91Ó°ÊÓ

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

Distribution M: Normal with ${\mathrm{μ}}_M=69.3and{\mathrm{σ}}_M=2.8$

Distribution W: Normal with ${\mathrm{μ}}_W=64.5and{\mathrm{σ}}_W=2.5$

$n_M=16$

$n_W=9$

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

Properties mean and standard deviation

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

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

Then we get

localid="1650515095538" $$\begin{gathered} {\mathrm{μ}}_{{\stackrel{¯}{x}}_M-{\stackrel{¯}{x}}_W}={\mathrm{μ}}_{{\stackrel{¯}{x}}_M}-{\mathrm{μ}}_{{\stackrel{¯}{x}}_W} \\ =69.3-64.5 \\ =4.8 \end{gathered}$$

localid="1650515126305" $$\begin{gathered} {\mathrm{σ}}_{{\stackrel{¯}{x}}_M-{\stackrel{¯}{x}}_W}=\sqrt{{\mathrm{σ}}_{{\stackrel{¯}{x}}_M}^2+{\mathrm{σ}}_{{\stackrel{¯}{x}}_W}^2} \\ =\sqrt{\frac{2.8^2}{16}+\frac{2.5^2}{9}} \\ ≈1.0883 \end{gathered}$$

Normal with${\mathrm{μ}}_{x_M-{\stackrel{¯}{x}}_W}=4.8$and${\mathrm{σ}}_{{\stackrel{¯}{x}}_M-{\stackrel{¯}{x}}_W}≈1.0883$.

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

From part (a)

${\mathrm{μ}}_{x_M-{\stackrel{¯}{x}}_W}=4.8$

${\mathrm{σ}}_{{\stackrel{¯}{x}}_M-{\stackrel{¯}{x}}_W}≈1.0883$

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

localid="1650515152026" $$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{2-4.8}{1.0883} \\ =-2.57 \end{gathered}$$

Find the probability using table A

localid="1650515176292" $$\begin{gathered} P\left({\stackrel{¯}{x}}_M-{\stackrel{¯}{x}}_Wâ‰\yen 2\right)=P(Zâ‰\yen -2.57) \\ =P(Z<2.57) \\ =0.9949 \end{gathered}$$

$P\left({\stackrel{¯}{x}}_M-{\stackrel{¯}{x}}_Wâ‰\yen 2\right)=0.9949.$

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

From part (b)

$$\begin{gathered} P\left({\stackrel{¯}{x}}_M-{\stackrel{¯}{x}}_Wâ‰\yen 2\right)=0.9949 \\ =99.49\% \end{gathered}$$

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.