Perform Chi-squared test of independence. The significance level alpha is 5%.
Columns: category 1
Rows: categoty 2
Contingency table
67 26 16
128 63 46
a)
degree of freedom(df) =(rows-1)*(columns-1)= | 2 |
b)
Applying chi square test of independence: |
Expected | Ei=row total*column total/grand total | c | d | e | Total |
a | 61.431 | 28.038 | 19.532 | 109 | |
b | 133.569 | 60.962 | 42.468 | 237 | |
total | 195 | 89 | 62 | 346 | |
chi square χ2 | =(Oi-Ei)2/Ei | c | d | e | Total |
a | 0.505 | 0.148 | 0.639 | 1.2916 | |
b | 0.232 | 0.068 | 0.294 | 0.5940 | |
total | 0.7371 | 0.2162 | 0.9323 | 1.8857 | |
test statistic X2= | 1.886 |
c)
p value = | 0.390 | from excel: chidist(1.8857,2) |
d)
since p value >0.05
option C is correct
Do not reject that they are independent at 5% significance level and reserve judgement. We may accept independence, but with some unknown probability.
Perform Chi-squared test of independence. The significance level alpha is 5%. Columns: category 1 Rows: categoty...
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