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A new fertilizer was applied to the soil of 36 bean plants. 44% showed increased growth. Find the margin of error and 95% con
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Answer #1

sample proportion, = 0.44
sample size, n = 36
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.44 * (1 - 0.44)/36) = 0.0827

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 * 0.0827
ME = 0.162

Margin of error = 16.2%


CI = (pcap - z*SE, pcap + z*SE)
CI = (0.44 - 1.96 * 0.0827 , 0.44 + 1.96 * 0.0827)
CI = (0.278 , 0.602)

CI = 27.8% to 60.2%

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