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a) A random sample of 15 employees of airtel call Centre was taken and each employee took a competency test. The mean of the

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Answer #1

a)

sample mean, xbar = 56.3
sample standard deviation, s = 7.1
sample size, n = 15
degrees of freedom, df = n - 1 = 14

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.145


ME = tc * s/sqrt(n)
ME = 2.145 * 7.1/sqrt(15)
ME = 3.932

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (56.3 - 2.145 * 7.1/sqrt(15) , 56.3 + 2.145 * 7.1/sqrt(15))
CI = (52.37 , 60.23)


b)

sample mean, xbar = 67.45
sample standard deviation, σ = 2.93
sample size, n = 100


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


ME = zc * σ/sqrt(n)
ME = 1.96 * 2.93/sqrt(100)
ME = 0.57

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (67.45 - 1.96 * 2.93/sqrt(100) , 67.45 + 1.96 * 2.93/sqrt(100))
CI = (66.88 , 68.02)


c)

sample proportion, = 0.2818
sample size, n = 110
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.2818 * (1 - 0.2818)/110) = 0.0429

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 * 0.0429
ME = 0.084

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.2818 - 1.96 * 0.0429 , 0.2818 + 1.96 * 0.0429)
CI = (0.198 , 0.366)

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