Question

The following table contains information on matched sample values whose differences are normally distributed. (You may find it useful to reference the appropriate table: z table or t table)

Number	Sample 1	Sample 2
1	17	20
2	12	12
3	21	22
4	21	20
5	16	21
6	14	16
7	17	18
8	17	20

a. Construct the 90% confidence interval for the mean difference μ_D. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.)

Answer 1

The following table is obtained:

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	17	20	-3
	12	12	0
	21	22	-1
	21	20	1
	16	21	-5
	14	16	-2
	17	18	-1
	17	20	-3
Average	16.875	18.625	-1.75
St. Dev.	3.091	3.249	1.909
n	8	8	8

and the sample size is n = 8. For the score differences, we have

$\bar D = -1.75$$s_D = 1.909$

We need to construct the 90% confidence interval for the population mean μ. The following information is provided:

Sample Mean	-1.75
Sample Standard Deviation	1.909
Sample Size	8

The critical value for α = 0.1 and df = n-1 = 7 degrees of freedom is $t_c = z_{1-\alpha/2; n-1}$ = 1.895. The corresponding confidence interval is computed as shown below:

$CI = \displaystyle \left( \bar X - t_c \times \frac{s}{\sqrt{n}}, \bar X + t_c \times \frac{s}{\sqrt{n}} \right)$

$CI = \displaystyle \left( -1.75 - 1.895 \frac{1.909}{\sqrt{8}} , -1.75 + 1.895 \frac{1.909}{\sqrt{8}} \right)$

CI = (-3.029, -0.471)

Therefore, based on the data provided, the 90% confidence interval for the population mean is -3.029 < μ < −0.471, which indicates that we are 90% confident that the true population mean μ is contained by the interval (-3.029, -0.471)