Answer
a. .80
b. We must consider the normal chance that the population is skewed.
Work Step by Step
a. First, as the book recommends, we find the value of $\hat{p}$. To do this, we first must find the standard deviation:
$\sigma = \sqrt{\frac{(.5)(.5)}{64}}=.0625$
Thus, it follows:
$\hat{p}=z\sigma+\mu=.6028$
We now need to find the value of z so that we can find the power of the test:
$z=\frac{\hat{p}-p}{\sqrt{\frac{-p^2+p}{n}}}$
Plugging in the known values and then using the table of z-scores, we find that the sample proportion is:
$=1-.1977=.80$
b. The reason we consider the shaded area is that we only care about the area of the new curve that is not included in the typical population curve. In other words, we must consider the fact that there may be an abnormally high number of girls in the normal population of a study.
