Question

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 21, 14, 13, 24, 17, 22, 25, 12. [You may find it useful to reference the t table.]

a. Calculate the sample mean and the sample standard deviation. (Round intermediate calculations to at least 4 decimal places. Round "Sample mean" to 3 decimal places and "Sample standard deviation" to 2 decimal places.)

b. Construct the 90% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)

c. Construct the 95% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)

d. What happens to the margin of error as the confidence level increases from 90% to 95%?

As the confidence level increases, the margin of error becomes larger.
As the confidence level increases, the margin of error becomes smaller.

Answer 1

a)

sample mean = 18.500
std.dev = 5.15

b)

sample mean, xbar = 18.5
sample standard deviation, s = 5.15
sample size, n = 8
degrees of freedom, df = n - 1 = 7

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.895

ME = tc * s/sqrt(n)
ME = 1.895 * 5.15/sqrt(8)
ME = 3.45

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (18.5 - 1.895 * 5.15/sqrt(8) , 18.5 + 1.895 * 5.15/sqrt(8))
CI = (15.05 , 21.95)

c)

ssample mean, xbar = 18.5
sample standard deviation, s = 5.15
sample size, n = 8
degrees of freedom, df = n - 1 = 7

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.365

ME = tc * s/sqrt(n)
ME = 2.365 * 5.15/sqrt(8)
ME = 4.306

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (18.5 - 2.365 * 5.15/sqrt(8) , 18.5 + 2.365 * 5.15/sqrt(8))
CI = (14.19 , 22.81)

d)

As the confidence level increases, the margin of error becomes larger.