Question

A psychiatrist is interested in finding a 90% confidence interval for the tics per hour exhibited by children with Tourette syndrome. The data below show the tics in an observed hour for 12 randomly selected children with Tourette syndrome.

0

12

0

1

0

1

10

8

1

11

7

3

a. To compute the confidence interval use a

distribution.

b. With 90% confidence the population mean number of tics per hour that children with Tourette syndrome exhibit is between and days.

c. If many groups of 12 randomly selected children with Tourette syndrome are observed, 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 tics per hour and about percent will not contain the true population mean number of tics per hour.

Hint: Hints

Answer 1

a)

use a t distribution

b)

sample mean, xbar = 4.5
sample standard deviation, s = 4.7386
sample size, n = 12
degrees of freedom, df = n - 1 = 11

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.796

ME = tc * s/sqrt(n)
ME = 1.796 * 4.7386/sqrt(12)
ME = 2.5

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (4.5 - 1.796 * 4.7386/sqrt(12) , 4.5 + 1.796 * 4.7386/sqrt(12))
CI = (2.04 , 6.96)

With 90% confidence the population mean number of tics per hour that children with Tourette syndrome exhibit is between 2.04 and days 6.96.

c. If many groups of 12 randomly selected children with Tourette syndrome are observed, then a different confidence interval would be produced from each group. About 90 percent of these confidence intervals will contain the true population mean number of tics per hour and about 10 percent will not contain the true population mean number of tics per hour.