50c What is your conclusion if alpha = .05? Find the p-value or place bounds on

Question

50c What is your conclusion if alpha = .05? Find the p-value or place bounds on it. Consider a Mann-Whitney test with n_1 = n_2 = 6. 51a Assuming no ties, what is the smallest test statistic that could possibly be observed? Describe what this scenario would look like and justify your answer. 51b Assuming the null hypothesis is true and there will be no ties, what is the probability this smallest possible test statistic will be observed? A certain car part is manufactured in three different factories named A, B. and C. Randomly selected weeks of the year were randomly selected and the number of defective car parts during that week were recorded, n_1 = 5. n_2 = n_3 = 7 A: {2, 4, 7, 13, 15} B: {0, 0, 0, 2, 3, 3, 8} C: {0, 1, 2, 3, 4, 5, 11}

Explanation / Answer

Problem 51

(a) For a mann - whitney test where n1 = n2 = 6 , minimum test statistic would be when there are all minimum ranks belong to one sample group and all maximum ranks will be assigned tp second sample group.

That means sample 1 will have ranks ( 1,2,3,4,5,6) and the other sample will have ranks ( 7,8,9,10,11,12)

R1 = sum of all minimum ranks = 21

R2 = sumof all maximum ranks = 57

so U - value for the given ranks

U1 = R1 - n1 (n1 +1)/2 = 21- 21 = 0

U2 = R2 - n2 (n2 +1)/2 = 57 - 21 = 36

Minimum of two U will be the test statistic U which is 0

so smallest test statistic would be 0

(b) Here Null hypothesis is true and there are no ties, so

Probability of getting this test statisitc that means probability of getting U value = 0 is =?

mU = n1n2/2 = 6 * 6/2 = 18

U = sqrt [ n1n2(n1 + n2 + 1)/12] = sqrt [ 36 * 13/12] = 6.245

Z = ( U - mU )/U = ( 0-18)/ 6.245= - 2.88

so Probability of getting minimum test statistic = = 0.0040 ( for two tailed)

mU and U are the mean and standard deviation of U

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