Question

A survey was conducted to estimate the proportion of California workers who would rather live in a different state.  In a random sample of 100 California workers, 28% indicated that they would rather living in another state.  A 95% confidence interval for the proportion of all Californian workers who would rather live in another state is:

Group of answer choices

(0.164, 0.396)

(0.192, 0.368)

(0.207, 0.353)

(0.230, 0.330)

Answer 1

Solution :

Given that,

n = 100

= 0.28

1 - = 1 - 0.28 = 0.72

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

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

= 1.96 * (((0.28 * 0.72) / 100)

= 0.088

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.28 - 0.088 < p < 0.28 + 0.088

0.192 < p < 0.368

(0.192,0.368)