Homework SourseAssuming I want to run a goodness of fit Chi squared test ho
Assuming I want to run a goodness of fit Chi squared test, how would I do this based on data above? Sample taken from 20 bags of each. Are the colors evenly distributed?
| blue | brown | green | orange | red | yellow | total |
|---|---|---|---|---|---|---|
| 214 | 152 | 197 | 209 | 149 | 163 | 1084 |
| 20% | 14% | 18% | 19% | 14% | 15% | 100% |
Solution
the test statistc is
chi square=sum(O-E)^2/E
where E=(rowtotal*column total)/N
Since p value is less than 0.5, we reject the null hypothesis and conclude that the colors are not evenly distributed.
| Chi-square Contingency Table Test for Independence | |||||
| Col 1 | Col 2 | Total | |||
| Row 1 | Observed | 214 | 95 | 309 | |
| Expected | 224.80 | 84.20 | 309.00 | ||
| O - E | -10.80 | 10.80 | 0.00 | ||
| (O - E)² / E | 0.52 | 1.39 | 1.91 | ||
| Row 2 | Observed | 152 | 32 | 184 | |
| Expected | 133.86 | 50.14 | 184.00 | ||
| O - E | 18.14 | -18.14 | 0.00 | ||
| (O - E)² / E | 2.46 | 6.56 | 9.02 | ||
| Row 3 | Observed | 197 | 73 | 270 | |
| Expected | 196.43 | 73.57 | 270.00 | ||
| O - E | 0.57 | -0.57 | 0.00 | ||
| (O - E)² / E | 0.00 | 0.00 | 0.01 | ||
| Row 4 | Observed | 209 | 90 | 299 | |
| Expected | 217.53 | 81.47 | 299.00 | ||
| O - E | -8.53 | 8.53 | 0.00 | ||
| (O - E)² / E | 0.33 | 0.89 | 1.23 | ||
| Row 5 | Observed | 149 | 40 | 189 | |
| Expected | 137.50 | 51.50 | 189.00 | ||
| O - E | 11.50 | -11.50 | 0.00 | ||
| (O - E)² / E | 0.96 | 2.57 | 3.53 | ||
| Row 6 | Observed | 163 | 76 | 239 | |
| Expected | 173.88 | 65.12 | 239.00 | ||
| O - E | -10.88 | 10.88 | 0.00 | ||
| (O - E)² / E | 0.68 | 1.82 | 2.50 | ||
| Total | Observed | 1084 | 406 | 1490 | |
| Expected | 1084.00 | 406.00 | 1490.00 | ||
| O - E | 0.00 | 0.00 | 0.00 | ||
| (O - E)² / E | 4.95 | 13.23 | 18.18 | ||
| 18.18 | chi-square | ||||
| 5 | df | ||||
| .0027 | p-value |
