Question

You are interested in finding a 98% 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.

5

4

10

8

1

2

6

5

8

11

a. To compute the confidence interval use a

distribution.

b. With 98% confidence the population meannumber of days of class that college students miss is between and 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 percent of these confidence intervals will contain the true population mean number of missed class days and about percent will not contain the true population mean number of missed class days.

Answer 1

a)

use a t distribution

b)

sample mean, xbar = 6
sample standard deviation, s = 3.266
sample size, n = 10
degrees of freedom, df = n - 1 = 9

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.821

ME = tc * s/sqrt(n)
ME = 2.821 * 3.266/sqrt(10)
ME = 2.9

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (6 - 2.821 * 3.266/sqrt(10) , 6 + 2.821 * 3.266/sqrt(10))
CI = (3.09 , 8.91)

With 98% confidence the population meannumber of days of class that college students miss is between 3.09 and days.8.91

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 98 percent of these confidence intervals will contain the true population mean number of missed class days and about 2percent will not contain the true population mean number of missed class days.