Question

A random sample of the closing stock prices in dollars for a company in a recent year is listed below. Assume that σ is ​$2.36.

Construct the 90​% and 99​% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.

22.34	17.02	21.44	16.52	22.13	20.58	19.08	16.08
16.09	19.24	19.58	22.67	17.62	19.96	15.18	21.32

The​ 90% confidence interval is ($ ? , $ ?)

​(Round to two decimal places as​ needed.)

The​ 99% confidence interval is ($ ? , $ ?)

​(Round to two decimal places as​ needed.)

Which statement below interprets the results​ correctly?

A. The probability that the mean closing stock price is in the​ 90% confidence interval is about​ 90% and the probability that the mean closing stock price is in the​ 99% confidence interval is about​ 99%.

B. The​ 90% confidence interval contains the mean closing stock price​ 90% of the time and the​ 99% confidence interval contains the mean closing stock price​ 99% of the time.

C. There is​ 90% confidence that the mean closing stock price is in the​ 90% confidence interval and​ 99% confidence that the mean closing stock price is in the​ 99% confidence interval.

D. ​90% of the mean closing stock prices are in the​ 90% confidence interval and​ 99% of the mean closing stock prices are in the​ 99% confidence interval.

Which interval is​ wider?

A. The 99% Confidence Interval

B. The 90% Confidence Interval

Answer 1

a)

sample mean, xbar = 19.1781
sample standard deviation, σ = 2.36
sample size, n = 16

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 * 2.36/sqrt(16)
ME = 0.97

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (19.1781 - 1.6449 * 2.36/sqrt(16) , 19.1781 + 1.6449 * 2.36/sqrt(16))
CI = (18.21 , 20.15)

b)

sample mean, xbar = 19.1781
sample standard deviation, σ = 2.36
sample size, n = 16

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.5758

ME = zc * σ/sqrt(n)
ME = 2.5758 * 2.36/sqrt(16)
ME = 1.52

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (19.1781 - 2.5758 * 2.36/sqrt(16) , 19.1781 + 2.5758 * 2.36/sqrt(16))
CI = (17.66 , 20.7)

c)

C. There is​ 90% confidence that the mean closing stock price is in the​ 90% confidence interval and​ 99% confidence that the mean closing stock price is in the​ 99% confidence interval.

d)

A. The 99% Confidence Interva