Question

1. The waiting times​ (in minutes) of a random sample of 21 people at a bank have a sample standard deviation of 3.5 minutes. Construct a confidence interval for the population variance sigma squared and the population standard deviation sigma. Use a 99 % level of confidence. Assume the sample is from a normally distributed population. What is the confidence interval for the population variance sigma squared​?

2.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. If​ convenient, use technology to construct the confidence intervals. A random sample of 55 home theater systems has a mean price of ​$113.00. Assume the population standard deviation is ​$15.20.

Answer 1

1)

Here s = 3.5 and n = 21
df = 21 - 1 = 20

α = 1 - 0.99 = 0.01
The critical values for α = 0.01 and df = 20 are Χ^2(1-α/2,n-1) = 7.4338 and Χ^2(α/2,n-1) = 39.9968

CI = (20*3.5^2/39.9968 , 20*3.5^2/7.4338)
CI = (6.1255 , 32.9576)

Here s = 3.5 and n = 21
df = 21 - 1 = 20

α = 1 - 0.99 = 0.01
The critical values for α = 0.01 and df = 20 are Χ^2(1-α/2,n-1) = 7.4338 and Χ^2(α/2,n-1) = 39.9968

CI = (sqrt(20*3.5^2/39.9968) , sqrt(20*3.5^2/7.4338))
CI = (2.475 , 5.7409)

2)

sample mean, xbar = 113
sample standard deviation, σ = 15.2
sample size, n = 55

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

CI = (113 - 1.64 * 15.2/sqrt(55) , 113 + 1.64 * 15.2/sqrt(55))
CI = (109.64 , 116.36)

sample mean, xbar = 113
sample standard deviation, σ = 15.2
sample size, n = 55

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

CI = (113 - 1.96 * 15.2/sqrt(55) , 113 + 1.96 * 15.2/sqrt(55))
CI = (108.98 , 117.02)