Question

In a recent survey of stats classes, 70 of the 274 students reported having dependent children.

(a) Find a 99% confidence interval for the sample proportion and state your interval with conclusion in a statistical sentence.

Answer 1

sample proportion, = 0.2555
sample size, n = 274
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.2555 * (1 - 0.2555)/274) = 0.0263

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

Margin of Error, ME = zc * SE
ME = 2.58 * 0.0263
ME = 0.0679

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.2555 - 2.58 * 0.0263 , 0.2555 + 2.58 * 0.0263)
CI = (0.1876 , 0.3234)
Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.1876 < p < 0.3234 , which indicates that we are 99% confident that the true population proportion p is contained by the interval (0.1876 , 0.3234)