Answer:
<em>0.5306</em><em>0.5694</em>
Step-by-step explanation:
USing the formuls for calculating the confidence interval for the population proportion;
CI = p±Z*√[p(1-p)/n]
p is the percentage proportion of the population 55%
Z is the z-score at 99% confidence interval = 2.576
n is the sample size = 1079
CI = 0.55 ± 2.576*[0.55(1-0.55)/√1079]
CI = 0.55 ± 2.576*[0.55(0.45)/√1079]
CI = 0.55 ± 2.576*[0.2475/√1079]
CI = 0.55 ± 2.576*[0.2475/32.85]
CI = 0.55 ± 2.576*[0.00753]
CI = 0.55 ±0.0194
CI =(0.55-0.0194, 0.55+0.0194)
CI = (0.5306, 0.5694)
<em>Hence, a 99% confidence interval of the proportion of the population that will support such a law is 0.5306</em><em>0.5694</em>