Chapter 7.4, Problem 25P

(a)

To determine

Find the 85% confidence interval for ${\mu }_1-{\mu }_2$.

Answer to Problem 25P

The 85% confidence interval for ${\mu }_1-{\mu }_2$ is $-1.35<{\mu }_1-{\mu }_2<2.39$.

Explanation of Solution

Confidence interval:

The confidence interval for ${\mu }_1-{\mu }_2$ when both ${\sigma }_1$ and ${\sigma }_2$ are unknown is,

$\left({\overline{x}}_1-{\overline{x}}_2\right)-E<{\mu }_1-{\mu }_2<\left({\overline{x}}_1-{\overline{x}}_2\right)+E$

In the formula, $s_1$ and $s_2$ are sample standard deviations of populations 1 and 2, ${\overline{x}}_1$ and ${\overline{x}}_2$ are sample means from populations 1 and 2, $n_1$ and $n_2$ are sample sizes of population 1 and 2, $E=t_c\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$, c is confidence level, and $t_c$ is the critical value for confidence level c.

Let ${\mu }_1$ denotes the mean measure of self-esteem through competence, ${\mu }_2$ denotes the mean measure of self-esteem through social acceptance, ${\mu }_3$ denotes the mean measure of self-esteem through attractiveness.

The confidence level is 85%.

Critical value:

Substitute 15 for $n_1$, 15 for $n_2$ in the degrees of freedom formula.

$\begin{array}{c}d.f.=smaller\left(15-1,15-1\right)\\ =smaller\left(14,14\right)\\ =14\end{array}$

The degrees of freedom are 14.

Use the Appendix II: Tables, Table 6: Critical Values for Student’s t Distribution:

• In d.f. column locate the value 14.
• In c row of locate the value 0.85.
• The intersecting value of row and columns is 1.523.

The critical value is 1.523.

Substitute 15 for $n_1$, 15 for $n_2$, 3.07 for $s_1$, 3.62 for $s_2$, 1.523 for $t_c$ in the margin of error formula (E).

$\begin{array}{c}E=1.523\sqrt{\frac{{3.07}^2}{15}+\frac{{3.62}^2}{15}}\\ =1.523\sqrt{1.50195}\\ =1.523\left(1.2255\right)\\ =1.87\end{array}$

The margin of error E is 1.87.

Substitute 19.84 for ${\overline{x}}_1$, 19.32 for ${\overline{x}}_2$, 1.87 for E in the confidence formula.

$\begin{array}{c}\left(19.84-19.32\right)-1.87<{\mu }_1-{\mu }_2<\left(19.84-19.32\right)+1.87\\ 0.52-1.87<{\mu }_1-{\mu }_2<0.52+1.87\\ -1.35<{\mu }_1-{\mu }_2<2.39\end{array}$

Hence, the 85% confidence interval for ${\mu }_1-{\mu }_2$ is $-1.35<{\mu }_1-{\mu }_2<2.39$.

(b)

To determine

Find the 85% confidence interval for ${\mu }_1-{\mu }_3$.

Answer to Problem 25P

The 85% confidence interval for ${\mu }_1-{\mu }_3$ is $0.06<{\mu }_1-{\mu }_2<3.86$.

Explanation of Solution

From part (a), the critical value is 1.523.

Substitute 15 for $n_1$, 15 for $n_2$, 3.07 for $s_1$, 3.74 for $s_2$, 1.523 for $t_c$ in the margin of error formula (E).

$\begin{array}{c}E=1.523\sqrt{\frac{{3.07}^2}{15}+\frac{{3.74}^2}{15}}\\ =1.523\sqrt{1.56083}\\ =1.523\left(1.2493\right)\\ =1.90\end{array}$

The margin of error E is 1.90.

Substitute 19.84 for ${\overline{x}}_1$, 17.88 for ${\overline{x}}_2$, 1.90 for E in the confidence formula.

$\begin{array}{c}\left(19.84-17.88\right)-1.90<{\mu }_1-{\mu }_3<\left(19.84-17.88\right)+1.90\\ 1.96-1.90<{\mu }_1-{\mu }_3<1.96+1.90\\ 0.06<{\mu }_1-{\mu }_3<3.86\end{array}$

Hence, the 85% confidence interval for ${\mu }_1-{\mu }_3$ is $0.06<{\mu }_1-{\mu }_3<3.86$.

(c)

To determine

Find the 85% confidence interval for ${\mu }_2-{\mu }_3$.

