Answer to Question #109498 in Statistics and Probability for Ibrahim

Question

Mean Standard Deviation
Men 797 482
Women 660 414

Use these estimates as the mean m and standard deviation s for the U.S. populations for these age
groups. If we take a random sample of 40 men and 35 women, what is the probability of obtaining a
difference between sample means of 100 mg or more?

Answer 1

$\mu_m=797\\ \mu_w=660\\ \sigma_m=482\\ \sigma_w=414\\ n_m=40\\ n_w=35$

We should find $P\{\overline{x}_m-\overline{x}_w\geq 100\}=1-P\{\overline{x}_m-\overline{x}_w< 100\}.$

$\overline{x}=\overline{x}_m-\overline{x}_w\text{ is random variable}.\\ \mu=M(\overline{x}_m-\overline{x}_w)=M(\overline{x}_m)-M(\overline{x}_w)=\mu_m-\mu_w=\\ =797-660=137\text{ (here we used the theorem } \mu_{\overline{x}}=\mu).$

Random variables $\overline{x}_m$ and $\overline{x}_w$ are independent. So $D(\overline{x}_m-\overline{x}_w)=D(\overline{x}_m)+D(\overline{x}_w)=482^2+414^2=\\ =403720.\\ \sigma=\sigma(\overline{x}_m-\overline{x}_w)=\sqrt{403720}.\\ n=n_m+n_w=40+35=75.\\ \text{As } n\geq 30\text{ we can apply the consequence of Central Limit Theorem.}\\ \text{We will consider random variable } Z=\frac{\overline{x}-\mu}{\sigma/\sqrt{n}}.\\ Z\text{ will be asymptotically normal:}\\ \lim_{n\rightarrow\infty}P\{Z\leq z\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^ze^{-\frac{t^2}{2}}dt.\\ \text{So } P\{\overline{x}<100\}=P\{Z<\frac{100-\mu}{\sigma/\sqrt{n}}\}=P\{Z<\frac{100-137}{\sqrt{403720}/\sqrt{75}}\}=P\{Z<-0.5043\}\approx 0.3070.\\ P\{\overline{x}\geq 100\}=1-P\{\overline{x}<100\}\approx 1-0.3070=0.693.$