AP Stats Final ProjectBy Josh Spiezle, Emy Chinen, Emily Lopez, Reid Beloff

Our Question…

Do peanut M&M’s have the same % of color proportions as milk chocolate M&M’s?

Why did we choose this as our final project…?

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!

What will be the target population…?

Milk Chocolate M&M’s and Peanut M&M’s

Planned Sample Size…

56 ounces of each kind!

Where and how will you collect our data…?

We will be counting out each color in the bag then divide by the total number of M&M’s in the bag.

Our hypotheses:

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.

Our Conditions:

1.) Randomness – bags of each type of M&M’s

chosen randomly

Our Conditions:

2.)Expected values are greater than 5

Observed (Regular) Peanut (Becomes E) Expected (O-E)² /(E)

Red 249 139 141.71 81.2301

Orange 391 314 320.125 15.6916

Yellow 195 384 391.49 98.6189

Green 267 348 354.79 21.7229

Blue 462 296 301.77 85.0769

Brown 213 262 267.11 10.9614

Total 1777 1743 1777 χ² = 313.3018

Our Conditions:

3.) Independence

Since we are sampling without replacement, must check 10% condition.Np = 10np » Np = 10(1743) » Np = 17430 NR = 10nR » NR = 10(1777) » NR = 17770 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.

The test we shall perform…

X²-Homogeneity test!

df = (r-1)(c-1) = (6-1)(2-1) = 5

X² cdf (313.3018, 1000, 5) X² -value large # df

P-value = 1.38131x 10^-65

Color of M& M

(O-E)² /(E)

Red 81.2301

Orange 15.6916

Yellow 98.6189

Green 21.7229

Blue 85.0769

Brown 10.9614

∑ = χ² = 313.3018

Our graph:

Red Orange Yellow Green Blue Brown0

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Observed (Regular)Peanut (Becomes E)Expected(O-E)² /(E)

Type I and Type II errors, which one is worse…?

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 Conclusion:

Our Chi-Squared Test of Homogeneitygives a p-value of 1.382 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.

Actual Percentages

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