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A hospital claims that the proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%. In a random sample of 170 babies born in this hospital, 56 weighed over 7 pounds. Is there enough evidence to reject the hospital's claim at the level of significance?

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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

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