Childhood obesity: A national health survey weighed a sample of 543 boys aged $6-11$ and found that 83 of them were overweight. They weighed a sample of 533 girls aged 6-11 and found that 73 of them were overweight. Can you condude that the proportion of boys who are overweight differs from the proportion of girls who are overweight? let $p_1$ denote the proportion of boys who are overweight and let $p_2$ denote the proportion of girls who are overneight. Use the $\alpha =0.10$ level of significance and the P-value method with the TI-84 Plus calculator.
Part 1 of 4
State the appropriate null and alternate hypotheses.
$$\begin{array}{l}{H}_0:p_1-p_2\\ H_1:p_1\ne p_2\end{array}$$

This is a two-talled $\mathbf{v}$ test.

Part: $1/4$

Part 2 of 4

Find the P-value. Round the answer to four decimal places.
$$\text{ p-value - }◻$$

Step 1 ：State the appropriate null and alternate hypotheses. The null hypothesis is $H_0:p_1=p_2$ and the alternate hypothesis is $H_1:p_1\ne p_2$. This is a two-tailed test.

Step 2 ：Calculate the sample proportions. For boys, ${\hat{p}}_1=\frac{83}{543}=0.15285451197053407$. For girls, ${\hat{p}}_2=\frac{73}{533}=0.13696060037523453$.

Step 3 ：Calculate the pooled sample proportion. $\hat{p}=\frac{83+73}{543+533}=0.1449814126394052$.

Step 4 ：Calculate the test statistic using the formula \(Z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{543} + \frac{1}{533})}\). The test statistic is approximately 0.740362196455531.

Step 5 ：Find the P-value by looking up the test statistic in the standard normal distribution table. Since this is a two-tailed test, the P-value is twice the area to the right of the test statistic (or to the left if the test statistic is negative). The P-value is approximately 0.4590802507376859.

Step 6 ：Compare the P-value to the significance level. The P-value is greater than the significance level of 0.10, so we do not reject the null hypothesis. This means that we do not have enough evidence to conclude that the proportion of boys who are overweight differs from the proportion of girls who are overweight.

Step 7 ：The final answer is: The P-value is approximately $\boxed{{\displaystyle 0.4591}}$.