Question

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

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

Answer 1

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