Question

A sample of 24 observations is selected from a normal population where the sample standard deviation is 4.45. The sample mean is 16.45.  

a. Determine the standard error of the mean. (Round the final answer to 2 decimal places.)

The standard error of the mean is.

b. Determine the 90% confidence interval for the population mean. (Round the t-value to 3 decimal places. Round the final answers to 3 decimal places.)

The 90% confidence interval for the population mean is between and.

c. If you wanted a narrower interval, would you increase or decrease the confidence level?

(Click to select)  Increase  Decrease

Answer 1

a)
std. error. = 4.45/sqrt(24) = 0.91

b)
sample mean, xbar = 16.45
sample standard deviation, s = 4.45
sample size, n = 24
degrees of freedom, df = n - 1 = 23

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.714

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (16.45 - 1.714 * 4.45/sqrt(24) , 16.45 + 1.714 * 4.45/sqrt(24))
CI = (14.893 , 18.007)

c)
Decrease