Answer
a) For $H_0$, $p=0.5$, For $H_1$, $p\gt0.5$
b) $0.05$
c) Normal
d) Two-tailed
e) 1.00
f) $0.3174$
g) $1.96$
h) $0.05$
Work Step by Step
a. The null hypothesis is the average fraction of girls, which is $0.5$, and the alternative hypothesis is that the proportion is not 50 percent meaning that $p\ne0.5$.
b) $α$ is the significance level, which the problem says is $0.05$.
c) We can 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, because 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 exactly $50$ percent.
e) The problem states that the sample statistic is $1.00$.
f) To solve this problem, we check the table of negative z-scores, and go to column $0.00$ and row $-1.0$ to get the value of $0.1587$. Then we go to column $0.00$ and row $1.0$ on the table of positive z-scores to get $0.8413$. Hence we find:
$P=1−0.8413+0.1587=0.3174$
g) Using the table of z-scores, we can find that the critical value is $1.96$.
h) The significance level is $0.05$, hence the area of the critical region is $0.05$.
