Researchers want to determine whether all bags of M&M’s® have the same proportion of colors regardless of the flavor of M&M’s®. To test this, they sampled randomly king-size bags of each flavor and recorded their findings in the table. Flavor M&M’s® Color Red Orange Yellow Green Blue Brown Original 24 11 29 17 9 14 Peanut 15 20 30 25 15 19 Almond 22 17 21 12 28 7 Part A: What are the correct degrees of freedom for this table? (2 points) Part B: Calculate the expected count for the number of green peanut M&M’s®. Show your work. (3 points) Part C: Is there sufficient evidence that there is a difference in the proportion of colors for the different flavors of M&M’s®? Provide a statistical justification for your conclusion.
The Correct Answer and Explanation is :
To determine whether all bags of M&M’s® have the same proportion of colors regardless of flavor, we can conduct a Chi-Squared Test for Independence. Below are the answers to the questions provided:
Part A: Degrees of Freedom
The degrees of freedom (df) for a Chi-Squared test for independence can be calculated using the formula:
[
df = (r – 1) \times (c – 1)
]
where (r) is the number of rows (categories of M&M’s® flavors) and (c) is the number of columns (categories of colors).
In this case:
- There are 3 flavors (Original, Peanut, Almond) → (r = 3)
- There are 6 colors (Red, Orange, Yellow, Green, Blue, Brown) → (c = 6)
Thus, the degrees of freedom are:
[
df = (3 – 1) \times (6 – 1) = 2 \times 5 = 10
]
Part B: Expected Count for Green Peanut M&M’s®
To calculate the expected count for the number of green peanut M&M’s®, we use the formula:
[
E = \frac{(Row\ Total \times Column\ Total)}{Grand\ Total}
]
- Row Total for Peanut Flavor:
- Red: 15, Orange: 20, Yellow: 30, Green: 25, Blue: 15, Brown: 19
- Total for Peanut = 15 + 20 + 30 + 25 + 15 + 19 = 124
- Column Total for Green Color:
- Original: 17, Peanut: 25, Almond: 12
- Total for Green = 17 + 25 + 12 = 54
- Grand Total:
- Original Total: 24 + 11 + 29 + 17 + 9 + 14 = 104
- Peanut Total: 15 + 20 + 30 + 25 + 15 + 19 = 124
- Almond Total: 22 + 17 + 21 + 12 + 28 + 7 = 107
- Grand Total = 104 + 124 + 107 = 335
- Calculating the Expected Count:
[
E = \frac{(124 \times 54)}{335} \approx 20.079
]
Thus, the expected count for green peanut M&M’s® is approximately 20.08.
Part C: Statistical Justification
To determine if there is sufficient evidence that the proportions of M&M’s® colors differ by flavor, we will perform a Chi-Squared test using the observed and expected counts. The null hypothesis ((H_0)) states that there is no difference in the proportions of colors across the different flavors, while the alternative hypothesis ((H_a)) states that at least one flavor differs in color proportion.
- Calculate Chi-Squared Statistic:
The Chi-Squared statistic is calculated as:
[
\chi^2 = \sum \frac{(O – E)^2}{E}
]
where (O) is the observed count and (E) is the expected count. Each flavor’s observed counts and corresponding expected counts would need to be calculated, and the statistic would be summed across all cells. - Compare to Critical Value:
The calculated Chi-Squared value will be compared to the critical value from the Chi-Squared distribution table with (df = 10) at a significance level (e.g., (\alpha = 0.05)). If the calculated (\chi^2) is greater than the critical value, we reject the null hypothesis. - Conclusion:
If we find that the (\chi^2) statistic is significantly high, it suggests that the proportions of colors are indeed different among the flavors. If the p-value is less than the significance level, we have enough evidence to conclude that there is a statistically significant difference in the color distribution of M&M’s® based on flavor.
In summary, if the Chi-Squared test indicates a significant difference, we conclude that the proportion of colors in M&M’s® does vary with the flavor, indicating a non-uniform distribution of colors across different M&M’s® flavors.