Question

4. A recent poll of 1079 adults finds that 55% of Americans support a more stringent immigration law. Construct a 99% confidence interval of the proportion of the population that will support such a law.

Answer 1

Answer:

0.5306$<\mu<$0.5694

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)

Hence, a 99% confidence interval of the proportion of the population that will support such a law is 0.5306$<\mu<$0.5694