For a Level of Significance in a contingency table is performed using chi-square test. So, based on that, the given values are considered as Observed Frequency.
Expected Frequency = ( Row Total X Column Total ) / N
Observed Frequency (Given Table)
| Before | After | Total | |
| Stop | 62 | 58 | 120 |
| Near Stop | 30 | 20 | 50 |
| No Stop | 20 | 12 | 32 |
| Total | 112 | 90 | 202 |
Expected Frequency
| Before | After | |
| Stop | 67 | 53 |
| Near Stop | 28 | 22 |
| No Stop | 18 | 14 |
Calculating ( O - E )2
| Before | After | |
| Stop | 25 | 25 |
| Near Stop | 4 | 4 |
| No Stop | 4 | 4 |
Calculating (O-E)2 / E:
| Before | After | |
| Stop | 0.373134 | 0.471698 |
| Near Stop | 0.142857 | 0.181818 |
| No Stop | 0.222222 | 0.285714 |
Sum of (O-E)2/E =
| 1.677444 |
Degrees of Freedom = 6 -1 = 5
For a Degree of Freedom 5, chi value should be less than 11.7 for it significant upto 0.05.
So, 1.67 < 11.7. Hence, the answer is right! That the observance is statistically significant at 0.05 level of significance.
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