Mary is takinga k day crash course in statistics. Therecord shows that she was a

Question

Mary is takinga k day crash course in statistics. Therecord shows that she was absnet on days numberd n1, adifaiosdf nr.If the instruction feels that B is the largest unbeliavlbeprobabiltiy, will he or she conclude that Mary was losing intrestmroe towrad the end of the course? Does this work also
k=5
m=3

n=2

Same approach as number 4.

x=1,3,5   xbar=3
y=2,4     ybar=3

xbar-ybar=0

Total = C (5 3) = 10

            x bar '                y bar'                     
1 3 5     3                2 4 3                       0
1 3 2     2                4 5        9/2               -2.5    
1 3 4     8/3             2 5        7/2               -0.83
3 5 2     10/3           1 4       5/2                0.83
3 5 4      12/3           12        3/2       2.5
2 4 1      7/3             3 5      8/2                 -1.67
2 4 3      9/3             1 5       3                    0
2 4 5      11/3           1 3       2                    1.67
5 1 2      8/3              3 4       7/2                 -0.83
5 1 4      10/3            2 3       5/2                  0.83

xbar'-ybar'> xbar-ybar (Six times)

6/10=0.6=p
Mary is takinga k day crash course in statistics. Therecord shows that she was absnet on days numberd n1, adifaiosdf nr.If the instruction feels that B is the largest unbeliavlbeprobabiltiy, will he or she conclude that Mary was losing intrestmroe towrad the end of the course? Does this work also
k=5
m=3

n=2

Same approach as number 4.

x=1,3,5   xbar=3
y=2,4     ybar=3

xbar-ybar=0

Total = C (5 3) = 10

            x bar '                y bar'                     
1 3 5     3                2 4 3                       0
1 3 2     2                4 5        9/2               -2.5    
1 3 4     8/3             2 5        7/2               -0.83
3 5 2     10/3           1 4       5/2                0.83
3 5 4      12/3           12        3/2       2.5
2 4 1      7/3             3 5      8/2                 -1.67
2 4 3      9/3             1 5       3                    0
2 4 5      11/3           1 3       2                    1.67
5 1 2      8/3              3 4       7/2                 -0.83
5 1 4      10/3            2 3       5/2                  0.83

xbar'-ybar'> xbar-ybar (Six times)

6/10=0.6=p

Explanation / Answer

I believe , that you areassuming that it is a (m+n) day course , and Mary was absent on 1 , 3and 5 th day , present on the others. Then it does work. This testing procedure is called WilcoxonRank Sum Test . After you get the probability , p . We nowcampare it with B . If p

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