Question

3.a) As of the year 2000, rural and farm residents made up 10.9% of the Florida population. In 2010, a random sample of 150 Florida residents is taken and 14 of them are rural or farm residents. Find a 95% confidence interval for the percentage of rural and farm residents. Does this give evidence that the percentage of rural and farm residents has decreased since the year 2000? Explain carefully what is meant by 95% confidence

Answer 1

sample proportion, = 0.0933
sample size, n = 150
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.0933 * (1 - 0.0933)/150) = 0.0237

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

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0237
ME = 0.0465

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.0933 - 1.96 * 0.0237 , 0.0933 + 1.96 * 0.0237)
CI = (0.0468 , 0.1398)

There is not sufficient evidence that the percentage of rural and farm residents has decreased since the year 2000 because confidence interval contains 10.9%