Question

2. With 80% confidence, for sample proportion 0.40 and sample size 25, what is the upper confidence limit with 2 decimal places?

Answer 1

sample proportion, = 0.4
sample size, n = 25
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.4 * (1 - 0.4)/25) = 0.098

Given CI level is 80%, hence α = 1 - 0.8 = 0.2
α/2 = 0.2/2 = 0.1, Zc = Z(α/2) = 1.28

Margin of Error, ME = zc * SE
ME = 1.28 * 0.098
ME = 0.1254

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.4 - 1.28 * 0.098 , 0.4 + 1.28 * 0.098)
CI = (0.27 , 0.53)

For two tailed , upper limit= 0.53

for one tailed,

z value at 80% = 0.84

Upper limit = p + z *sqrt(p *(1-p)/n)
= 0.40 + 0.84 * sqrt(0.40 *(1-0.40)/25)
= 0.48