Refer to Exercise 36. Suppose we select independent SRSs of 16 young men and 9 young women and calculate the sample mean heights ${\overline{x}}_{M}and{\overline{x}}_W$

(a) Describe the shape, center, and spread of the sampling distribution of ${\overline{x}}_M-{\overline{x}}_W$

(b) Find the probability of getting a difference in sample means ${\overline{x}}_M-{\overline{x}}_W$that’s greater than or equal to $2$inches. Show your work.

(c) Should we be surprised if the sample mean height for the young women is more than $2$inches less than the sample mean height for the young men? Explain.

Select independent SRSs of $16$young men and $9$young women and calculate the sample mean heights.

Distribution M: Normal with ${\mu }_M=69.3and{\sigma }_M=2.8$

Distribution W: Normal with ${\mu }_W=64.5and{\sigma }_W=2.5$

$n_M=16$

$n_W=9$

The sample mean $\overline{x}$is normally distributed with mean $\mu$and standard deviation $\sigma /\sqrt{n}$(if the population distribution is normal with mean $\mu$and standard deviation $\sigma$).

Properties mean and standard deviation

${\mu }_{aX+bY}=a{\mu }_X+b{\mu }_Y$

${\sigma }_{aX+bY}=\sqrt{a^2{\sigma }_X^2+b^2{\sigma }_Y^2}$

Then we get

localid="1650515095538" $$\begin{gathered} {\mu }_{{\overline{x}}_M-{\overline{x}}_W}={\mu }_{{\overline{x}}_M}-{\mu }_{{\overline{x}}_W} \\ =69.3-64.5 \\ =4.8 \end{gathered}$$

localid="1650515126305" $$\begin{gathered} {\sigma }_{{\overline{x}}_M-{\overline{x}}_W}=\sqrt{{\sigma }_{{\overline{x}}_M}^2+{\sigma }_{{\overline{x}}_W}^2} \\ =\sqrt{\frac{2.8^2}{16}+\frac{2.5^2}{9}} \\ \approx 1.0883 \end{gathered}$$

Normal with${\mu }_{x_M-{\overline{x}}_W}=4.8$and${\sigma }_{{\overline{x}}_M-{\overline{x}}_W}\approx 1.0883$.

Select independent SRSs of $16$ young men and $9$ young women and calculate the sample mean heights.

From part (a)

${\mu }_{x_M-{\overline{x}}_W}=4.8$

${\sigma }_{{\overline{x}}_M-{\overline{x}}_W}\approx 1.0883$

The z-value is the value decreased by the population mean, divided by the standard deviation:

localid="1650515152026" $$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{2-4.8}{1.0883} \\ =-2.57 \end{gathered}$$

Find the probability using table A

localid="1650515176292" $$\begin{gathered} P\left({\overline{x}}_M-{\overline{x}}_W\ge 2\right)=P(Z\ge -2.57) \\ =P(Z<2.57) \\ =0.9949 \end{gathered}$$

$P\left({\overline{x}}_M-{\overline{x}}_W\ge 2\right)=0.9949.$

Select independent SRSs of $16$ young men and $9$ young women and calculate the sample mean heights.

From part (b)

$$\begin{gathered} P\left({\overline{x}}_M-{\overline{x}}_W\ge 2\right)=0.9949 \\ =99.49\% \end{gathered}$$

Since the probability is more than $5\%,$it is likely that the sample mean for men exceeds the sample mean for women and thus we should not be surprised.