Answer to Problem 25P

The 85% confidence interval for ${\mu }_2-{\mu }_3$ is $-0.61<{\mu }_2-{\mu }_3<3.49$.

Explanation of Solution

From part (a), the critical value is 1.523.

Substitute 15 for $n_1$, 15 for $n_2$, 3.62 for $s_1$, 3.74 for $s_2$, 1.523 for $t_c$ in the margin of error formula (E).

$\begin{array}{c}E=1.523\sqrt{\frac{{3.62}^2}{15}+\frac{{3.74}^2}{15}}\\ =1.523\sqrt{1.8061}\\ =1.523\left(1.3439\right)\\ =2.05\end{array}$

The margin of error E is 2.05.

Substitute 19.32 for ${\overline{x}}_1$, 17.88 for ${\overline{x}}_2$, 2.05 for E in the confidence formula.

$\begin{array}{c}\left(19.32-17.88\right)-2.05<{\mu }_2-{\mu }_3<\left(19.32-17.88\right)+2.05\\ 1.44-2.05<{\mu }_2-{\mu }_3<1.44+2.05\\ -0.61<{\mu }_2-{\mu }_3<3.49\end{array}$

Hence, the 85% confidence interval for ${\mu }_2-{\mu }_3$ is $-0.61<{\mu }_2-{\mu }_3<3.49$.

(d)

To determine

Interpret the confidence intervals found in (a), (b), (c) in the context of the problem.

Explain the relationship between average differences in influence on self-esteem between competence and social acceptance.

Explain the relationship between average differences in influence on self-esteem between competence and attractiveness.

Explain the relationship between average differences in influence on self-esteem between social acceptance and attractiveness.

Explanation of Solution

From part (a), the confidence interval for ${\mu }_1-{\mu }_2$ is $-1.35<{\mu }_1-{\mu }_2<2.39$. The interval can be interpreted as, there is 85% confidence that the difference in population means of self-esteem between competence and social acceptance lies within the interval –1.35 to 2.39.

The 85% confidence interval calculated for difference of means $\left({\mu }_1-{\mu }_2\right)$ includes negative and positive values, then it cannot be determined whether ${\mu }_1-{\mu }_2>0$ or ${\mu }_1-{\mu }_2<0$. The relation between two population means cannot be either ${\mu }_1<{\mu }_2$ or ${\mu }_1>{\mu }_2$. This shows that, there would be 85% sure that there is no difference between the average of first population $\left({\mu }_1\right)$ and average of second population $\left({\mu }_2\right)$. There is no difference in average influence on self-esteem of competence and social acceptance.

From part (b), the confidence interval for ${\mu }_1-{\mu }_3$ is $0.06<{\mu }_1-{\mu }_3<3.86$. The interval can be interpreted as, there is 85% confidence that the difference in population means of self-esteem between competence and attractiveness lies within the interval 0.06 to 3.86. It can be observed that, the 85% confidence interval calculated for difference of means $\left({\mu }_1-{\mu }_3\right)$ includes all positive values. The average difference in influence on self-esteem of competence is greater than attractiveness.

The 85% confidence interval calculated for difference of means $\left({\mu }_1-{\mu }_3\right)$ includes all the positive values, then ${\mu }_1-{\mu }_3>0$. The relation between two population means is ${\mu }_1>{\mu }_3$. This shows that, there would be 85% sure that the average of first population $\left({\mu }_1\right)$ was greater than the average of third population $\left({\mu }_3\right)$.

From part (c), the confidence interval for ${\mu }_2-{\mu }_3$ is $-0.61<{\mu }_2-{\mu }_3<3.49$. The interval can be interpreted as, there is 85% confidence that the difference in population means of self-esteem between social acceptance and attractiveness lies within the interval –0.61 to 3.49.

The 85% confidence interval calculated for difference of means $\left({\mu }_2-{\mu }_3\right)$ includes negative and positive values, then it cannot be determined whether ${\mu }_2-{\mu }_3>0$ or ${\mu }_2-{\mu }_3<0$. The relation between two population means cannot be either ${\mu }_2<{\mu }_3$ or ${\mu }_2>{\mu }_3$. This shows that, there would be 85% sure that there is no difference between the average of first population $\left({\mu }_2\right)$ and average of second population $\left({\mu }_3\right)$. There is no difference in average influence on self-esteem of social acceptance and attractiveness.