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.