Suppose that we have a simple random sample of 600 M&M; candies with the followi
ID: 3222810 • Letter: S
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: chi^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
expected frequency =100
X2 =235.4200
p val;ue=0.0000
there is evidence ......
blue
color is over....
observed Expected Chi square Probability O E=total*p =(O-E)^2/E blue 1/6 212.00 100.00 125.44 orange 1/6 147.00 100.00 22.09 green 1/6 103.00 100.00 0.09 red 1/6 50.00 100.00 25.00 yellow 1/6 46.00 100.00 29.16 brown 1/6 42.00 100.00 33.64 600 600 235.4200