Question

Plz attach formula and step by step to create confidence interval

A random poll of 800 working men found that 9% had taken on a second job to help pay the bills. a) Estimate the true percentage of men that are taking on second jobs by constructing a 95% confidence interval. b) A pundit on a TV news show claimed that only 6% of working men had a second job. Use your confidence interval to test whether his claim is plausible given the poll data. a) Construct a 95% confidence interval for the population proportion. The 95% confidence interval is ( ). (Round to three decimal places as needed.)

Answer 1

a)
sample proportion, pcap = 0.09
sample size, n = 800
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.09 * (1 - 0.09)/800) = 0.0101

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.09 - 1.96 * 0.0101 , 0.09 + 1.96 * 0.0101)
CI = (0.070 , 0.110)

CI (Proportion) = (o-z-p60, P), p+zay/B1, @)

b)
Reject the null hypothesis
because 0.06 is not included in the CI (0.070 , 0.110)