Chapter 18, Problem 18.26E

(a)

To determine

To explain:

The shapes of the distribution of the given sample.

Answer to Problem 18.26E

  $S=17.84$

Explanation of Solution

Given:

  $\begin{array}{l}{n}_1=25,s_1=14\\ n_2=25,s_2=21\end{array}$

Concept used:

Formula

Standard error

  $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Degree of freedom

  $=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_2-1}}$

Significance level

  $\alpha =1-\text{confidence}$

Calculation:

The $95\%$ confidence interval for difference

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ S^2=\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}\\ S^2=\frac{\left(25-1\right){\left(14\right)}^2+\left(25-1\right){\left(21\right)}^2}{25+25-2}\\ S^2=\frac{15288}{48}\\ S^2=318.5\\ S=17.84\end{array}$

(b)

To determine

To explain:

The appropriate test of the given sample.

Answer to Problem 18.26E

  $25.0258$

Explanation of Solution

Given:

  $\begin{array}{l}{n}_1=25,s_1=14\\ n_2=25,s_2=21\end{array}$

Concept used:

Formula

Standard error

  $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Degree of freedom

  $=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_2-1}}$

Significance level

  $\alpha =1-\text{confidence}$

Calculation:

To test is significance evidence of a different in elimination half-life depending

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_0:{\mu }_1\ne {\mu }_2\\ \alpha =0.05\end{array}$

The $95\%$ confidence interval for difference

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ S^2=\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}\\ S^2=\frac{\left(25-1\right){\left(14\right)}^2+\left(25-1\right){\left(21\right)}^2}{25+25-2}\\ S^2=\frac{15288}{48}\\ S^2=318.5\\ S=17.84\end{array}$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

(c)

To determine

To explain:

The difference in mean plasma testosterone level between adolescent males of the given sample.

Answer to Problem 18.26E

  $\left(-0.2721,-0.1023\right)$

Explanation of Solution

Given:

  $\begin{array}{l}{n}_1=25,s_1=14\\ n_2=25,s_2=21\end{array}$

Concept used:

Formula

Standard error

  $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Degree of freedom

  $=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_2-1}}$

Significance level

  $\alpha =1-\text{confidence}$

Calculation:

To test is significance evidence of a different in elimination half-life depending

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_0:{\mu }_1\ne {\mu }_2\\ \alpha =0.05\end{array}$

The $95\%$ confidence interval for difference

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ S^2=\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}\\ S^2=\frac{\left(25-1\right){\left(14\right)}^2+\left(25-1\right){\left(21\right)}^2}{25+25-2}\\ S^2=\frac{15288}{48}\\ S^2=318.5\\ S=17.84\end{array}$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

The $95\%$ confidence interval for difference in mean plasma testosterone level between adolescent males

A healthy weight is

  $\left(-0.2721,-0.1023\right)$