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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? $$ \begin{array}{lcrccc} \hline & \begin{array}{l} \text { Green } \\ \text { (Lime) } \end{array} & \begin{array}{c} \text { Purple } \\ \text { Orange } \end{array} & \begin{array}{c} \text { Red } \\ \text { (Grape) } \end{array} & \begin{array}{c} \text { Yellow } \\ \text { (Strawberry) } \end{array} & \text { (Lemon) } \\ \hline \text { Color } & 18 & 9 & 15 & 13 & 11 \\ \text { Flavor } & 13 & 16 & 19 & 34 & 9 \end{array} $$

Step by step solution

01

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.

02

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.

03

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.

04

Count the Degrees of Freedom

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

05

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.

06

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.

07

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis testing

Hypothesis testing is a critical part of statistical analysis. It is used to determine if there is enough evidence to reject a conjecture about a population parameter. In the context of the Skittles exercise, we start by defining two competing statements: the null hypothesis ( \( H_0 \) ) and the alternative hypothesis ( \( H_a \) ).

• The null hypothesis ( \( H_0 \) ): This posits that the method of choice (color vs. flavor) and the actual type of Skittles chosen are independent events. In other words, these two variables do not influence each other.
• The alternative hypothesis ( \( H_a \) ): This suggests a dependency between the method of choice and the type of Skittles. Essentially, these two factors are related, implying their independence is not guaranteed.

To proceed, we perform a chi-square test on the data. This test checks how much the observed data differ from expected data under the null hypothesis. If the calculated test statistic is large, it suggests that the null hypothesis may not hold, pointing towards the alternative hypothesis being true. It is this rigor in hypothesis testing that underpins the validity of the conclusions drawn from statistical analyses.

Contingency table

A contingency table is a pivotal tool in presenting and analyzing categorical data. It outlines how variables are related by showing their frequency distribution through a matrix format. For our Skittles project, the contingency table has 2 rows (color and flavor) and 5 columns (the different Skittles types).
This structure helps in understanding the relationship between the categories by displaying the observed data. In addition to observed frequencies, we calculate expected frequencies. These reflect the frequencies we would expect under the null hypothesis, assuming the variables are independent.
The expected frequency for each cell is calculated using the formula: \[ E = \frac{( ext{Row Total} \times ext{Column Total})}{ ext{Grand Total}} \]
By comparing observed and expected frequencies, we identify discrepancies that suggest dependency between variables. This transformation from raw data to a structured form enables clear analysis and conclusion drawing, especially when performing tests like the chi-square test.

Degrees of freedom

Degrees of freedom (df) are essential in statistical tests as they allow us to understand the balance of the data. They help determine the precise distribution of the test statistic under the null hypothesis. In chi-square tests, the degrees of freedom are crucial in assessing the variability among variables.
For a contingency table, the degrees of freedom are calculated using the formula:\[ ( ext{number of rows} - 1) \times ( ext{number of columns} - 1) \]
In the Skittles exercise, we have 2 rows (color and flavor) and 5 columns (types of Skittles), giving:\[ (2 - 1) \times (5 - 1) = 4 \]
Understanding degrees of freedom helps in interpreting the chi-square statistic. It aids in finding the critical value from statistical tables or software, which in turn is used to ascertain the p-value. This understanding is imperative for accurately determining whether or not to reject the null hypothesis. Recognizing how many degrees can "move" in accounting both limits and assures the accuracy of a statistical measure across different tests.