Question

need help help with the contingency table ( two tables below) have no idea how to set them up with 3x2 variables.

6. An ice cream company wants to determine if people's income is refated to the type of flavor they purchase. Therefore, they collected a survey of920 people who visited their store. They categorized people as being high or lowin income and whether or not they chose chocolate, vanilla, or strawberry ice cream. For people who were of low income, 100 chose chocolate, 120 chose vanilla, and 60 chose strawberry. For people who were of high income, 350 chose chocolate, 200 chose vanilla, and 90 chose strawberry. a. What is the null and alternative hypothesis? Ot9e is no dicteen e/velalronsnp petnen peoplesihene They choose ice-crean hee is a dliceeencenolalonswe behen peaples A mcone and ice aean faor hen dhoos Which statistical test should be calculated? b. Chi test of indeperdlen c Calculate that test. c. 120 (6ps) ontingency Table: oh'Phoe be Vla lo inc0me O:(00 Choc 153.3 C VIbh inone Total V2a O: E:158.50 Е: O: 120 Е: V2b O: E: 153 3 Total E 2 (0-E) (0-В) E Chacelale 100 (О-Е) 120 vanila Sum=

Answer 1

Solution: Part-A: Ho: There is no relationship between people income and type of flavor they purchase. HA: There is a relationship between people income and type of flavor they purchase. Part-B: Chi-square test of independence will be used. Part-C: The data given can be tabulated as below: Low High Total Chocolate 100 L 350 450 Vanilla 120 200 320 Strawberry 60 90 150 / Total 280 640 920 Observed Frequency Expected Frequency Income Flavor fo-fe (f. - f.)2 (f. -f.)/fe Low Low 100 120 60 1 Low High High High Chocolate Vanilla Strawberry Chocolate Vanilla Strawberry 450/920 x 280 = 136.957 -36.957 320/920 x 280 = 97.391 22.609 150/920 x 280 = 45.652 14.348 450/920 x 640 = 313.043 | 36.957 320/920 x 640 = 222.609 -22.609 150/920 x 640 = 104.348 -14.348 1365.784 511.153 205.860 1365.784 511.153 205.860 9.972 5.248 4.509 4.363 2.296 1.973 350 200 90 Ef. = 920 fe = 920 Ef.-f.)?/fe = 28.362

Test Statistic: x2 = 5 forfe' = 28.362 Degrees of Freedom = df = (number of categories in row variable - 1) x (number of categories in column variable - 1) Hence, df = (2-1) (3 - 1) = 2 P-value: From the xdistribution table; the area in the upper tail to the right of (x2 = 28.362) and for (df = 2), is given by: 0.0000 Hence, P-value = 0.0000 Let us assume, level of significance = a = 0.05 Since p-value < a, we reject the null hypothesis. There is sufficient evidence to conclude that there is a relationship between people income and type of flavor they purchase.