5) Expected frequency = (row total*column total)/grand total
where row total for never late = 109+20 , column total for "have smoked marijuana" = 92
and grand total = 500 (given in table)
so, expected frequency = (129*92)/500 = 11868/500= 23.736 or 23.74
6) At 0.05 significance level, the linear by linear association and nominal by nominal p values shows that the test results are significant for linear by linear association but not for nominal by nominal because the p value for linear by linear association is 0.14 and for nominal to nominal is 0.052.
so, its clear that p value for linear by linear association is significant, but the p value for nominal by nominal, pearson chi square and likelihood ratio are not significance.
so, there is no statistically significant relationship between being late for class and smoking marijuana
7) Fails to reject the null hypothesis because the p value is significant or the p value is more than the significance )level
8) The test result shows that there is no statistically significant relationship between being late for class and smoking marijuana, this means that there might be some other reasons for being late for class and some other reasons for smoking marijuana and we can't expect a student to be late for class due to smoking marijuana.
9) If a student is late for class, then there is no reason for relating it with smoking marijuana because result showed that there is no statistically significant relationship between smoking marijuana and being late for class.
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