Question

QUESTION 10 A survey conducted by the U.S. department of Labor found the 48 out of 500 heads of households were unemployed. Construct a 90% confidence interval estimate of the proportion of unemployed heads of households in the population. (0.091, 0.101) (0.095, 0.097) (0.083, 0.109) (0.074, 0.118) 10 points

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 48 / 500 = 0.096

1 - = 1 - 0.096 = 0.904

Z/2 = 1.645

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

= 1.645 * (((0.096 * 0.904) / 500)

Margin of error = E = 0.022

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.096 - 0.022 < p < 0.096 + 0.022

0.074 < p < 0.118

The 90% confidence interval for the population proportion p is : (0.074 , 0.118)