Refer exercise $35.$Suppose we select independent SRSs of $25$men aged $20$to $34$and $36$boys aged $14$and calculate the sample mean heights ${\overline{x}}_M$and ${\overline{x}}_B$.

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

(b) Find the probability of getting a difference in sample means ${\overline{x}}_M-{\overline{x}}_B$that’s less than$0$mg/dl. Show your work.

(c) Should we be surprised if the sample mean cholesterol level for the $14$-year-old boys exceeds the sample mean cholesterol level for the men? Explain.

Select independent SRSs of $24$men aged $20$to $34$and $36$boys aged $14$ and calculate the sample mean heights.

From exercise $35,$

Distribution M: Normal with ${\mu }_M=188and{\sigma }_M=41$

Distribution B: Normal with ${\mu }_B=170and{\sigma }_B=30$

$n_M=25$

$n_B=36$

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

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

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

Therefore we get,

localid="1650363687592" $$\begin{gathered} {\mu }_{{\overline{x}}_M-{\overline{x}}_B}={\mu }_{{\overline{x}}_M}-{\mu }_{{\overline{x}}_B} \\ =188-170 \\ =18 \end{gathered}$$

localid="1650363704750" $$\begin{gathered} {\sigma }_{{\overline{x}}_M-{\overline{x}}_B}=\sqrt{{\sigma }_{{\overline{x}}_M}^2+{\sigma }_{{\overline{x}}_B}^2} \\ =\sqrt{\frac{{41}^2}{25}+\frac{{30}^2}{36}} \\ \approx 9.6042 \end{gathered}$$

The distribution is normal with${\mu }_{{\overline{x}}_M-{\overline{x}}_B}=18$and${\sigma }_{{\overline{x}}_M-{\overline{x}}_B}\approx 9.6042.$

Select independent SRSs of $24$men aged $20$to $34$and $36$boys aged $14$ and calculate the sample mean heights.

From the result of part (a),

${\mu }_{{\overline{x}}_M-{\overline{x}}_B}=18$and ${\sigma }_{{\overline{x}}_M-{\overline{x}}_B}\approx 9.6042$

The z-value is the difference between the population mean and the standard deviation, divided by the population mean:

localid="1650514978192" $$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0-18}{9.6042} \\ =-1.87 \end{gathered}$$

Find the probability using table A

localid="1650363838861" $$\begin{gathered} P\left({\overline{x}}_M-{\overline{x}}_B<0\right)=P(Z<-1.87) \\ =0.0307 \end{gathered}$$

$P\left({\overline{x}}_M-{\overline{x}}_B<0\right)=0.0307.$

Select independent SRSs of $24$men aged $20$to $34$and $36$ boys aged $14$ and calculate the sample mean heights.

From part (b)

$$\begin{gathered} P\left({\overline{x}}_M-{\overline{x}}_B<0\right)=0.0307 \\ =3.07\% \end{gathered}$$

Because the chance is less than , it's improbable that the sample mean for boys will be higher than the sample mean for men, therefore we shouldn't be astonished.