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%
A new fertilizer was applied to the soil of 36 bean plants. 44% showed increased growth....
A new fertilizer was applied to the soil of 81 bean plants. 52% showed increased growth. Find the margin of error and 95% confidence interval for this statistic. (Round all answers to 1 decimal place.) Margin of Error (as a percentage):_?_% Confidence Interval: _?_% to _?_%
A new fertilizer was applied to the soil of 81 bean plants. 52% showed increased growth. Find the margin of error and 95% confidence interval for this statistic. (Round all answers to 1 decimal place.) Margin of Error (as a percentage):_?_% Confidence Interval: _?_% to _?_%
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It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A sample of 240 such companies showed that 101 of them provide such facilities on site. Construct a 96% confidence interval for the percentage of all such companies that provide such facilities on site. What is the margin of error for this estimate? Round your answers to one...