Answer:
Claim :The proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%.
n = 170
x = 56
We will use one sample proportion test
[tex]\widehat{p}=\frac{x}{n}[/tex]
[tex]\widehat{p}=\frac{56}{170}[/tex]
[tex]\widehat{p}=0.3294[/tex]
The proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%.
[tex]H_0:p \neq 0.36 \\H_a:p= 0.36[/tex]
Formula of test statistic =[tex]\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]=\frac{0.3294-0.36}{\sqrt{\frac{0.36(1-0.36)}{170}}}[/tex]
=−0.8311
Now refer the p value from the z table
P-Value is .202987 (Calculated by online calculator)
Level of significance α = 0.05
Since p value < α
So we reject the null hypothesis .
Hence the claim is true
