Question

Favorite Skittles Flavor? Exercise 7.13 on page 518 discusses a sample of people choosing their favorite Skittles flavor by color (green, orange, purple, red, or yellow). A separate poll sampled 91 people, again asking them their favorite Skittles flavor, but rather than by color they asked by the actual flavor (lime, orange, grape, strawberry, and lemon, respectively). \(^{19}\) Table 7.32 shows the results from both polls. Does the way people choose their favorite Skittles type, by color or flavor, appear to be related to which type is chosen? (a) State the null and alternative hypotheses. (b) Give a table with the expected counts for each of the 10 cells. (c) Are the expected counts large enough for a chisquare test? (d) How many degrees of freedom do we have for this test? (e) Calculate the chi-square test statistic. (f) Determine the p-value. Do we find evidence that method of choice affects which is chosen? $$ \begin{array}{lcrccc} \hline & \begin{array}{l} \text { Green } \\ \text { (Lime) } \end{array} & \begin{array}{c} \text { Purple } \\ \text { Orange } \end{array} & \begin{array}{c} \text { Red } \\ \text { (Grape) } \end{array} & \begin{array}{c} \text { Yellow } \\ \text { (Strawberry) } \end{array} & \text { (Lemon) } \\ \hline \text { Color } & 18 & 9 & 15 & 13 & 11 \\ \text { Flavor } & 13 & 16 & 19 & 34 & 9 \end{array} $$

Short answer

The test involves setting up two hypotheses, computing expected frequencies and chi-square statistic, and then drawing conclusion based on the computed p-value.

Step-by-step solution

Set up the hypotheses

The null hypothesis (\( H_0 \)) is: Type of choice and chosen type aren't related. Alternative Hypothesis (\( H_a \)): Type of choice and chosen type are related.

Compute Expected Frequencies

The expected frequency for each cell in a contingency table is \(E = (Row Total * Column Total) / Grand Total \). Calculate this for all 5*2 cells in the given table.

Verify Chi-Square Test Conditions

All expected counts should be at least 5 for Chi-Square test validity. Check if all the calculated expected counts meet this condition.

Count the Degrees of Freedom

\ Degrees of freedom =(no. of rows - 1)*(no. of columns - 1)=(2 - 1) * (5 - 1) = 4.

Compute Test Statistic

Chi-square test statistic is \(\chi^2 = \sum ((Observed - Expected)^2 / Expected)\). Observed is the given data, and Expected is from Step 2. Sum this ratio for all cells.

Find p-value

The p-value is the probability of getting a chi-square as extreme as the test statistic, given the null hypothesis is true. Use chi-square distribution table with df=4 (from step 4), or software to find this.

Draw Conclusions

If the p-value is less than the significance level (typically 0.05), reject the null hypothesis and conclude that the way people choose their favorite skittles flavor, by color or by actual flavor, does appear to be related to which type is chosen.