Chapter 18, Problem 18.39E

(a)

To determine

To find:

The two experimental groups are not significant different at the beginning of the study of the given sample.

Answer to Problem 18.39E

  $p-$ Value $=0.9921$

Explanation of Solution

Given:

  $\begin{array}{l}{n}_1=10,\overline{x_1}=109.31,\\ n_2=10,\overline{x_2}=109.33,\end{array}$

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 population mean weight of female rats at week $0$ of two groups by ${\mu }_1,{\mu }_2$.

Mean weight is significantly different in two groups in the beginning or not

  $$

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

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_0:{\mu }_1\ne {\mu }_2\\ \alpha =0.5\end{array}$

The $95\%$ confidence interval for difference

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ S^2=\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}\\ S^2=\frac{\left(10-1\right)14.305+\left(10-1\right)24.1378}{18}\\ S^2=\frac{9\times 14.305+9\times 24.1378}{18}\\ S=4.38425\end{array}$

Test statistic

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ t=\frac{109.31-109.33}{4.38425\sqrt{\frac{2}{10}}}\\ t=-0.010\end{array}$

  $p-$ Value $=0.9921$

(b)

To determine

To find:

The difference between the mean weight gains of two populations of the given sample.

Answer to Problem 18.39E

  $t=3.7627$

Explanation of Solution

Given:

  $\begin{array}{l}{n}_1=10,\overline{x_1}=194.63,\\ n_2=10,\overline{x_2}=183.24,\end{array}$

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 population mean weight of female rats at week $13$ of two groups by ${\mu }_1,{\mu }_2$.

Mean weight is significantly different in two groups in the beginning or not

  $$

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

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_0:{\mu }_1\ne {\mu }_2\\ \alpha =0.5\end{array}$

The $95\%$ confidence interval for difference

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ S^2=\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}\\ S^2=\frac{\left(10-1\right)60.95+\left(10-1\right)30.67}{18}\\ S^2=\frac{9\times 60.95+9\times 30.67}{18}\\ S=6.7687\end{array}$

Test statistic

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ t=\frac{194.63-183.24}{6.7687\sqrt{\frac{2}{10}}}\\ t=3.7627\end{array}$

  $p-$ Value $=0.00071$