In order to compare this test statistic to the critical value X α 2 , we must solve for the degrees of freedom and remember our significance level is =0.05. According to the text, the degrees of freedom is solved as follows: df = ( r − 1 ) ( c − 1 ) Where r= # of rows and c= # of columns (Donnelly, 2012, p. 613). df = ( 4 − 1 ) ( 3 − 1 ) = ( 3 ) ( 2 ) = 6 Using the chi-square right-tail distribution chart in appendix A: X α 2 = 12.592 (Donnelly, 2012) Since X 2 > X α 2 , it is safe to say that we must reject the Null hypothesis. This tells John and Maria that there is a dependency between types of meat and presentation types. In summary, we have helped John and Maria understand that there is no difference between the types of meats and types of presentations. We also used the Chi-square method to determine that the two factors are indeed dependent of one another. This means that of all their items, no particular type of meat or presentation is ordered more frequently, on average, than the others. Knowing this, John and Maria should be careful not to decrease their options of any one category too much. If they are to add more items, they should also spread out the variety, as their customers order from all over the menu, not just in one type of meat or presentation. Finally, The Chi-square method did tell us that the factors are dependent, so John and Maria need to keep in mind that as they change one item, say adding more pork items, this may significantly change the demand of the types of presentations. References Donnelly, R. A. (2012). Business statistics plus MyStatLab. Upper Saddle River, NJ: Pearson. Zaiontz, C. (2016). Two Factor ANOVA without Replication. Retrieved July 09, 2016, from
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