Question

Bag	Blue	Orange	Green	Yellow	Red	Brown	Total Number of Candies
1	9	13	14	10	7	7	60
2	13	10	6	9	9	8	55
3	13	12	4	10	9	6	54
4	16	13	8	6	6	8	57
5	10	10	12	5	15	4	56
6	9	18	3	6	12	12	60
7	11	13	6	15	8	6	59
8	12	18	5	9	6	5	55
9	12	10	8	15	10	1	56
10	12	14	8	12	7	9	62
11	11	10	9	11	8	8	57
12	13	11	11	7	10	5	57
13	10	15	7	10	8	8	58
14	13	14	6	8	8	5	54
15	10	14	5	6	13	10	58
16	15	14	5	8	5	10	57
17	11	16	9	8	5	9	58
18	14	9	8	7	9	6	53
19	13	7	8	8	9	8	53
20	10	10	7	10	11	11	59
21	8	7	7	13	14	10	59
22	8	11	9	10	5	11	54
23	7	8	9	8	8	13	53
24	10	11	6	7	12	10	56
25	9	13	7	8	9	8	54
26	10	6	13	9	9	8	55
27	10	12	9	8	10	7	56
28	11	12	7	12	10	8	60
29	9	14	7	9	8	6	53
30	10	10	7	10	11	11	59

1. Mars Inc. indicates that the M&M Minis^®colors are distributed as follows:

Color               Proportion

Blue                        24%

Orange                    20%

Green                     16%

Yellow                   14%

Red                         13%

Brown                    13%

Conduct a hypothesis test (α=0.05) to see if the proportion of brownis the same as the company states.  Specify the hypothesis of the test, critical value, test statistic, p-value, decision (reject the null/ fail to reject the null), and the conclusion of what the decision means.

2. At the 0.01 level of significance, test if the true mean total number of candies is greater than 54.  Specify the hypotheses of the test, critical value, test statistic, p-value, decision (reject the null/ fail to reject the null), and the conclusion of what the decision means.

Answer 1

A)

Bag	Blue	Orange	Green	Yellow	Red	Brown	Total Number of Candies
1	9	13	14	10	7	7	60
2	13	10	6	9	9	8	55
3	13	12	4	10	9	6	54
4	16	13	8	6	6	8	57
5	10	10	12	5	15	4	56
6	9	18	3	6	12	12	60
7	11	13	6	15	8	6	59
8	12	18	5	9	6	5	55
9	12	10	8	15	10	1	56
10	12	14	8	12	7	9	62
11	11	10	9	11	8	8	57
12	13	11	11	7	10	5	57
13	10	15	7	10	8	8	58
14	13	14	6	8	8	5	54
15	10	14	5	6	13	10	58
16	15	14	5	8	5	10	57
17	11	16	9	8	5	9	58
18	14	9	8	7	9	6	53
19	13	7	8	8	9	8	53
20	10	10	7	10	11	11	59
21	8	7	7	13	14	10	59
22	8	11	9	10	5	11	54
23	7	8	9	8	8	13	53
24	10	11	6	7	12	10	56
25	9	13	7	8	9	8	54
26	10	6	13	9	9	8	55
27	10	12	9	8	10	7	56
28	11	12	7	12	10	8	60
29	9	14	7	9	8	6	53
30	10	10	7	10	11	11	59
Total	329	355	230	274	271	238	1697

Let P : Population proportion of brown candies.

N = total number of candies = 1697

X = number of brown candies = 238

p = sample proportion of brown candies= X/N = 238 /1697 = 0.1402.

We have to test the hypothesis that

The proportion of brown candies is 13%.

i.e. Null Hypothesis - Ho : P= 0.13

against

Alternative Hypothesis - Ha: P # 0.13 ( two-tailed test).

We used one sample proportion test.

The value of test statistic is

The value of test statistic is Z = 1.2494.

Alpha: Level of significance = 0.05

At 5% level of significance, critical values are -Z_alpha/2 and Z_alpha/2.

Critical values are -1.96 and 1.96

Since the test is two-tailed and value of test statistic is 1.2494, p-value is obtained by

p-value = 2* P ( Z > 1.2494)

From normal probability table

p-value = 2 * 0.1058

= 0.2116

Decision: Since p-value > alpha, we failed to reject Ho at 5% level of significance.

Conclusion : There is sufficient evidence support to claim that the proportion of brown candies is 13%.

B) Let X denotes the total number of candies.

mu : Mean total number of candies.

We have to test the hypothesis that

Mean total number of candies is greater than 54.

i.e Null Hypothesis -

against

Alternative Hypothesis - ( right-tailed test).

Since population variance is unknown, we used one sample t-test for testing population mean.

The value of test statistic is

Where

Total Number of Candies (X)	(Xi- Xbar)^2
60	11.7875
55	2.4545
54	6.5879
57	0.1877
56	0.3211
60	11.7875
59	5.9209
55	2.4545
56	0.3211
62	29.5207
57	0.1877
57	0.1877
58	2.0543
54	6.5879
58	2.0543
57	0.1877
58	2.0543
53	12.7213
53	12.7213
59	5.9209
59	5.9209
54	6.5879
53	12.7213
56	0.3211
54	6.5879
55	2.4545
56	0.3211
60	11.7875
53	12.7213
59	5.9209
1697	181.3667

Under Ho the value of test statistic is

The value of test statistic is t = 5.6215.

Alpha: level of significance = 0.01

Since the test is right-tailed, at 1% level of significance, critical value is

Value of test statistic is 5.6215 and test is right-tailed, p-value is obtained by

Decision: Since p-value is small as compared to level of significance alpha, we reject Ho at 1% level of significance.

Conclusion: There is sufficient evidence support to claim that the mean total number of candies is greater than 54.