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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
Procedure1. Pour one bag of Skittles
onto a paper towel2. Sort the Skittles by color3. Count the # of each
color and record4. Calculate total # of
Skittles in individual bag5. Place skittles in a
cup/bowl6. Repeat steps 1-5 for the
remaining 4 bags
Intro & Research (cont.)Weakness StrengthThe population size
could have been largerThe 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
Data CollectionData collected by:1. Sorting the colors in a 2.17 oz. bag of
Original Skittles2. Counting them & recording the total of each
color3. 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 percentageEX. 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
Green
Purp
le
Yello
wRed
Ora
nge
02468
101214
Total # of each Color
Color Count %
GREEN 11 19
PURPLE 13 22
YELLOW 13 22
RED 9 16
ORANGE 12 21
BAG TWO
02468
10121416
Total # of each Color
Green Purple YellowRed Orange
Color Count %
GREEN 14 23
PURPLE 14 23
YELLOW 13 21
RED 11 18
ORANGE 9 15
BAG THREE
Color Count %
GREEN 11 19
PURPLE 16 27
YELLOW 10 17
RED 12 20
ORANGE 10 17
Green
Purp
le
Yello
wRed
Ora
nge
02468
1012141618
Total # of each Color
BAG FOUR
Green
Purp
le
Yello
wRed
Ora
nge
02468
1012141618
Total # of each Color
Color Count %
GREEN 12 20
PURPLE 10 16
YELLOW 11 18
RED 17 28
ORANGE 11 18
BAG FIVE
Green
Purp
le
Yello
wRed
Ora
nge
0
4
8
12
16
20
Total # of each Color
Color Count %
GREEN 18 30
PURPLE 14 23
YELLOW 4 6.7
RED 13 22
ORANGE 11 18.3
Cumulative AverageColor Count
GREEN 66
PURPLE 67
YELLOW 51
RED 62
ORANGE 53
0
20
40
60
80
Color Total
Green Purple YellowRed Orange
5-number summary:Min- 51 Mean: 59.8Q1- 52 σ: 6.62Med- 62Q3- 66.5Max- 67
Shape: the graph is roughly symmetricOutliers: there are no outliersCenter: 62Spread:51- 67
The graph to the right shows the sum of each color within the sample population
Inference ProcedureNull 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 TestHo: 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 usedCheck Conditions:1. The data does not come from a SRS therefore, I may not
be able to generalize about the population2. 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 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.