Question

A recent study of 3100 children randomly selected found 20​% of them deficient in vitamin D.

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

left parenthesis nothing comma nothing right parenthesis,

​(Round to three decimal places as​ needed.)

Answer 1

Solution :

Given that,

n = 3100

Point estimate = sample proportion = =0.20

1 - = 1 -0.20 =0.80

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

= 2.326 (((0.20*0.80) / 3100)

E= 0.0167

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.20 -0.0167 < p <0.20+ 0.0167

0.1833< p < 02167

The 98% confidence interval for the population proportion p is : 0.1833,02167