Question

Given that n= 120 and X = 9, construct a 98% confidence interval estimate for the population proportion (π). What is the sample proportion (compute)? What is the standard error of the proportion (compute)? What are the critical values? From what distribution? What is the confidence interval (compute)? State the interval values to at least 3 decimal places. If an expert asserted that the population proportion was 0.10, at a 98% level of confidence, would the data gathered in this sample tend to support the expert's opinion.

Answer 1

sample proportion, pcap = 9/120 = 0.075
sample size, n = 120

Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.075 * (1 - 0.075)/120) = 0.024

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, Zc = Z(α/2) = 2.33

Margin of Error, ME = zc * SE
ME = 2.33 * 0.024
ME = 0.0559

z-distribution

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.075 - 2.33 * 0.024 , 0.075 + 2.33 * 0.024)
CI = (0.019 , 0.131)

Therefore, based on the data provided, the 98% confidence interval for the population proportion is 0.019 < p < 0.131 , which indicates that we are 98% confident that the true population proportion p is contained by the interval (0.019 , 0.131)

the data gathered in this sample tend to support the expert's opinion