1. The Distribution of Blue M&Ms By Samantha Boccard & Adrienne Umali

2. Project Description The M&M/Mars Company claims that each 1.69 oz bag of plain M&Ms it manufactures contains a proportion of.24 blue M&Ms. Two very concerned, M&M-loving stats students believe that this is not the actual proportion and design an experiment to test the claim.

3. The null/alternative hypotheses The M&M/Mars Company claims that the proportion of blue M&Ms in each 1.69 oz milk chocolate bag is.24. We will be testing the claim to be false.

4. Data Collection 1. Gather 20 bags of 1.6 oz milk chocolate M&Ms (preferably from different stores) 2. Find the claimed proportions of colors of M&Ms from www.mms.com 3. Then write the color of M&Ms on separate pieces of paper and place in a hat. Mix the pieces of paper and randomly choose a piece of paper with a color on it. This will be the color you are testing for, in our case it was blue.

5. Data Collection Con’t 4. 4. Open the 20 bags, making sure to keep them separated from each other (and that no one eats then), and count chosen color of M&Ms and the total number of M&Ms in each bag. 5. Divide the number of the chosen color M&Ms in the bag over the total number to reach the proportion. Record all data 6. To find the overall proportion, divide all of the overall chosen color with the overall number of M&Ms.

6. The Data Bag 1.0926 Bag 2.1754 Bag 3.2069 Bag 4.0545 Bag 5.1296 Bag 6.0714 Bag 7.1071 Bag 8.2241 Bag 9.2222 Bag 10.1230 Bag 11.2931 Bag 12.1379 Bag 13.2982 Bag 14.0877 Bag 15.2456 Bag 16.2982 Bag 17.1786 Bag 18.1525 Bag 19.2727 Bag 20.3036 Overall proportion: 208/1128=.184397

7. Summery Of Proportions of Blue M&Ms The distribution of blue M&Ms is slightly skewed right with a 5 number summary of.0545,.11505,.177,.2915,.3036, with no apparent outliers.

8. Our Lovely Test (Part 1) Ho: p =.24 Ha: p =.24 Where p is the proportion of blue milk chocolate M&Ms present in one 1.69 oz bag. We will use a single sample proportion problem for this experiment. This includes using 2 tail-z procedures with.24 as an estimate for p. We chose this experiment because we are only testing for one variable and do not have a hypothesis as to whether the claim is higher or lower than the actual.

9. Assumptions SRS (while the designated color was chosen using an SRS, the actual bags were not. And though we were very precise at buying bags from various locations throughout the island, the M&Ms in our experiment may not sufficiently represent the entire M&M population) 10n≤N (assume milk chocolate M&M bag population is ≥11280) √ np≥10 (1128(.24) ≥10;) √ n(1-p) ≥10 (1128(1-.24) ≥10) √

10. Our Lovely Test (Part 2) n=1128 p-hat=.18 alpha=.05 z= p-hat –p √p(1-p)/n) =.18 -.24 √.24(.76)/1128) = -4.718 P-hat =.18 p=.24 P-value = 0(2) = essentially 0

11. Conclusion If the true population proportion of milk chocolate blue M&Ms in a 1.69 oz bag were truly.24, then the probability of obtaining a sample proportion of.18 or more extreme is essentially 0. Therefore there is sufficient evidence to show that the claim is false at 5%.

12. Problems With Our Experiment Though we tried our best to select M&Ms from various locations, it is possible that our sample did not accurately represent the entire population of M&Ms since they were all purchased from the same relative location. Another problem was deformed M&Ms, as the original claim could have not counted the deformities as an M&M.