Question

A recent study of 2600 children randomly selected found 21​% 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 3 decimal places)

b. Explain carefully what the interval means?

A. We are 98% confident that the percent of people deficient in vitamin D is 24%

B. We are 98% confident that the percent of children deficient in vitamin D is 24%

C. We are 98% confident that the interval contains the true proportion of children deficient in Vitamin D.

D. We are 98% confident that the interval contains the true proportion of people deficient in Vitamin D.

Answer 1

Part a

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

n = 2600

P = 0.21

Confidence level = 98%

Critical Z value = 2.3263

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.21 ± 2.3263* sqrt(0.21*(1 – 0.21)/ 2600)

Confidence Interval = 0.21 ± 2.3263* 0.0080

Confidence Interval = 0.21 ± 0.0186

Lower limit = 0.21 - 0.0186 = 0.1914

Upper limit = 0.21 + 0.0186 = 0.2286

Confidence interval = (0.191, 0.229)

Part b

C. We are 98% confident that the interval contains the true proportion of children deficient in Vitamin D.

[…Because confidence interval is about the true population proportion, that is, it is about the proportion of the all children deficient in Vitamin D.]