Question

For questions 1-3:

For this question use the following values: n = 14, s_D = 0.80, and d = -0.33.

1)

Find the P-value for the test

H₀: μ_D = 0

H₁: μ_D ≠ 0

0.035

0.070

0.106

0.147

0.161

0.222

0.260

0.300

0.324

0.379

2)

Compute the lower limit of the 94% confidence interval on the differencebetween population means.

-0.946

-0.770

-0.581

-0.482

-0.136

-0.112

0.047

0.542

0.608

0.928

3)

Compute the upper limit of the 94% confidence interval on the differencebetween population means.

		-0.779
		-0.699
		-0.620
		-0.382
		0.004
		0.110
		0.137
		0.313
		0.529
		0.972 For questions 4-6: For this question use the following values: n = 13, sD = 0.70, and d = 0.33.

QUESTION 4

Find the P-value for the test

H₀: μ_D = 0

H₁: μ_D ≠ 0

		0.006
		0.034
		0.067
		0.115
		0.231
		0.246
		0.276
		0.311
		0.325
		0.337

Question 5

Compute the lower limit of the 89% confidence interval on the differencebetween population means.

		-0.915
		-0.807
		-0.742
		-0.689
		-0.651
		-0.158
		-0.005
		0.154
		0.226
		0.736

  

Question 6

Compute the upper limit of the 89% confidence interval on the differencebetween population means.

		-0.827
		-0.450
		0.125
		0.208
		0.477
		0.665
		0.702
		0.750
		0.809
		0.863

Answer 1

1)

t = (-0.33/(0.8/sqrt(14))
= -1.5434

p value is calculated using t = -1.5434 and df = 13

p value = 0.147

2)
t value at 94% = 2.06

CI = d - t *(s/sqrt(n))
= -0.33 - 2.06 *(0.8/sqrt(14))
= -0.770

3)

t value at 94% = 2.06

CI = d + t *(s/sqrt(n))
= -0.33 + 2.06 *(0.8/sqrt(14))
= 0.110

4)

1)

t = (0.33/(0.7/sqrt(13))
= 1.7

p value is calculated using t = 1.7 and df = 12

p value = 0.115

5)
t value at 89% = 1.726

CI = d - t *(s/sqrt(n))
= 0.33 - 1.726 *(0.7/sqrt(13))
= -0.005

6)

t value at 89% = 1.726

CI = d + t *(s/sqrt(n))
= 0.33 + 1.726 *(0.7/sqrt(13))
= 0.665