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

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

Step-by-step explanation:

The (1 - α)% confidence interval for population proportion p is:

`CI=\hat p\pm z_(\alpha/2)* \sqrt{(\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)* \sqrt{(\hat p(1-\hat p))/(n)}`

`=0.20\pm 2.33* \sqrt{(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).