Suppose that we have a simple random sample of 600 M&M; candies with the followi

Question

Suppose that we have a simple random sample of 600 M&M; candies with the following distribution: 212 of the candies are blue. 147 of the candies are orange. 103 of the candies are green. 50 of the candies are red. 46 of the candies are yellow. 42 of the candies are brown. What is the expected count for each color if we assume that the colors are evenly distributed in a bag? Test if there is evidence that the colors are NOT evenly distributed: X^2 = p-value = Conclusion: There (is/is not evidence at the .05 level that the m&m; colors are evenly distributed. What color contributes the MOST to the test statistic: and why? The color is (over/under) represented than expected.

Explanation / Answer

null hypothesis will be that all colors occur in the same proportion. More formally, if p1 is the population proportion of red candies, p2 is the population proportion of orange candies, and so on, then the null hypothesis is that p1= p2 = .

. . = p6 = 1/6.

The alternative hypothesis is that at least one of the population proportions is not equal to 1/6.

the expected counts for each of these colors would be (1/6) x 600 = 100.

E(BLUE)=100

E(0RANGE)=100

E(GREEN)=100

E(RED)=100

E(YELLOW)=100

E(BROWN)=100

e calculate the contribution to our statistic from each of the colors. Each is of the form (Actual – Expected)2/Expected.:

We then total all of these contributions and determine that our chi-square statistic is 125.44 + 22.09 + 0.09 + 25 +29.16 + 33.64 =235.42

CHI SQ=235.42

The number of degrees of freedom for a goodness of fit test is simply one less than the number of levels of our variable.

Since there were six colors, we have 6 – 1 = 5 degrees of freedoM

We find that CHI SQ test statistic with five degrees of freedom has a p-value of 7.29 x 10-49

P VALUE= 7.29 x 10-49 =0.0000

p<0.05

reject null hypothesis

Accept alternative hypothesis

We conclude that M&Ms are not evenly distributed among the six different colors.

there is evidence at the 0.05 level that the m&m colors are evenly distributed

blue colour contributes most to the test statisic

blue we have (212 – 100)2/100 = 125.44

since observed frequency is more for blue.

the color green is under represented than expected.

actuallky expected =1/6=0.166

observed for green =0.09

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.