Chapter 10.2, Problem 38E

a.

To determine

To discuss: The sampling distribution of ${\overline{x}}_M-{\overline{x}}_W$

Answer to Problem 38E

It is approximately normally distributed.

Explanation of Solution

Given:

  $\begin{array}{l}{\mu }_M=69.3,{\sigma }_M=2.8,n_M=16\\ {\mu }_W=170,{\sigma }_W=2.5,n_B=9\end{array}$

Formula used:

The properties for mean and standard deviation are:

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

Calculation:

The calculation for mean and standard deviation is done as:

  $\begin{array}{l}{\mu }_{aX-bY}=a{\mu }_X-b{\mu }_X=69.3-64.5=4.8\\ {\sigma }_{aX+bY}=\sqrt{a^2{\sigma }_X^2+b^2{\sigma }_Y^2}=\sqrt{\frac{{2.8}^2}{16}+\frac{{2.5}^2}{9}}=1.0883\end{array}$

Thus, it is normally distributed with mean = 4.8 and standard deviation = 1.0883

b.

To determine

To find: The probability that difference between ${\overline{x}}_M-{\overline{x}}_W$ is at least 2.

Answer to Problem 38E

The probability is 0.9949.

Explanation of Solution

Calculation:

The probability that difference between ${\overline{x}}_M-{\overline{x}}_W$ is at least 2 can be calculated as:

  $\begin{array}{c}P\left({\overline{x}}_M-{\overline{x}}_W\ge 2\right)=P\left(Z\ge \frac{2-4.8}{1.0883}\right)\\ =P\left(Z\ge -2.57\right)\\ =0.9949\end{array}$

Thus, the probability of difference is 0.9949.

c.

To determine

To explain: If it is surprising that the sample mean height of female is more than 2 inches than the sample mean height of male.

Answer to Problem 38E

No.

Explanation of Solution

The above part shows that the probability is 0.9949 which is above 0.05. It implies that the event is not unusual. So, it is not surprising that the sample mean height of female is more than 2 inches than the sample mean height of male.