In: Statistics and Probability
A bake shop owner wants to know how to make his Bakalava (sweet walnut and honey pastry) to the maximum appeal for his patrons. He thinks that the major factors that determine appeal are the amounts honey in the filling and the number of layers of phillo dough used to make the pastry. During an eight week period he makes nine batches of Bakalva using a low, medium and high mount of honey and with 10, 15 and 20 layers of phillo dough. He then counts how many of each type he sells. The table below summarizes his sales of each variation on the basic (medium honey 15 layers) recipe.
Layers/ Honey |
Low |
Medium |
High |
10 |
100 |
105 |
105 |
15 |
98 |
100 |
123 |
20 |
118 |
122 |
59 |
Calculate a 2-factor χ2 test to answer the following questions. Show enough of your work for part credit if you get the number wrong and use an α level of
0.05.
Observed |
||||
Layers/Honey |
Low |
Medium |
High |
Total |
10 |
100 |
105 |
105 |
310 |
15 |
98 |
100 |
123 |
321 |
20 |
118 |
122 |
59 |
299 |
Total |
316 |
327 |
287 |
930 |
Expected = Row Total* Column Total / Grand Total |
||||
Layers/Honey |
Low |
Medium |
High |
|
10 |
105.3333 |
109 |
95.66667 |
|
15 |
109.071 |
112.8677 |
99.06129 |
|
20 |
101.5957 |
105.1323 |
92.27204 |
|
Chi-square= Sum[(Exp-Obs)^2/Exp] |
||||
Layers/Honey |
Low |
Medium |
High |
|
10 |
0.270042 |
0.146789 |
0.910569 |
|
15 |
1.12373 |
1.467016 |
5.784922 |
|
20 |
2.648745 |
2.706312 |
11.99745 |
|
CHI_SQUARE = 27.0555713058413 |
Degrees of freedom = (#Rows – 1)(#Cols – 1) = 2*2 = 4 Critical value = X20.05;4 = 9.487729
Since Observed Chi-Square > critical value, we reject the null hypothesis of no association between the recipes and sales and conclude that there is significant association between recipes and sales.
Honey and layers interact significantly to produce a different appeal.