Answer:
[tex]z=\frac{0.815 -0.5}{\sqrt{\frac{0.5(1-0.5)}{211}}}=9.151[/tex]
We can calculate the p value using the fact that we are conducting a right tailed test:
[tex]p_v =P(z>9.151)\approx 0[/tex]
Since the p value i approximately 0 we can conclude that we have enough evidence to say that with this method, the probability of a baby being a boy is significantly greater than 0.5
Step-by-step explanation:
Information given
n=172+39=211 represent the sample size
X=172 represent the number of boys in the sample
[tex]\hat p = \frac{172}{211}= 0.815[/tex] estimated proportion of of boys
[tex]p_o=0.5[/tex] is the value to verify
z would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to test
The ides is verify if with this method, the probability of a baby being a boy is greater than 0.5, so then the system of hypothesis are.:
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
The statitic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Replacing the info we got:
[tex]z=\frac{0.815 -0.5}{\sqrt{\frac{0.5(1-0.5)}{211}}}=9.151[/tex]
We can calculate the p value using the fact that we are conducting a right tailed test:
[tex]p_v =P(z>9.151)\approx 0[/tex]
Since the p value i approximately 0 we can conclude that we have enough evidence to say that with this method, the probability of a baby being a boy is significantly greater than 0.5