Question

Vitamin​ D, whether ingested as a dietary supplement or produced naturally when sunlight falls upon the​ skin, is essential for​ strong, healthy bones. The bone disease rickets was largely eliminated during the​ 1950s, but now there is concern that a generation of children more likely to watch TV or play computer games than spend time outdoors is at increased risk. A recent study of 3400 children randomly selected found 2020​% of them deficient in vitamin D. Find a 98% confidence interval.

Answer 1

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:

$CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The information provided is:

$n=3400\\\hat p=0.20\\\alpha =0.02$

Compute the critical value of z for 98% confidence interval as follows:

$z_{\alpha/2}=z_{0.02/2}=z_{0.01}=2.33$

*Use a z-table for the value.

Compute the 98% confidence interval for the population proportion as follows:

$CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$

     $=0.20\pm 2.33\times \sqrt{\frac{0.20(1-0.20)}{3400}}$

     $=0.20\pm 0.0161\\=(0.1839,\ 0.2161)\\\approx (0.18,\ 0.22)$

Thus, the 98% confidence interval for children deficient in vitamin D is (0.18, 0.22).