Question

You wish to test the following claim (HaHa) at a significance level of α=0.002α=0.002. For the context of this problem, μd=PostTest−PreTestμd=PostTest-PreTest where the first data set represents a pre-test and the second data set represents a post-test. (Each row represents the pre and post test scores for an individual. Be careful when you enter your data and specify what your μ1μ1 and μ2μ2 are so that the differences are computed correctly.)

      Ho:μd=0Ho:μd=0
      Ha:μd≠0Ha:μd≠0

You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:

pre-test	post-test
38.5	44.5
34.8	42
39.9	6.6
51.5	38.9
30.5	29.2
47	121.8
34	20
40.1	79.7
45.4	-9
38.5	70.9
35	23.7
32.6	-27.8
44.3	150.3
38.2	66.6
44.1	-19.7
48	126.2

What is the test statistic for this sample?
test statistic =  (Report answer accurate to 4 decimal places.)

What is the p-value for this sample?
p-value =  (Report answer accurate to 4 decimal places.)

Answer 1

H0:μd = 0

Ha:μd ≠ 0

Pre-test	Post-test	Difference (d)	(d-Mean)^2
38.5	44.5	-6	2.54
34.8	42	-7.2	0.16
39.9	6.6	33.3	1672.30
51.5	38.9	12.6	407.79
30.5	29.2	1.3	79.10
47	121.8	-74.8	4516.68
34	20	14	466.29
40.1	79.7	-39.6	1024.40
45.4	-9	54.4	3843.23
38.5	70.9	-32.4	615.35
35	23.7	11.3	356.97
32.6	-27.8	60.4	4623.15
44.3	150.3	-106	9683.79
38.2	66.6	-28.4	432.90
44.1	-19.7	63.8	5097.07
48	126.2	-78.2	4985.24
	Total	-121.5	37806.95

Σd = -121.5

n = 16

d̅ = Σd/n = -121.5/16=-7.59

sd = (Σ(d-d̅)2/(n-1))^0.5

= (37806.95/15)^0.5

= 50.2

Test statistic: t = (d̅-D)/(sd/n^0.5) = (-7.59-0)/(50.2/16^0.5) = -0.61

Critical value (Using Excel function T.INV.2T(probability,n-1)) = T.INV.2T(0.05,15) = 2.13

Since test statistic is less than critical value, we do not reject the null hypothesis and conclude that μd = 0.