Question

12. (8 pts) A survey of 1,000 randomly chosen U.S. adults asked whether the respondents thought that the manna eaten by the Israelites during their forty years of wandering in the desert was really popcorn. There were 70 people who said yes. Calculate a 95% confidence interval for the proportion of U.S. adults who believe the manna was really popcorn. Show your work.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 70/1000=0.07

1 -   = 1- 0.07 =0.93

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.07*0.93) / 1000)

E = 0.016

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.07-0.016 < p < 0.07+0.016

(0.054 ,0.086)