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# Mann-whitney U Test.... watch

1. I am getting somewhat confused.

so i have my 2 lots of data:

GROUP 1
14
13
9
11
12
7
19
9
21
19

GROUP 2
5
16
9
18
14
22
16
10
18
18

10 participants per group. This was looking at how many words can be recalled correctly, and was to do with categorization (psychology). Group 1's words were not categorized in any way, and Group 2's were categorized under headings that applied to them.
How do i go about applying the mann-whitney u test on this? I know its something to do with ranking, but i just cant really get my head around it....

and i assumed it belonged in maths because its a statistical test...
2. (Original post by Fandellos)
I am getting somewhat confused.

so i have my 2 lots of data:

GROUP 1
14
13
9
11
12
7
19
9
21
19

GROUP 2
5
16
9
18
14
22
16
10
18
18

10 participants per group. This was looking at how many words can be recalled correctly, and was to do with categorization (psychology). Group 1's words were not categorized in any way, and Group 2's were categorized under headings that applied to them.
How do i go about applying the mann-whitney u test on this? I know its something to do with ranking, but i just cant really get my head around it....

and i assumed it belonged in maths because its a statistical test...

First step: is to rank all of the scores from both groups togeather, with a rank of 1 being given to the smallest score, 2 to the second smallest and so on. For tied scores, the mean of the ranks involved are given e.g two scores tied for 7th and 8th both are given 7.5

2nd step: work out sum of ranks in the smaller sample or pick one if samples are the same. this value is known as T

3rd: Calculate U from fomula in which Na is one sample size and Nb is the other sample size

U= NaNb + (Na (Na+1) )
------------ - T
2
4th:

Calculate U' (U prime)

U'=NaNb - U

5th: Compare U and U' picking the smaller one.

Thn you will need to look up the relevant table of signinficane for the Mann-whitney test.

The observed value must be equal to, or smaller than the tabled value to be signifincant - then you can reject the nul hypothesis
3. I remember this. You poor, poor thing.
4. If you have a cardwell book for A2 psychology it explains everything. Its really good and it has the table in the back
5. ok, ive done the rankings, are they correct?:

Group 1 Ranks
14 = 7
13 = 6
9 = 2.5
11 = 4
12 = 5
7 = 1
19 = 8.5
9 = 2.5
21 = 10
19 = 8.5

Group 2 Ranks
5 = 1
16 = 5.5
9 = 2
18 = 8
14 = 4
22 = 10
16 = 5.5
10 = 3
18 = 8
18 = 8
6. Bump.

1st) Ok im pretty sure the rankings ive done are correct (see above)
2nd) Now the sum of ranks for group 1 = 55, and group 2 = 55, so there is no smaller one so i will use 55 as T, right?

3rd) so using the formula:

U = (NaNb + (Na(Na+1))/2) - T

U = (10x10 + (10x(10+1))/2) - 55

U = (100 + (110/2)) - 55

therefore U = 100

am i correct up to here?

4th) using U'=NaNb - U i get:

U' = 100 - 100

which equals 0.

5th) 0 is the smaller value so i will use it instead of 100.

it seems a little low but that's what i have calculated....

BUT i cant find any table of critical values!!! HELP!!

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Updated: November 9, 2008
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