In a study conducted by the U.S. Department of Health and Human Services, a sample of 546 boys aged 6–11 was weighed, and it was determined that 91 of them were overweight. A sample of 508 girls aged 6–11 was also weighed, and 74 of them were overweight. Can you conclude that the proportion of boys who are overweight differs from the proportion of girls who are overweight? Find the P-value and state a conclusion. Round the answer to four decimal places.

So the p value is a very high value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of boys who are overweight is not significantly different from the proportion of girls who are overweight

Step-by-step explanation:

1) Data given and notation

[tex]X_{M}=91[/tex] represent the number of boys that were overweight.

[tex]X_{W}=74[/tex] represent the number of girls that were overweight.

[tex]n_{M}=546[/tex] sample of boys selected

[tex]n_{W}=508[/tex] sample of girls selected

[tex]p_{M}=\frac{91}{546}=0.167[/tex] represent the proportion of boys that were overweight.

[tex]p_{W}=\frac{74}{508}=0.146[/tex] represent the proportion of girls that were overweight.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the value for the test (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion of boys who are overweight differs from the proportion of girls who are overweight , the system of hypothesis would be:

Null hypothesis:[tex]p_{M} = p_{W}[/tex]

Alternative hypothesis:[tex]p_{M} \neq p_{W}[/tex]

We need to apply a z test to compare proportions, and the statistic is given by:

[tex]z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}[/tex] (1)

Where [tex]\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{91+74}{546+508}=0.157[/tex]

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

[tex]z=\frac{0.167-0.146}{\sqrt{0.157(1-0.157)(\frac{1}{546}+\frac{1}{508})}}=0.936[/tex]

4) Statistical decision

For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test.

Since is a two tailed test the p value would be:

[tex]p_v =2*P(Z>0.936)=0.3493[/tex]

So the p value is a very high value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of boys who are overweight is not significantly different from the proportion of girls who are overweight