Question

A nutritionist is interested in determining 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. Find a 95% confidence interval for the true proportion of American adults who eat salad at least once a week. Due to the different methods, round OUT and give 2 decimal places. Select the best possible answer.

Answer 1

Solution :

n = 200

x = 176

= x / n = 176 / 200 = 0.880

1 - = 1 - 0.880 = 0.120

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

=1.960 * (((0.880 * 0.120) / 200)

= 0.045

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.880 - 0.045 < p < 0.880 + 0.045

0.835 < p < 0.925