|
Chi-Square Test of independence |
|
|
|
|
|
|
Observed Frequencies |
|
Y |
|
|
0 |
Yes |
No |
|
|
|
|
Total |
|
LS1 |
28 |
10 |
|
|
|
|
38 |
|
LS2 |
48 |
114 |
|
|
|
|
162 |
|
Total |
76 |
124 |
|
|
|
|
200 |
|
|
|
|
|
|
|
|
|
Expected frequency of a cell = sum of row*sum of column /
total sum |
|
Expected Frequencies |
|
|
Yes |
No |
|
|
|
|
Total |
|
LS1 |
76*38/200=14.44 |
124*38/200=23.56 |
|
|
|
|
38 |
|
LS2 |
76*162/200=61.56 |
124*162/200=100.44 |
|
|
|
|
162 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Total |
76 |
124 |
|
|
|
|
200 |
|
|
|
|
|
|
|
|
|
(fo-fe)^2/fe |
|
| LS1 |
12.734 |
7.804 |
|
|
|
|
|
| LS2 |
2.987 |
1.831 |
|
|
|
|
|
Chi-Square Test Statistic,?² = ?(fo-fe)^2/fe =
25.3557
Level of Significance = 0.05
Number of Rows = 2
Number of Columns = 2
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 2- 1 )
= 1
p-Value = 0.0000 [Excel function:
=CHISQ.DIST.RT(?²,df) ]
Decision: p-value < ? , Reject Ho
conclusion: there is enough evidence to conlcude
that learning gains were dependent of
learning style
