Question

Question 1: A student in an introductory statistics course investigated if there is evidence that the proportion of milk chocolate M&M's that are green differs from the proportion of dark chocolate M&M's that are green. She purchased a bag of each variety, and her data are summarized in the following table.

	Green	Not Green	Total
Milk Chocolate	8	33	41
Dark Chocolate	4	38	42
Total	12	71	83

c. Estimate the p-value using the randomization distribution and show how you found it on the plot.

Answer 1

n = 83 Sample Size    
     
Null and Alternative hypothesis are      
Ho : Chocolate Types and Color are independent     
Ha : Chocolate types and color are not independent     
     
Let α = 0.05 Significance Level    
To find Expected frequencies     
Expected Frequency = (Row Total * Column Total)/Grand Total     
Grand Total = 83     
Expected Frequency Matrix (rounding up the frequencies to integer)     

Expected Frequencies (E )	Green	Non green	Total
Milk Chocolate	5.9277	35.0723	41
Dark Chocolate	6.0723	35.9277	42
Total	12	71	83

    
Calculate value of χ2 test statistic =     
    
Following table gives the value of (Observed - Expected)2 / Expected  

	Green	Non green	Total
Milk Chocolate	0.7245	0.1224	0.8469
Dark Chocolate	0.7072	0.1195	0.8267
Total	1.4317	0.2420	1.6736

Calculate value of χ2 test statistic =     

χ2 test statistic = 1.6736      
      
Degrees of freedom = df = (number of rows - 1) * (number of columns - 1)      
df = (2-1) * (2-1) = 1      
df = Degrees of freedom = 1      
      
From chi square table or CHISQ.DIST.RT(X, df) function of Excel      
p-value for χ2 = CHISQ.DIST.RT(1.6736, 1) = 0.1958      
p-value = 0.1958      
      
α = 0.05      
0.1958 > 0.05      
Since p-value > α, we DO NOT reject the null hypothesis      
      
Conclusion      
There does not exist enough statistical evidence at 5% level of significance to show that      
the proportion of milk chocolate M&M's that are green differs from the proportion      
of dark chocolate M&M's that are green