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: 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

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