Question

As schools moved to the remote online training model some faculty were requested to learn new online skills. The effectivity of the training was evaluated with a “before and after” test of the teacher’s evaluation of their skills. The scale was a 10-point Likert scale and the teacher was asked to give a “pre” before and “post” after the training. A small sample of 10 teachers is in your data files at the end of this test. Use an alpha of .05. Run the proper analysis of “before and after” and answer the following questions.

Give the following:

1- Mean pre-test.  

2- Variance pre-test   

3- Mean post-test   

4- Variance post-test

5- Number of observations   

6- Number of teachers surveyed in this test

7- tStat    

8- P(T<=t) one

9- Conclusion reached:

. . .

Pre Train	Post Train
4	5
0	1
6	5
3	7
4	9
3	7
2	7
3	6
4	4
9	8

Answer 1

The given data is tabulated as under.

Pre Train	Post Training
4	5
0	1
6	5
3	7
4	9
3	7
4	7
3	6
4	9
9	8

We know that the mean and Variance are calculated as

and

We need to calculate the sum and the sum of squares.

1- Mean pre-test.  

We form a table of squares and the sum of each as follows:

	Pre Train	Post Training	x^2	y^2
	4	5	16	25
	0	1	0	1
	6	5	36	25
	3	7	9	49
	4	9	16	81
	3	7	9	49
	4	7	16	49
	3	6	9	36
	4	9	16	81
	9	8	81	64
Total	40	64	208	460

Mean pre test is  

The mean pre-test=4.

2- Variance pre-test

Therefore variance pre-test=5.3333

3- Mean post-test

4- Variance post-test

Variance post-test=5.6

5- Number of observations:

Thre are 10 in pre-test and 10 in post test totaling to 20.

6- Number of teachers surveyed in this test

There are 10 teachers surveyed.

7 t-test:

W shall be using a paired t-test for the difference between pre- and post test. We expect an improvement after the training and hence the difference Pre test-post test scores should be negative. Let for the ith teacher.

Let be the population mean difference

We calculate the difference between pre and post training of each teacher and take the mean and Variance for those Differences.

First we calculate the differences and the mean and variance of the differences by using the table below:

	Pre Train	Post Training	Diff	Diff^2
	4	5	-1	1
	0	1	-1	1
	6	5	1	1
	3	7	-4	16
	4	9	-5	25
	3	7	-4	16
	4	7	-3	9
	3	6	-3	9
	4	9	-5	25
	9	8	1	1
Total	40	64	-24	104

Mean difference

7. Test statistic:

will have a t-distribution with 10-1=9 df.

8. P(T<t) = 0.004312. By EXCEL function T.DIST.RT(3.3425,9).

9. Conclusion: Since the p-value <0.05, we reject the null hypothesis. Hence, we conclude that there is enough evidence that the training is effective in teachers.