Researchers want to determine whether all bags of M&Ms have the same proportion of colors regardless of the flavor of M&Ms. To test this, they sampled randomly king-size bags of each flavor and recorded their findings in the table.
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&Ms®. 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&Ms® ? Provide a statistical justification for your conclusion. (5 points) (10 points)
The Correct Answer and Explanation is :
Part A: Degrees of Freedom
To calculate the degrees of freedom for a chi-squared test, we use the formula:
[
\text{Degrees of Freedom} = (r – 1) \times (c – 1)
]
Where:
- (r) = number of rows in the contingency table
- (c) = number of columns in the contingency table
Looking at the table from the link, assume there are two flavors of M&Ms (e.g., peanut and regular) and several colors. If the table has 2 rows (one for each flavor) and 6 columns (one for each color), the degrees of freedom would be calculated as:
[
\text{Degrees of Freedom} = (2 – 1) \times (6 – 1) = 1 \times 5 = 5
]
So, the degrees of freedom are 5.
Part B: Expected Count for Green Peanut M&Ms
To calculate the expected count for a category, use the formula:
[
E = \frac{(R_{\text{total}} \times C_{\text{total}})}{N}
]
Where:
- (R_{\text{total}}) = total count for the row (e.g., total number of peanut M&Ms)
- (C_{\text{total}}) = total count for the column (e.g., total number of green M&Ms)
- (N) = total number of M&Ms in the entire table (all rows and columns)
If we assume:
- The total number of peanut M&Ms is (R_{\text{total}}),
- The total number of green M&Ms is (C_{\text{total}}),
- The grand total (N) is the total number of all M&Ms in the table.
For instance, if:
- The total number of peanut M&Ms ((R_{\text{total}})) is 200,
- The total number of green M&Ms ((C_{\text{total}})) is 50,
- The grand total (N) is 1000,
Then the expected count for green peanut M&Ms would be:
[
E = \frac{200 \times 50}{1000} = 10
]
So, the expected count for green peanut M&Ms would be 10.
Part C: Testing for Proportional Differences
To determine if there is sufficient evidence that the proportion of colors is different across flavors of M&Ms, we perform a chi-squared test for independence. The null hypothesis ((H_0)) is that the proportion of M&M colors is the same across the different flavors. The alternative hypothesis ((H_A)) is that the proportion of M&M colors differs across the flavors.
Steps:
- Calculate the chi-squared statistic:
The chi-squared statistic is calculated as:
[
\chi^2 = \sum \frac{(O_i – E_i)^2}{E_i}
]
Where:
- (O_i) = observed count for each category (e.g., the number of green M&Ms in peanut bags),
- (E_i) = expected count for each category (calculated as shown above).
- Determine the p-value:
Once we have the chi-squared statistic, we compare it to the chi-squared distribution with 5 degrees of freedom to find the p-value. - Make a conclusion:
If the p-value is less than the significance level ((\alpha = 0.05)), we reject the null hypothesis and conclude that there is sufficient evidence to suggest a difference in the proportion of colors across M&M flavors. Otherwise, we fail to reject the null hypothesis.
For example, if the chi-squared statistic calculated is 12.5 and the critical value for 5 degrees of freedom at (\alpha = 0.05) is 11.07, since 12.5 > 11.07, we reject the null hypothesis. This indicates that there is a statistically significant difference in the proportion of colors across the different M&M flavors.
In conclusion, based on the statistical test, we would assess whether there is sufficient evidence to conclude that the color proportions are different across M&M flavors.