Answer :
Answer:
(0.062, 0.130)
Step-by-step explanation:
Sample size = n = 500
Number of heads that were unemployed = x = 48
Proportion of heads that were unemployed = p = [tex]\frac{x}{n}=\frac{48}{500}=0.096[/tex]
Proportion of heads that were not unemployed = q = 1 - p = 1 - 0.096 = 0.904
Confidence Level = 99%
z-value for 99% confidence level = z = 2.58
The confidence interval about a population proportion is calculated as:
[tex](p-z\sqrt{\frac{pq}{n}} , p+z\sqrt{\frac{pq}{n}})[/tex]
Using the values, we get:
[tex](0.096-2.58\sqrt{\frac{0.096 \times 0.904}{500}},0.096+2.58\sqrt{\frac{0.096 \times 0.904}{500}})\\\\ = (0.062,0.130)[/tex]
Thus, 99% confidence interval for the proportion of unemployed heads of households in the population is (0.062, 0.130)