Question

A simple random sample of size n is drawn. The sample mean is found to be 35.1, and the sample standard deviation is found to be 8.7.

a. Construct the 90% confidence interval if the sample size is 40.

b. Construct the 98% confidence interval if the sample size is 40.

c.How does increasing the level of confidence affect the size of the margin of error?

d. Would you be able to compute the confidence intervals if the sample size is 18? Explain.

Answer 1

a)
sample mean, xbar = 35.1
sample standard deviation, s = 8.7
sample size, n = 40
degrees of freedom, df = n - 1 = 39

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.68

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (35.1 - 1.68 * 8.7/sqrt(40) , 35.1 + 1.68 * 8.7/sqrt(40))
CI = (32.79 , 37.41)

b)
Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.43

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (35.1 - 2.43 * 8.7/sqrt(40) , 35.1 + 2.43 * 8.7/sqrt(40))
CI = (31.76 , 38.44)

c)
Margin of error increases

d)
Yes, because the assumption is data is normally distributed.