Question

The distribution of M&M colors is 24% blue, 20% orange, 16% green, 14% yellow, 13% red, and 13% brown.

You have a small bag of M&M's and count the colors. Your bag has 9 blue, 9 orange, 4 green, 12 yellow, 3 red, and 6 brown M&M's.

Before you begin, how many M&M's do you need in your sample (minimum) to be able to do this test? Remember, your assumption for chi-square is that every expected frequency has at least 5 in the cell, that is, if the calculated E is less than 5, you cannot do the problem.

The assumption here is E >/= 5, where E = n(p).

Find the Test statistic

What is the critical value? CV(chi-square)

What was your decision regarding the null?

What does your data show? In words.

Answer 1

Assuming level of significance to be 5%.

Chi-Square Goodness of Fit test

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H0: p1 =0.24, p2 =0.2, p3​=0.16, p4​=0.14, p5 =0.13, p6=0.13 Ha​: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. (2) Degrees of Freedom The number of degrees of freedom is df=n-1=6-1=5 (3) Test Statistics The Chi-Squared statistic is computed as follows: (4)Critical Value and Rejection Region Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df=n-1=6-1=5, so the critical value becomes 11.0705. Then the rejection region for this test is R={χ2:χ2>11.0705}. (5)P-value The P-value is the probability that a chi-square statistic having 5 degrees of freedom is more extreme than 11.0705. The p-value is p=Pr(χ2>8.5634)=0.1278 (6) The decision about the null hypothesis Since it is observed that χ2=8.5634≤χc2​=11.0705, it is then concluded that the null hypothesis is not rejected. (7) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.05 significance level.

Let me know in the comments if anything is not clear. I will reply ASAP! Please do upvote if satisfied!