Chapter 21, Problem 21.28E

(a)

To determine

To find: The two sample standard deviations.

Answer to Problem 21.28E

The two sample standard deviations are 0.25 and 0.15.

Explanation of Solution

Given info:

In the given information, $n_{Alcohol}=25$, $n_{Placebo}=25$, ${\overline{x}}_{Alcohol}=0.25$, ${\overline{x}}_{Placebo}=0.12$, $SE{M}_{Alcohol}=0.05$, and $SE{M}_{Placebo}=0.03$.

Calculation:

The sample standard deviation for alcohol is,

$\begin{array}{c}SE{M}_{Alcohol}=\frac{s_{Alcohol}}{\sqrt{n_{Alcohol}}}\\ s_{Alcohol}=SE{M}_{Alcohol}\sqrt{n_{Alcohol}}\\ =0.05\left(\sqrt{25}\right)\\ =0.25\end{array}$

The sample standard deviation for placebo is,

$\begin{array}{c}SE{M}_{Placebo}=\frac{s_{Placebo}}{\sqrt{n_{Placebo}}}\\ s_{Placebo}=SE{M}_{Placebo}\sqrt{n_{Placebo}}\\ =0.03\left(\sqrt{25}\right)\\ =0.15\end{array}$

Thus, the two sample standard deviations are 0.25 and 0.15.

(b)

To determine

To find: The degrees of freedom by using conservative Option 2 for two-sample t procedures.

Answer to Problem 21.28E

The degrees of freedom are 24.

Explanation of Solution

Calculation:

The degrees of freedom for the two-sample t procedures with the conservative Option 2 is,

$\begin{array}{c}\text{df}=\text{Smaller of }{n}_{Alcohol}-1\text{ and }{n}_{Placebo}-1\\ =25-1\\ =24\end{array}$

Thus, the degrees of freedom are 24.

(c)

To determine

To find: The 90% confidence interval for the mean difference between the two groups.

Answer to Problem 21.28E

The 90% confidence interval for the mean difference between the two groups is $\left(\text{0}\text{.0302},\text{0}\text{.2298}\right)$.

Explanation of Solution

Calculation:

Critical value:

From the table of “t Distribution critical values”, the critical value for 24 degrees of freedom with 90% confidence level is 1.711.

Confidence interval:

The 90% confidence interval for the mean difference between the two groups is,

$\begin{array}{c}CI=\left({\overline{x}}_{Alcohol}-{\overline{x}}_{Placebo}\right)\pm t*\sqrt{\frac{s_{Alcohol}^2}{n_{Alcohol}}+\frac{s_{Placebo}^2}{n_{Placebo}}}\\ =\left({\overline{x}}_{Alcohol}-{\overline{x}}_{Placebo}\right)\pm t*\sqrt{SE{M}_{Alcohol}^2+SE{M}_{Placebo}^2}\\ =0.25-0.12\pm 1.711\sqrt{{0.05}^2+{0.03}^2}\\ =0.13\pm 0.0998\end{array}$

$\begin{array}{c}=\left(0.13-0.0998,0.13+0.0998\right)\\ =\left(\text{0}\text{.0302},\text{0}\text{.2298}\right)\end{array}$

Thus, the 90% confidence interval for the mean difference between the two groups is $\left(\text{0}\text{.0302},\text{0}\text{.2298}\right)$.