Thirty sixth graders were randomly selected from a school di
Solution
4.
Here are the entries for the Difference column:
-2
-4
-2
-3
-3
-2
5
4
3
2
1
0
-1
-5
-4
Let ud = u2 - u1.
u1 = with training
u2 = without training
Formulating the null and alternative hypotheses,
Ho: ud = 0
Ha: ud =/ 0
At level of significance = 0.05
As we can see, this is a two tailed test.
Calculating the standard deviation of the differences (4th column):
s = 3.104528183
Thus, the standard error of the difference is sD = s/sqrt(n):
sD = 0.80158573
Calculating the mean of the differences (third column):
XD = -0.733333333
As t = [XD - uD]/sD, where uD = the hypothesized difference = 0 , then
t = -0.914853279
As df = n - 1 = 14
Then the critical value of t is
tcrit = +/- 2.144786688
As |t| < 2.1448, WE FAIL TO REJECT THE NULL HYPOTHESIS.
Thus, there is no significant difference the means of the matched pairs. [CONCLUSION]

