Using excel<data<megastat<chi square<contigency table
| Chi-square Contingency Table Test for Independence | |||||
| Prefer to eat before workout | Prefer to eat after workout | Total | |||
| Runner | Observed | 68 | 121 | 189 | |
| Expected | 68.33 | 120.67 | 189.00 | ||
| O - E | -0.33 | 0.33 | 0.00 | ||
| (O - E)² / E | 0.00 | 0.00 | 0.00 | ||
| Swimmer | Observed | 73 | 128 | 201 | |
| Expected | 72.67 | 128.33 | 201.00 | ||
| O - E | 0.33 | -0.33 | 0.00 | ||
| (O - E)² / E | 0.00 | 0.00 | 0.00 | ||
| Total | Observed | 141 | 249 | 390 | |
| Expected | 141.00 | 249.00 | 390.00 | ||
| O - E | 0.00000 | 0.00000 | 0.00000 | ||
| (O - E)² / E | 0.00311 | 0.00176 | 0.004866 | ||
| .004866 | chi-square | ||||
| 1 | df | ||||
| .9444 | p-value |
Answer: Chi square test = 0.004866
4. (4pts) Calculate the test statistic, x^(2), for a chi-squared test for association using the following...