Question

You wish to test the following claim (HaHa) at a significance level of α=0.10α=0.10. For the context of this problem, μd=PostTest−PreTestμd=PostTest-PreTest. Each row gives the scores of a single individual.

      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
57.2	60.1
45.5	30.1
52.9	70.5
45.3	104.9
42.9	41.8
48.1	11.6
65.3	23.8
46.5	51.5
47.6	20.3
58.8	40.8
41.1	-30.1
39.6	10.5
56.9	35.6
52.3	9.7
37.5	43.6
55.1	74.3
42.3	27.7
55.1	51.1
54.6	-23.2
59.9	62.8

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.)

The p-value is...

• less than (or equal to) αα
• greater than αα

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is less than 0.
• There is not sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is less than 0.
• The sample data support the claim that the mean difference of post-test from pre-test is less than 0.
• There is not sufficient sample evidence to support the claim that the mean difference of post-test from pre-test is less than 0.

Answer 1

Number	Post Test	Pre Test	Difference
	60.1	57.2	2.9	297.735025
	30.1	45.5	-15.4	1.092025
	70.5	52.9	17.6	1021.12203
	104.9	45.3	59.6	5469.34203
	41.8	42.9	-1.1	175.695025
	11.6	48.1	-36.5	490.401025
	23.8	65.3	-41.5	736.851025
	51.5	46.5	5	374.616025
	20.3	47.6	-27.3	167.573025
	40.8	58.8	-18	13.286025
	-30.1	41.1	-71.2	3231.35403
	10.5	39.6	-29.1	217.415025
	35.6	56.9	-21.3	48.233025
	9.7	52.3	-42.6	797.780025
	43.6	37.5	6.1	418.407025
	74.3	55.1	19.2	1125.93803
	27.7	42.3	-14.6	0.060025
	51.1	55.1	-4	107.226025
	-23.2	54.6	-77.8	4025.26803
	62.8	59.9	2.9	297.735025
Total	717.4	1004.5	-287.1	19017.1295

Test Statistic :-





t = -2.0292

P - value = P ( t > 2.0292 ) = 0.0283

Looking for the value t = 2.0292 in t table across n - 1 = 20 - 1 = 19 degree of freedom.

Reject null hypothesis if P value < level of significance
P - value = 0.0283 < 0.1, hence we reject null hypothesis
Conclusion :- Reject null hypothesis

P value is  less than (or equal to) αα

reject the null

There is not sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is less than 0.