A survey conducted by the U.S. department of Labor found the 48 out of 500 heads of households were unemployed. Compute a 99% confidence interval for the proportion of unemployed heads of households in the population. Round to three decimal places.

Answer:99% Confidence interval is given as;Lower Limit = 0.062Upper Limit = 0.130 Step-by-step explanation:Given the data in the question;x = 48n = 500sample proportion p = x/n = 48/500 = 0.096with 99% confidencesignificance level ∝ = 1 - 99% = 1 - 0.99 = 0.01∝/2 = 0.01 / 2 = 0.005Critical Z-value = [tex]Z_{\alpha /2[/tex] = [tex]Z_{0.005[/tex]  = 2.576 Now, Standard error of p : SE = √[ (p × ( 1 - p) / n ) ]we substituteSE = √[ (0.096 × ( 1 - 0.096) / 500 ) ]SE = √[ (0.096 × 0.904) / 500  ]SE = √[ 0.086784 / 500 ) ]SE = 0.0131745soMargin of Error E = [tex]Z_{\alpha /2[/tex] × SE E = 2.576  × 0.0131745E = 0.0339Now 99% confidence interval will be⇒ 0.096 ± 0.0339⇒ (0.096 - 0.0339, 0.096 + 0.0339 )⇒ ( 0.062, 0.130 )Lower Limit = 0.062Upper Limit = 0.130