Question

# Is there enough evidence that the average number of PURPLE Skittles in a package is greater...

1. Is there enough evidence that the average number of PURPLE Skittles in a package is greater than the average number of ORANGE Skittles in a package? Use the attached Skittles sample and test at the significance level of .01
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 Package Red Orange Yellow Green Purple Total Dominant Color 1 15 15 16 12 14 72 Y 2 13 6 11 16 15 61 G 3 9 9 10 18 15 61 G 4 7 9 8 18 15 57 G 5 13 9 14 10 7 53 Y 6 15 20 8 10 9 62 O 7 12 11 11 16 8 58 G 8 14 14 15 5 12 60 Y 9 14 14 15 5 12 60 Y 10 16 10 13 10 10 59 R 11 14 11 12 13 9 59 R 12 10 11 16 8 15 60 Y 13 7 15 12 12 13 59 O 14 7 15 12 12 13 59 O 15 18 10 10 9 13 60 R 16 10 9 12 15 9 55 G 17 8 13 11 6 21 59 P 18 12 14 12 10 11 59 O 19 12 8 11 15 12 58 G 20 14 5 13 14 13 59 R 21 16 10 11 15 10 62 R 22 10 9 9 9 21 58 P 23 15 6 12 12 12 57 R 24 10 12 17 9 8 56 Y 25 16 7 16 10 11 60 R 26 12 10 12 8 16 58 P 27 7 13 5 11 15 51 P 28 8 11 18 12 11 60 Y 29 14 9 10 11 13 57 R 30 12 12 12 7 14 57 P 31 16 7 10 7 13 53 R 32 13 14 13 7 12 59 O 33 10 9 9 9 21 58 P 34 13 7 14 14 10 58 Y 35 11 9 19 6 13 58 Y 36 10 11 14 12 10 57 Y 37 11 9 19 6 13 58 Y 38 8 13 15 14 9 59 Y 39 14 13 13 6 11 57 R 40 12 10 14 6 11 53 Y 41 8 11 15 14 12 60 Y

Here we use two sample t test to test :

Test results from MINITAB :

Two-Sample T-Test and CI: Purple, Orange

Two-sample T for Purple vs Orange

N Mean StDev SE Mean
Purple 41 12.49 3.26 0.51
Orange 41 10.73 3.02 0.47

Difference = mu (Purple) - mu (Orange)
Estimate for difference: 1.756
99% lower bound for difference: 0.110
T-Test of difference = 0 (vs >): T-Value = 2.53 P-Value = 0.007 DF = 80
Both use Pooled StDev = 3.1391

as p value = 0.007 < 0.01 , we reject the null hypothesis .

Or in other words , average number of purple skittles are greater than average number of orange skittles