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 23% of them deficient in vitamin D.

a.) Construct a 98% confidence interval for the true proportion of children who are deficient in vitamin D.

(_____, ______)

(round to three decimal places as needed)

Answer 1

Solution :

Given that,

n = 3400

Point estimate = sample proportion = = 23% = 0.23

1 - = 1 - 0.23 = 0.77

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z _0.01 = 2.326

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.326 * (((0.23 * 0.77) / 3400)

= 0.017

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.23 - 0.017 < p < 0.23 + 0.017

0.213 < p < 0.247

The 98% confidence interval for the population proportion p is : (0.213 , 0.247)