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Random variables $X_1$ , $X_2$ , ... , $X_n$ form a random sample from the Normal distribution with mean 20 and standard deviation 2. Random variables $Y_1$ , $Y_2$ , ... , $Y_m$ form a random sample from the Normal distribution with mean 19 and standard deviation 3.

a) If $n$ = 10 and $m$ = 5, determine the probability that the sample mean of the $Y$’s exceeds the sample mean of the $X$’s.

b) Suppose that $n$ = 2$m$. What is the smallest aggregate sample size ($n$ + $m$ = 3$m$) for which the probability of the event described in part a is no greater than 0.001.

I figured out part "a" but part b is confusing me, I know it involves qnorm(.001) but I can't seem to find out what to do with the aggregate sample size. Any ideas?

1 Answers 1

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We can begin with:

\begin{equation} \bar{X} \sim N(20, 4/n) \end{equation}

\begin{equation} \bar{Y} \sim N(19, 9/m) \end{equation}

Then, it will be useful to define:

\begin{equation} \bar{Y} - \bar{X} = W \sim N\biggl(-1, \frac{4m + 2n}{mn}\biggl) \end{equation}

Hence, now we can write: $P(\bar{Y} > \bar{X}) = P(W > 0) = P\biggl(Z > \sqrt{\frac{mn}{4m + 2n}}\biggr)$

Where Z is a standard normal random variable. Hence $P(Event) = 1 - \Phi\biggl(\sqrt{\frac{mn}{4m + 2n}}\biggr)$

Now to answer part (b). Using the fact that $n = 2m$. Hence: $P(event) = 1 - \Phi\biggl(\sqrt{\frac{m}{4}}\biggr)$

Set this equal to .001, and solve for $m$! \begin{align*} .001 &=1 - \Phi\biggl(\sqrt{\frac{m}{4}}\biggr) \\ \Phi\biggl(\sqrt{\frac{m}{4}}\biggr) &= .999 \\ \sqrt{\frac{m}{4}}& = \Phi^{-1}(.999) \\ m &= 4[\Phi^{-1}(.999)]^2 \end{align*}

If you're using R, then $\Phi^{-1}(x) = \texttt{qnorm(x)}$