Question

9. A confidence interval for estimating the mean

An industrial/organizational psychologist wants to improve worker productivity for a client firm, but first he needs to gain a better understanding of the life of the typical white-collar professional. Fortunately, he has access to the 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by WorldOne Research, which surveyed a sample of 650 white-collar professionals (250 legal professionals and 400 other professionals).

One of the survey questions was, “During the average workday, how many hours do you spend attending meetings?” For the subsample of legal professionals (n = 250), the mean response was M = 2.0 hours, with a sample standard deviation of s = 4.2 hours.

The estimated standard error is sMM =   .

Use the following Distributions tool to develop a 90% confidence interval estimate of the mean number of hours legal professionals spend attending meetings during a typical workday.

0123Standard Normalt Distribution

Select a Distribution

The psychologist can be 90% confident that the interval from   to   includes the unknown population mean µ.

Normally the psychologist will not know the value of the population mean. But consider the (unrealistic) scenario that a census of legal professionals is conducted. The census reveals that the population mean is μ = 2.1.

How would the psychologist most likely react to the news?

The psychologist would be surprised that μ = 2.1, because that value is outside the confidence interval.

The psychologist would be surprised that μ = 2.1, because that value is inside the confidence interval.

The psychologist would not be surprised that μ = 2.1, because that value is inside the confidence interval.

The psychologist would not be surprised that μ = 2.1, because that value is outside the confidence interval.

Answer 1

sample mean, xbar = 2
sample standard deviation, s = 4.2
sample size, n = 250
degrees of freedom, df = n - 1 = 249

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.651

ME = tc * s/sqrt(n)
ME = 1.651 * 4.2/sqrt(250)
ME = 0.4

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (2 - 1.651 * 4.2/sqrt(250) , 2 + 1.651 * 4.2/sqrt(250))
CI = (1.56 , 2.44)

The psychologist would not be surprised that μ = 2.1, because that value is inside the confidence interval.