Question

You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 35 business days, the mean closing price of a certain stock was $105.84. Assume the population standard deviation is $10.29.

Answer 1

sample mean, xbar = 105.84
sample standard deviation, σ = 10.29
sample size, n = 35

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.6449

ME = zc * σ/sqrt(n)
ME = 1.6449 * 10.29/sqrt(35)
ME = 2.861

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (105.84 - 1.6449 * 10.29/sqrt(35) , 105.84 + 1.6449 * 10.29/sqrt(35))
CI = (102.99 , 108.69)
Therefore, based on the data provided, the 90% confidence interval for the population mean is 102.98 < μ < 108.7 which indicates that we are 90% confident that the true population mean μ is contained by the interval (102.98 , 108.7)

sample mean, xbar = 105.84
sample standard deviation, σ = 10.29
sample size, n = 35

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 * 10.29/sqrt(35)
ME = 3.409

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (105.84 - 1.96 * 10.29/sqrt(35) , 105.84 + 1.96 * 10.29/sqrt(35))
CI = (102.43 , 109.25)
Therefore, based on the data provided, the 95% confidence interval for the population mean is 102.43 < μ < 109.25 which indicates that we are 95% confident that the true population mean μ is contained by the interval (102.43 , 109.25)

Width of 95% confidence interal is more