Answer
a. $H_0$, $p=.5$; $H_1$, $p\ne .5$
Work Step by Step
a. The null hypothesis is the average fraction of girls, which is .5, and the alternative hypothesis (this time) is that the proportion is not 50 percent (rather than greater than 50 percent), meaning that $p\ne .5$.
b. $\alpha $ is the significance level, which the problem states is .05.
c. We see that the sample distribution of the sample statistic is a normal distribution.
d. In this problem, we are asked to show that the percentage of girls born is not 50 percent. Thus, the distribution is two tailed, for we do not care on what side of 50 percent it is: we are just trying to verify the percent of girls born is not 50 percent.
e. The problem states that the sample statistic is 1.00.
f. To solve this problem, go to the table of negative z-scores, and go to column .00 and row -1.0 to get a value of .1587. Then, go to column .00 and row 1.0 on the table of positive z-scores to get .8413. Thus, we find:
$P=1-.8413+.1587=.3174$
g. Using the table of z-scores, we find that the critical value is 1.96.
h. Since the significance level is .05, it follows that the area of the critical region is .05.
