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a)
sample proportion, = 0.075
sample size, n = 400
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.075 * (1 - 0.075)/400) = 0.0132
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 * 0.0132
ME = 0.0259
CI = (pcap - z*SE, pcap + z*SE)
CI = (0.075 - 1.96 * 0.0132 , 0.075 + 1.96 * 0.0132)
CI = (0.049 , 0.101)
b)
No, it would not be reasonable, because confidence interval
contains 6%
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