Question

You wish to test the following claim (HaHa) at a significance level of α=0.001α=0.001.  dd denotes the mean of the difference between pre-test and post-test scores.       

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:

post-test	pre-test
49.3	31.9
47.1	27.5
42.6	40.9
42.2	43.4
41.5	39.1
41.8	52.5
33.5	17
42.9	40.1
33.5	31.6
39.8	57.2
47.1	45.7
46.5	50.9
51.7	52.4
58	61.3
47.5	54.9

• Make sure to choose the appropriate column for Sample 1 and 2 so that you are estimating the improved post-test score: μ2−μ1μ2-μ1

1. What is the test statistic for this sample?
   
   test statistic =  Round to 3 decimal places.
   
2. What is the p-value for this sample? Round to 4 decimal places.
   
   p-value =
   
3. The p-value is...
   • less than (or equal to) αα
   • greater than αα
   
   
4. This test statistic leads to a decision to...
   • reject the null
   • accept the null
   • fail to reject the null
   
   
5. 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

Observation Table:

post-test	pre-test	d=pre- post
49.3	31.9	-17.4
47.1	27.5	-19.6
42.6	40.9	-1.7
42.2	43.4	1.2
41.5	39.1	-2.4
41.8	52.5	10.7
33.5	17	-16.5
42.9	40.1	-2.8
33.5	31.6	-1.9
39.8	57.2	17.4
47.1	45.7	-1.4
46.5	50.9	4.4
51.7	52.4	0.7
58	61.3	3.3
47.5	54.9	7.4
	Mean (d)=	AVERAGE(C2:C16)	-1.24
	Sd (d)=	STDEV(C2:C16)	10.2203
	P-value	TDIST(0.47,14,1)	0.3228

Test statistic,

Degrees of freedom = n - 1 = 15 - 1 = 14

P-value = 0.3228

The p-value is greater than α.

This test statistic leads to a decision to fail to reject the null hypothesis.

The final conclusion is that, There is not sufficient sample evidence to support the claim that the mean difference of post-test from pre-test is less than 0.