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BY: Kristin Taylor 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 1. 2. 3. 4. 5. 6. 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 The experiment setup have been larger The number of each color of Skittles could have been miscalculated, which would have skewed the sum in the bag 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 2. Counting them & recording the total of each color 3. Add up all the totals to get the total amount of Skittles in the bag 4. Then divide the # of each color by the total # of Skittles to get the percentage EX. 11/58 = .189 ≈ 19% 1. 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 Total # of each Color 14 12 10 8 6 4 2 0 Color Count % GREEN 11 19 PURPLE 13 22 YELLOW 13 22 RED 9 16 ORANGE 12 21 BAG TWO Color Count % GREEN 14 23 PURPLE 14 23 12 YELLOW 13 21 10 RED 11 18 ORANGE 9 15 Total # of each Color 16 14 8 6 4 2 0 Green Purple Red Orange Yellow BAG THREE Color Count % GREEN 11 19 18 PURPLE 16 27 16 YELLOW 10 17 14 RED 12 20 12 ORANGE 10 17 10 Total # of each Color 8 6 4 2 0 BAG FOUR Color Count % 18 GREEN 12 20 16 PURPLE 10 16 14 YELLOW 11 18 12 RED 17 28 10 ORANGE 11 18 Total # of each Color 8 6 4 2 0 BAG FIVE Total # of each Color 20 18 16 14 12 10 8 6 4 2 0 Color Count % GREEN 18 30 PURPLE 14 23 YELLOW 4 6.7 RED 13 22 ORANGE 11 18.3 Cumulative Average Color Total Color Count GREEN 66 PURPLE 67 60 YELLOW 51 40 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 80 20 0 Green Purple Red Orange Yellow 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 59.8 Yellow 51 59.8 Red 62 59.8 Orange 53 59.8 Step 2: The χ² GOF Test will be used Check Conditions: 1. The data does not come from a SRS therefore, I may not be able to generalize about the population 2. 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 = 3.66 59.8 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.