Chapter 6.5, Problem 232E

To determine

Find the best estimate for ${\mu }_1-{\mu }_2$.

Obtain the margin of error and confidence interval.

Explanation of Solution

The best estimate for the difference in the means is $\underset{\_}{{\overline{x}}_d=3.7}$.

The confidence interval for ${\mu }_d={\mu }_1-{\mu }_2$ using t-distribution is as follows:

$\text{C}\text{.I}={\overline{x}}_d\pm t^{*}\frac{s_d}{\sqrt{n_d}}$

Where ${\overline{x}}_d,s_d\text{ , and }{n}_d$ are the mean standard deviation and the sample size for the sample differences, respectively.

Degrees of freedom:

$\begin{array}{c}{n}_d-1=30-1\\ =29\end{array}$

Thus, the degrees of freedom is 29.

t^*-value:

Step-by-step procedure to obtain the t^*-value using StatKey:

• Choose t under Theoretical Distributions.
• In Edit Parameters, enter the degrees of freedom as 29.
• Click Two-Tail.
• Click on Custom Proportion, enter 0.95 as Proportion.
• Click OK.

Output obtained using StatKey is as follows:

Thus, the t-value for a two-tailed test is $t^{*}=\pm 2.045$.

The margin of error is obtained as follows:

$\begin{array}{c}\text{Margin of error}=t^{*}\frac{s_d}{\sqrt{n_d}}\\ =2.045\left(\frac{2.1}{\sqrt{30}}\right)\\ =2.045\left(0.38\right)\\ =0.78\end{array}$

Thus, the margin of error is 0.78.

The 95% confidence interval for ${\mu }_d={\mu }_1-{\mu }_2$ is obtained as follows:

$\begin{array}{c}\text{C}\text{.I}=3.7\pm 0.78\\ =\left(2.92,4.48\right)\end{array}$

Thus, the 95% confidence interval for the difference in the means is (2.92, 4.48).