1. Are Skittles Evenly Distributed?

2. 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

3. 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

4. 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

5. 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%

6. 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 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 z. bag would range from Skittles.

7. BAG ONE Color Count % GREEN 11 19 PURPLE 13 22 YELLOW RED 9 16 ORANGE12 21

8. BAG TWO Color Count % GREEN 14 23 PURPLE YELLOW 13 21 RED 11 18 ORANGE9 15

9. BAG THREE Color Count % GREEN 11 19 PURPLE 16 27 YELLOW 10 17 RED 1220
ORANGE

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

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

12. 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 Mean: 59.8
Q σ: 6.62
Med- 62 Q
Max- 67
Shape: the graph is roughly symmetric
Outliers: there are no outliers
Center: 62
Spread:51- 67

13. 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

14. 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

15. 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
= ( )² + ( )² + ( )² + ( )² + ( )²
= 3.66

16. 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.