By Josh Spiezle, Emy Chinen, Emily Lopez, Reid Beloff
Do peanut M&M’s have the same % of color proportions as milk chocolate M&M’s?
Well, we chose this as our final project because our group has Em and Em in it, and we all LOVE chocolate. Clearly, this made the most sense!
Milk Chocolate M&M’s and Peanut M&M’s
56 ounces of each kind!
We will be counting out each color in the bag then divide by the total number of M&M’s in the bag.
Ho: The proportions of colors in peanut M&M’s matches the proportions of the milk chocolate M&M colors. Ha: At least one of the color proportions between the 2 types of M&M’s are not equal.
1.) Randomness – bags of each type of M&M’s chosen randomly
2.)Expected values are greater than 5 Observed (Regular)Peanut (Becomes E)Expected(O-E)² /(E) Red Orange Yellow Green Blue Brown Total χ² =
3.) Independence Since we are sampling without replacement, must check 10% condition. N p = 10n p » N p = 10(1743) » N p = N R = 10n R » N R = 10(1777) » N R = Safe to assume at least 17,430 peanut M&M’s and 17,770 milk chocolate M&M’s in entire population so 10% condition is met.
X ² -Homogeneity test! df = (r-1)(c-1) = (6-1)(2-1) = 5 X² cdf ( , 1000, 5) X² -value large # df P-value = x 10^-65 Color of M& M (O-E)² /(E) Red Orange Yellow Green Blue Brown ∑ = χ² =
Type I-Statistical test rejects that proportions of colors are equal when they are. Type II-Statistical test fails to reject them being equal when they are not. We believe that neither error type is worse in this situation so we will be leaving our alpha level at 0.05.
Our Chi-Squared Test of Homogeneitygives a p-value of x 10^-65 which is significantly less than any reasonable alpha level. Thus we are able to reject our null hypothesis and conclude there is enough evidence to say that the proportion of the colors in peanut M&M’s and milk chocolate M&Ms are not equal.
Plain Variety: 30% brown 20% yellow 20% red 10% green 10% orange 10% blue Peanut Variety: 20% brown 20% yellow 10% red 10% green 10% orange 30% blue