \mathrm { hrs } \
\hline n _ { 1 } = 11 & n _ { 2 } = 9 \
\hline
\end{array}$
Construct a 98% confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ , the difference between the mean drying time for paint of type A And the mean drying time for paint of type B.
A) $5.85 \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. Do Not assume that the population standard deviations are equal. A paint manufacturer wished 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. $\begin{array} { | c | c | | } 
\hline \hline \text { Type A } & \text { Type B } \
\hline \bar { x } _ { 1 } = 75.7 \mathrm { hrs } & \bar { x } _ { 2 } = 64.3 \mathrm { hrs } \
\hline s _ { 1 } = 4.5 \mathrm { hrs } & s _ { 2 } = 5.1 \mathrm { hrs } \
\hline n _ { 1 } = 11 & n _ { 2 } = 9 \
\hline
\end{array}$
Construct a 98% confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ , the difference between the mean drying time for paint of type A And the mean drying time for paint of type B. 
A) $5.85 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 16.95 \mathrm { hrs }$
B) $5.78 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 17.02 \mathrm { hrs }$
C) $5.92 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 16.88 \mathrm { hrs }$
D) $6.08 \mathrm { hrs } < \mu _ { 1 } - \mu _ { 2 } < 16.72 \mathrm { hrs }$

Quizplus · Accepted Answer

We will use the formula for the confidence interval of the difference between two population means when the population standard deviations are unknown and potentially unequal:

$$(\bar{x}_{1}-\bar{x}_{2})\pm t_{\alpha/2,df}\sqrt{\frac{s_{1}^2}{n_1}+\frac{s_{2}^2}{n_2}}$$

where $df$ is the degrees of freedom, given by:

$$df=\frac{\left(\frac{s_{1}^2}{n_1}+\frac{s_{2}^2}{n_2}\right)^2}{\frac{(s_{1}^2/n_1)^2}{n_1-1}+\frac{(s_{2}^2/n_2)^2}{n_2-1}}$$

Substituting the given values, we get:

$$(75.7-64.3)\pm t_{0.01, 16.373}\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}$$

Calculating the degrees of freedom:

$$df=\frac{\left(\frac{4.5^2}{11}+\frac{5.1^2}{9}\right)^2}{\frac{(4.5^2/11)^2}{11-1}+\frac{(5.1^2/9)^2}{9-1}}\approx 16.373$$

Using a t-table with 16 degrees of freedom, and $\alpha=0.02$ (since we have a two-tailed test and want a 98% confidence interval), we get:

$$t_{0.01, 16.373}=2.921$$

Substituting these values, we get:

$$11.4\pm 2.921\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}$$

Simplifying and rounding to two decimal places, we get:

$$11.4\pm 5.62$$

Therefore, the 98% confidence interval for $\mu_{1}-\mu_{2}$ is:

$$5.78\mathrm{\ hrs}<\mu_{1}-\mu_{2}<17.02\mathrm{\ hrs}$$

So the best choice is (B) with the confidence interval rounded to two decimal places.

Construct the Indicated Confidence Interval for the Difference Between the Two