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BY: Kristin Taylor

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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

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Procedure 1. Pour one bag of Skittles onto a paper towel 2. Sort the Skittles by color 3. Count the # of each color and record 4. Calculate total # of Skittles in individual bag 5. Place skittles in a cup/bowl 6. Repeat steps 1-5 for the remaining 4 bags

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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

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Data Collection Data collected by: 1. 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%

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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.

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BAG ONE ColorCount% GREEN1119 PURPLE1322 YELLOW1322 RED916 ORANGE1221

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BAG TWO ColorCount% GREEN1423 PURPLE1423 YELLOW1321 RED1118 ORANGE915

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BAG THREE ColorCount% GREEN1119 PURPLE1627 YELLOW1017 RED1220 ORANGE1017

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BAG FOUR ColorCount% GREEN1220 PURPLE1016 YELLOW1118 RED1728 ORANGE1118

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BAG FIVE ColorCount% GREEN1830 PURPLE1423 YELLOW46.7 RED1322 ORANGE1118.3

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Cumulative Average ColorCount GREEN66 PURPLE67 YELLOW51 RED62 ORANGE53 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 The graph to the right shows the sum of each color within the sample population

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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

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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. ClassObservedExpected Green6659.8 Purple6759.8 Yellow5159.8 Red6259.8 Orange5359.8

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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: Χ ² = ∑(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

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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.

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