Answer:
[tex]0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644[/tex]
[tex]0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672[/tex]
We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672
Step-by-step explanation:
We can begin founding the estimated proportion of people overweight:
[tex]\hat p =\frac{2913}{4430}= 0.658[/tex]
We need to find a critical value for the confidence level using the normal standard distribution. We know that 95% is the confidence level, then the significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The confidence interval for the true population proportion of interest is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
Replacing the data provided we got:
[tex]0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644[/tex]
[tex]0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672[/tex]
We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672
