### 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

  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

Strength

 The experiment setup  The Skittles were all the same size  No half pieces

## Data Collection

 Data collected by: 1.

Sorting the colors in a 2.17 oz. bag of Original Skittles 2.

3.

Counting them & recording the total of each color 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%

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

6 4 2 0 14 12 10 8

Total # of each Color Color

GREEN PURPLE YELLOW RED ORANGE

Count

11 13 13 9 12

%

19 22 22 16 21

BAG TWO

16 14 12 10 8 6 4 2 0

Total # of each Color

Green Red Purple Orange Yellow

Color

GREEN PURPLE YELLOW RED ORANGE

Count

14 14 13 11 9

%

23 23 21 18 15

BAG THREE Color

GREEN PURPLE YELLOW RED ORANGE

Count

11 16 10 12 10

%

19 27 17 20 17 6 4 2 0 18 16 14 12 10 8

Total # of each Color

BAG FOUR Total # of each Color

6 4 2 0 18 16 14 12 10 8

Color

GREEN PURPLE YELLOW RED ORANGE

Count

12 10 11 17 11

%

20 16 18 28 18

BAG FIVE

20 18 16 14 12 10 8 6 4 2 0

Total # of each Color Color

GREEN PURPLE YELLOW RED ORANGE

Count

18 14 4 13 11

%

30 23 6.7

22 18.3

### Cumulative Average

Color

GREEN PURPLE YELLOW RED ORANGE

Count

66 67 51 62 53 The graph to the right shows the sum of each color within the sample population 5-number summary: Min- 51 Q1- 52 Mean: 59.8

σ: 6.62

Med- 62 Q3- 66.5

Max- 67

Color Total

80 60 40 20 0 Green Red Purple 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

Green Purple Yellow Red Orange

Observed

66 67 51 62 53

Expected

59.8

59.8

59.8

59.8

59.8

Step 2: The χ² GOF Test will be used  Check Conditions: 1.

2.

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: 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

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.