Question

ou have been tasked with reviewing product consistency regarding the packaging of plain M&M’s candy. Your job is to determine if the number of blue, green, brown, orange, and yellow M&M’s in each bag is relatively equal. Obviously, not every bag will contain the exact same number of M&M’s, divided into equal quantities. However, as part of your quality control responsibilities you want to determine if there is a significant difference in the number of colors in each bag. After all, you don’t want customers to open a bag and only receive primarily one or two colors. You could use a one variable chi-square test to analyze your research question: is the number of blue, green, brown, orange, and yellow M&M’s significantly different? Your stated null and alternative hypotheses would look like this: Ho: There is no significant difference between the number of blue, green, brown, orange, and yellow M&M’s per bag. H1: There is a significant difference between the number of blue, green, brown, orange, and yellow M&M’s per bag. Next, we would move through the steps required to complete the chi-square test. Class: How would you conduct this analysis? What are the steps that we must follow?

Answer 1

Question: How would you conduct this analysis? What are the steps that we must follow?

Step 1:

Ho: Null Hypothesis: There is no significant difference between the number of blue, green, brown, orange, and yellow M&M’s per bag.

H1: Alternative Hypothesis: There is a significant difference between the number of blue, green, brown, orange, and yellow M&M’s per bag.

Step 2:

Observed Frequencies:

The following Table is to be completed from Observed Data:

Color	Observed Frequency (O)
Blue
Green
Brown
Orange
Yellow
Total =	N

Step 3:

Since the Total number of M&M’s in each bag is relatively equal., the Expected number of each color = N/5 as follows:

Expected Frequencies:

Color	Expected Frequency (E)
Blue	N/5
Green	N/5
Brown	N/5
Orange	N/5
Yellow	N/5
Total =	N

Step 4:

The Test Statistic is calculated as follows:

,

where O_i, (i = 1, 2,...n) are the observed frequencies.

E_i, i = 1, 2,...,n are the Expected Frequencies.

Take = 0.05

ndf = n - 1 = 5 - 1 = 4

From Table, critical value of = 9.4877

If the Calculated value of is greater than critical value of = 9.4877, the difference is significant. Reject null hypothesis.

Conclusion:

The data do not support the clam that there is no significant difference between the number of blue, green, brown, orange, and yellow M&M’s per bag.

If the Calculated value of is less than critical value of = 9.4877, the difference is not significant. Fail to reject null hypothesis.

Conclusion:

The data support the clam that there is no significant difference between the number of blue, green, brown, orange, and yellow M&M’s per bag.