Question

A bag of M&MS was randomly selected from the grocery store shelf, and the color counts were: Brown 20, Red 5, Yellow 10, Orange 4, Green 5, Blue 6. Mars Inc. claims that they produce M&Ms with the following distributions Brown 30% Red 20% Yellow 20% Orange 10% Green 10% Blue 10%.

1. n = 50, is there evidence of difference between the random bag that was picked and Mars Inc.'s claims?

2. Graph the results.

Answer 1

Categories Observed Expected Brown 20 50*0.3=15 Red 5 50*0.2=10 Yellow 10 50*0.2=10 (fo-fe)2/fe (20-15)2/15 = 1.667 (5-10)2/10 = 2.5 (10-10)2/10 = 0 (4-5)2/5 = 0.2 (5-5)215 = 0 (6-5)2/5 = 0.2 Orange 4 50*0.1=5 Green 5 50*0.1=5 Blue 6 50*0.1=5 Sum = 50 50 4.567 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H : P1 = 0.3, p2 = 0.2, p3 = 0.2, P4 = 0.1, p5 = 0.1, p6 = 0.1 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) Rejection Region Based on the information provided, the significance level is a = 0.05, the number of degrees of freedom is df = 6 - 1 = 5, so then the rejection region for this test is R= {x? : x² > 11.07}.


(3) Test Statistics The Chi-Squared statistic is computed as follows: n X - į (O; – E;) E; = 1.667 + 2.5 + 0 + 0.2 + 0 + 0.2 = 4.567 i=1 (4) Decision about the null hypothesis = 4.567 < x2 = 11.07, it is then concluded that the null hypothesis is Since it is observed that x? not rejected. (5) 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 a = 0.05 significance level.

Therefore the claim by Mars Inc is accepted at level of significance 0.05.

p-value = P(X> 4.567), where X~chi-square with d.f =5

therefore p-value = 0.47096