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**

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