Question

In a study of school dining halls, a sample of 75 high schools in New Jersey showed that 15 provided gluten-free meal options. Develop a 98% confidence interval for the proportion of the studied New Jersey high schools that provide gluten-free meals.

Group of answer choices

10.9% to 29.1%

15.3%to 24.6%

9.2.% to 30.8%

None of the above

Answer 1

Solution :

Given that,

n = 75

x = 15

Point estimate = sample proportion = = x / n = 15/75=0.2

1 -   = 1-0.2 =0.8

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 (((0.2*0.8) / 75)

E = 0.1074

A 98% confidence interval for proportion p is ,

- E < p < + E

0.2 - 0.1074< p < 0.2+0.1074

0.092< p < 0.3080

9.2.% to 30.8%