Using the normal approximation, what is the $z$-value of a sample difference of $\hat p_1−\hat p_2=−0.18$?

Question

A few concepts from my textbook that I do not understand:

• If the Random, Normal, and Independent conditions are met, then is it true that $\hat p_1+2\hat p_2$ is approximately normally distributed?

My instinct is that it would remain normal, since both the mean and standard deviation of $\hat p_2$, which is itself normal, would be multiplied by a constant $-2$?

• When Anne and Carl play tennis, Anne has a $50\%$ chance of winning each game, and when Billy and Carl play, Billy has a $40\%$ chance of winning each game. Anne and Billy each play Carl $25$ times. If $\hat p_1$ is Anne's sample proportion of wins and $\hat p_2$ is Billy's sample proportion of wins, then $\hat p_1−\hat p_2$ may be approximated by a normal distribution. Using this approximation, what is the $z$-value of a sample difference of $\hat p_1−\hat p_2=−0.18$?

I'm not sure how to approach this; should I use the probabilities in the two-sample Z formula and use the result of approximately $-1.286$?

Thanks!

Answer 1

I don't remember statistics quite well. But I think we can proceed as follows.

Using this approximation, what is the $z$-value of a sample difference of $\hat p_1-\hat p_2 = -0.18$?

Denote the sample difference as $$\hat d = \hat p_1-\hat p_2.$$

Then we are being asked to find the $z$ score of $\hat d$.

We see that $$\mu = E[\hat d] = E[\hat p_1]-E[\hat p_2] = .5-.4 = .1$$ and that $$\sigma^2 = \text{Var}(\hat d) = \text{Var}(\hat p_1)+\text{Var}(\hat p_2) = \frac{1}{25}(.5)(1-.5)+\frac{1}{25}(.4)(1-.6) = 0.0164.$$

Thus, under the normal and independent assumptions, we see that $$\hat d \sim N(\mu = .1, \sigma^2 = 0.0164)$$ and hence $$z = \frac{-0.18-\mu}{\sigma} = -2.186433$$