Chapter 10.2, Problem 37E

(a)

To determine

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

Answer to Problem 37E

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

  $\begin{array}{l}{\mu }_{M-B}=18\\ {\sigma }_{M-B}=50.8035\end{array}$

Explanation of Solution

It is given that there are two distributions that is,

  $\begin{array}{l}\text{Distribution }M:\text{ Normal with }{\mu }_M=188,{\sigma }_M=41\\ \text{Distribution }B:\text{ Normal with }{\mu }_B=170,{\sigma }_B=30\end{array}$

Now, if M and B are normally distributed then there difference $M-B$ 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:

  $\begin{array}{l}{\mu }_{M-B}={\mu }_M-{\mu }_B\\ =188-170\\ =18\\ {\sigma }_{M-B}=\sqrt{{\sigma }^2{}_M+{\sigma }^2{}_B}\\ =\sqrt{{41}^2+{30}^2}\\ =\sqrt{2581}\\ =50.8035\end{array}$

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

  $\begin{array}{l}{\mu }_{M-B}=18\\ {\sigma }_{M-B}=50.8035\end{array}$

(b)

To determine

To calculate the mean of the sampling distribution.

Answer to Problem 37E

The mean of the sampling mean is $18$ mg/dl.

Explanation of Solution

Given:

  $\begin{array}{l}{\mu }_1=188\\ {\mu }_2=170\\ n_1=25\\ n_2=36\\ {\sigma }_1=41\\ {\sigma }_2=30\end{array}$

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

  $\begin{array}{l}{\mu }_{{\overline{x}}_1-{\overline{x}}_2}={\mu }_1-{\mu }_2\\ =188-170\\ =18\end{array}$

Thus, we conclude that the mean of the sampling mean is $18$ mg/dl.

(c)

To determine

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

Answer to Problem 37E

The difference in the sample means are expected to be vary by $9.6042$ mg/dl from the mean difference of $18$ mg/dl.

Explanation of Solution

Given:

  $\begin{array}{l}{\mu }_1=188\\ {\mu }_2=170\\ n_1=25\\ n_2=36\\ {\sigma }_1=41\\ {\sigma }_2=30\end{array}$

Since the sample $25$ men aged $20\text{ to 34}$ is less than $10\%$ of all men aged $20\text{ to 34}$ and since the sample of $36$ boys aged fourteen is less than $10\%$ of allboys aged fourteen. Thus, the standard deviation is as follows:

  $\begin{array}{l}{\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}\\ =\sqrt{\frac{{41}^2}{25}+\frac{{30}^2}{36}}\\ =9.6042\end{array}$

Thus, we conclude that the difference in the sample means are expected to be vary by $9.6042$ mg/dl from the mean difference of $18$ mg/dl.