Question

A random sample of 800 movie goers in Flagstaff found 328 movie goers who had bought popcorn on their last visit. Find a 95% confidence interval for the true percent of movie goers in Flagstaff who have bought popcorn on their last visit. Express your results to the nearest hundredth of a percent.

Answer 1

Solution :

Given that,

n = 800

x = 328

Point estimate = sample proportion = = x / n = 328/800=0.41

1 -   = 1- 0.41 =0.59

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z 0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.41*0.59) /800 )

E = 0.03408

A 95% confidence interval proportion p is ,

- E < p < + E

0.41-0.03408 < p < 0.41+0.03408

0.3759< p < 0.44408

37.59%<p< 44.41% rounded