Question

A psychiatrist is interested in finding a 98% confidence interval for the tics per hour exhibited by children with Tourette syndrome. The data below show the tics in an observed hour for 15 randomly selected children with Tourette syndrome.

9

1

9

9

2

2

12

3

4

12

6

6

5

6

3

a. To compute the confidence interval use a

distribution.

b. With 98% confidence the population mean number of tics per hour that children with Tourette syndrome exhibit is between and days.

c. If many groups of 15 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.

Answer 1

The sample size is n = 15 . The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:

	X	X2
	9	81
	1	1
	9	81
	9	81
	2	4
	2	4
	12	144
	3	9
	4	16
	12	144
	6	36
	6	36
	5	25
	6	36
	3	9
Sum =	89	707

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

df=15−1=14, and the significance level is α=0.02.

Based on the provided information, the critical t-value for α=0.02 and df = 14 degrees of freedom is t_c = 2.624

The 98% confidence for the population mean μ is computed using the following expression

Therefore, based on the information provided, the 98 % confidence for the population mean μ is

CI = (5.933 - 2.423, 5.933 + 2.423)

CI = (3.51, 8.356)

c. If many groups of 15 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

98%

and about percent will not contain the true population mean number of tics per hour.

2%