Question

A July 2015 Gallup poll interviewed by phone a random sample of 1009 American adults; 675 said they were regular coffee drinkers. Give a 99% confidence interval for the proportion of American adults who are regular coffee drinkers. detailed solution please and thank you.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 675 / 1009 = 0.669

1 - = 1 - 0.669 = 0.331

Z_/2 = 2.576

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

= 2.576 * (((0.669 * 0.331) / 1009)

= 0.038

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.669 - 0.038 < p < 0.669 + 0.038

0.631 < p < 0.707

The 99% confidence interval for the population proportion p is : 0.631 , 0.707