Question

You are interested in finding a 90% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 10 randomly selected college students.

11 0 10 11 3 11 8 9 4 3

a. To compute the confidence interval use a distribution.

b. With 90% confidence the population mean number of days of class that college students miss is answer ____ between and answer ____ days.

c. If many groups of 10 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About answer ___ percent of these confidence intervals will contain the true population mean number of missed class days and about answer ____ percent will not contain the true population mean number of missed class days.

Answer 1

a)
Student's t distribution

b)
sample mean, xbar = 7
sample standard deviation, s = 4.11
sample size, n = 10
degrees of freedom, df = n - 1 = 9

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.262

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (7 - 2.262 * 4.11/sqrt(10) , 7 + 2.262 * 4.11/sqrt(10))
CI = (4.06 , 9.94)

c)
About 90 percent of these confidence intervals will contain the true population mean number of missed class days and about 10 percent will not contain the true population mean number of missed class days.