Chapter 10.2, Problem 38E

(a)

To determine

To find out what is the shape of the sampling distribution of $\text{}\frac{305}{1526}=0.20=20\%{\overline{x}}_M-{\overline{x}}_W$ and explain why.

Answer to Problem 38E

The distribution of $M-W$ is normal with mean and standard deviation as:

  $\begin{array}{l}{\mu }_{M-B}=4.8\\ {\sigma }_{M-B}=3.7537\end{array}$

Explanation of Solution

It is given that there are two distributions that is,

  $\begin{array}{l}\text{Distribution }M:\text{ Normal with }{\mu }_M=69.3,{\sigma }_M=2.8\\ \text{Distribution }W:\text{ Normal with }{\mu }_W=64.5,{\sigma }_W=2.5\end{array}$

Now, if M and B are normally distributed then there difference $M-W$ is also normally distributed.

And we know that the properties of normal are:

  $\begin{array}{l}{\mu }_{aX+bY}=a{\mu }_X+b{\mu }_Y\\ {\sigma }_{aX+bY}=\sqrt{a^2{\sigma }^2{}_X+b^2{\sigma }^2{}_Y}\end{array}$

Then we obtain:

  $\begin{array}{l}{\mu }_{M-W}={\mu }_M-{\mu }_W\\ =69.3-64.5\\ =4.8\\ {\sigma }_{M-W}=\sqrt{{\sigma }^2{}_M+{\sigma }^2{}_W}\\ =\sqrt{{2.8}^2+{2.5}^2}\\ =3.7537\end{array}$

Thus, we conclude that the distribution of $M-W$ is normal with mean and standard deviation as:

  $\begin{array}{l}{\mu }_{M-B}=4.8\\ {\sigma }_{M-B}=3.7537\end{array}$

(b)

To determine

To calculate the mean of the sampling distribution.

Answer to Problem 38E

The mean of the sampling mean is $4.8$ inches.

Explanation of Solution

Given:

  $\begin{array}{l}{\mu }_1=69.3\\ {\mu }_2=64.5\\ n_1=16\\ n_2=9\\ {\sigma }_1=2.8\\ {\sigma }_2=2.5\end{array}$

Thus, the mean of the sampling distribution of the difference in sample means is the difference in the population means. This implies;

  $\begin{array}{l}{\mu }_{{\overline{x}}_1-{\overline{x}}_2}={\mu }_1-{\mu }_2\\ =69.3-64.5\\ =4.8\end{array}$

Thus, we conclude that the mean of the sampling mean is $4.8$ inches.

(c)

To determine

To calculate and interpret the standard deviation of the sampling distribution.

Answer to Problem 38E

The difference in the sample means are expected to be vary by $1.0883$ inches from the mean difference of $4.8$ inches.

Explanation of Solution

Given:

  $\begin{array}{l}{\mu }_1=69.3\\ {\mu }_2=64.5\\ n_1=16\\ n_2=9\\ {\sigma }_1=2.8\\ {\sigma }_2=2.5\end{array}$

Since the sample $16$ young men is less than $10\%$ of all young men and since the sample of $9$ young women is less than $10\%$ of all young women. Thus, the standard deviation is as follows:

  $\begin{array}{l}{\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}\\ =\sqrt{\frac{{2.8}^2}{16}+\frac{{2.5}^2}{9}}\\ =1.0883\end{array}$

Thus, we conclude that the difference in the sample means are expected to be vary by $1.0883$ inches from the mean difference of $4.8$ inches.