The following summer the institute wanted to test the succes

The following summer, the institute wanted to test the success of this new program in a different way. They enrolled 20 French language teachers in the 4-week program. At the beginning of this period, the teachers were given the listening test of Modern Language Association for the understanding of spoken French. After they completed the program in 4 weeks, the listening test was given again. (The actual French spoken in the two tests was different, so that simply taking the first test should not improve the score on the second test.) Each teacher’s pre-test and post-test results are listed in the following table. The maximum possible score on the test is 36.

Teacher

Pre-test score

Post-test score

1

32

34

2

31

31

3

29

35

4

10

16

5

30

33

6

33

36

7

22

24

8

25

28

9

32

26

10

20

26

11

30

36

12

20

26

13

24

27

14

24

24

15

31

32

16

30

31

17

15

15

18

32

34

19

23

26

20

23

26

a) Are these dependent or independent samples? Explain your answer.

b) Add another column to the table above for the difference between the pre-test and post-test scores. Find that difference for each teacher.

c) Find the mean and the standard deviation of the difference column.

xD = _________­­­­­______ sD = ____________________

d) Test the hypothesis that the new program is successful to improve the language skills of French language teachers at a 0.05 level of significance. Show all five steps of the hypothesis testing process including the interpretation. (Assume that all the scores from such tests will be normally distributed. Use of the calculator menu is fine. Points will be deducted for missing steps.)

Teacher	Pre-test score	Post-test score
1	32	34
2	31	31
3	29	35
4	10	16
5	30	33
6	33	36
7	22	24
8	25	28
9	32	26
10	20	26
11	30	36
12	20	26
13	24	27
14	24	24
15	31	32
16	30	31
17	15	15
18	32	34
19	23	26
20	23	26

Solution

They are dependent as same sample is tested two times before and after.

b) and c)

H0: muD = 0

Ha: mu D<0

One tailed test for mean difference paired

Std error = 0.647 and muD = -2.5

Test statistic = -3.86

df =19

p value = 0.0005

As p is less than 0.05 reject null hypothesis.

There is improvement.

Teacher	Pre-test score	Post-test score	Diff
1	32	34	-2
2	31	31	0
3	29	35	-6
4	10	16	-6
5	30	33	-3
6	33	36	-3
7	22	24	-2
8	25	28	-3
9	32	26	6
10	20	26	-6
11	30	36	-6
12	20	26	-6
13	24	27	-3
14	24	24	0
15	31	32	-1
16	30	31	-1
17	15	15	0
18	32	34	-2
19	23	26	-3
20	23	26	-3
		Mean	-2.5
		Std dev	2.89282233