Are Skittles Evenly Distributed?

Introduction & Research Question Question- Are the flavors in a 2.17 oz. bag of original Skittles evenly distributed? Population of interest- 5 bags of 2.17 ounce original Skittles

Procedure Pour one bag of Skittles onto a paper towel Sort the Skittles by color Count the # of each color and record Calculate total # of Skittles in individual bag Place skittles in a cup/bowl Repeat steps 1-5 for the remaining 4 bags

Intro & Research (cont.) Weakness Strength The population size could have been larger The number of each color of Skittles could have been miscalculated, which would have skewed the sum in the bag The experiment setup The Skittles were all the same size No half pieces

Data Collection Data collected by: Sorting the colors in a 2.17 oz. bag of Original Skittles Counting them & recording the total of each color Add up all the totals to get the total amount of Skittles in the bag Then divide the # of each color by the total # of Skittles to get the percentage EX. 11/58 = .189 ≈ 19%

I am confident that my sample represents the population because the total number of Skittles within the five bags were around the same total. The total ranged from 58-61. Therefore, I am confident that if a larger sample size was used then the total amount of Skittles would be within this range. Using the z-interval test on a TI-83, I’m 90% confident that the total amount of Skittles in a 2.17 0z. bag would range from 55-65 Skittles.

BAG ONE Color Count % GREEN 11 19 PURPLE 13 22 YELLOW RED 9 16 ORANGE 12 21

BAG TWO Color Count % GREEN 14 23 PURPLE YELLOW 13 21 RED 11 18 ORANGE 9 15

BAG THREE Color Count % GREEN 11 19 PURPLE 16 27 YELLOW 10 17 RED 12 20 ORANGE

BAG FOUR Color Count % GREEN 12 20 PURPLE 10 16 YELLOW 11 18 RED 17 28 ORANGE

BAG FIVE Color Count % GREEN 18 30 PURPLE 14 23 YELLOW 4 6.7 RED 13 22 ORANGE 11 18.3

Cumulative Average Color Count GREEN 66 PURPLE 67 YELLOW 51 RED 62 ORANGE 53 The graph to the right shows the sum of each color within the sample population 5-number summary: Min- 51 Mean: 59.8 Q1- 52 σ: 6.62 Med- 62 Q3- 66.5 Max- 67 Shape: the graph is roughly symmetric Outliers: there are no outliers Center: 62 Spread:51- 67

Inference Procedure Null Hypothesis- The flavors of Original Skittles in a 2.17 oz. bag are evenly distributed. Alternative Hypothesis- The flavors of Original Skittles in a 2.17 oz. bag are not evenly distributed. Significance level: α =.05 Sample size: 5 bags of 2.17 oz. Skittles

Chi-square Test Ho: The flavors of Original Skittles in a 2.17 oz. bag are evenly distributed. Ha: The color of Original Skittles in a 2.17 oz. bag are not evenly distributed. Class Observed Expected Green 66 59.8 Purple 67 Yellow 51 Red 62 Orange 53

Step 2: The χ² GOF Test will be used Check Conditions: The data does not come from a SRS therefore, I may not be able to generalize about the population The expected numbers are greater than 5 Step 3: Χ² = ∑(O-E)² E = (66-59.8)² + (67-59.8)² + (51-59.8)² + (62-59.8)² + (53-59.8)² 59.8 59.8 59.8 59.8 59.8 = 3.66

Step 4: Using a TI-84, the p-value was 0 Step 4: Using a TI-84, the p-value was 0.45 There is strong evidence to reject the null hypothesis at the α = .05 level because the p-value is greater than .05 (.45 ≥ .05). Therefore, the flavors in a 2.17 oz. bag of Original Skittles are not evenly distributed, which can be seen in the graphical displays of each individual bag. From reviewing my graphical displays and charts I noticed that within four of the bags of Skittles only two of the colors within the bag had equal amounts.