Question

A nutritionist is interested in calculating confidence intervals for the percentage of American adults who eat salad at least once a week.

She takes a random sample of 200 American adults and finds that 176 of them eat salad at least once a week.

What sample size would she need in order to have a confidence level of 99% and a margin of error of 0.03?

Answer 1

Solution:

Given that,

n =200

x =176

= x /n = 176 /200 =0.880

1 - = 1 - 0.880 = 0.120

margin of error = E = 0.03

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

Sample size = n = ((Z / 2) / E)2 * * (1 - )

= (2.576 / 0.03)2 * 0.880 * 0.120

= 778

n = sample size =778