Respuesta :
Answer:
The 98% confidence interval for children deficient in vitamin D is (0.18, 0.22).
Explanation:
The (1 - α)% confidence interval for population proportion p is:
[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The information provided is:
[tex]n=3400\\\hat p=0.20\\\alpha =0.02[/tex]
Compute the critical value of z for 98% confidence interval as follows:
[tex]z_{\alpha/2}=z_{0.02/2}=z_{0.01}=2.33[/tex]
*Use a z-table for the value.
Compute the 98% confidence interval for the population proportion as follows:
[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]=0.20\pm 2.33\times \sqrt{\frac{0.20(1-0.20)}{3400}}[/tex]
[tex]=0.20\pm 0.0161\\=(0.1839,\ 0.2161)\\\approx (0.18,\ 0.22)[/tex]
Thus, the 98% confidence interval for children deficient in vitamin D is (0.18, 0.22).
