Mann Whitney U For comparison data
Using Mann Whitney U Non-parametric i.e. no assumptions are made about data fitting a normal distribution Is used to compare the medians of two sets of data It measures the overlap between the two data sets You must have between 6 and 20 replicates of data The data sets can have unequal numbers of replicates
Example of normally distributed data
When to use Mann-Whitney U-test Curve not normally distributed ie. non parametric Compares overlap between two data sets
The Equation U1U1 = n 1 x n 2 + ½ n 2 (n 2 + 1) - R 2 U2U2 = n 1 x n 2 + ½ n 1 (n 1 + 1) - R 1
The Equation U1U1 = n 1 x n 2 + ½ n 2 (n 2 + 1) - R 2 U2U2 = n 1 x n 2 + ½ n 1 (n 1 + 1) - R 1 Where: U1U1 =Mann - Whitney U for data set 1 n1n1 =Sample size of data set 1 R1 R1 =Sum of the ranks of data set 1 U2U2 =Mann - Whitney U for data set 2 n2n2 =Sample size of data set 2 R2 R2 =Sum of the ranks of data set 2
1. Establish the Null Hypothesis H 0 (this is always the negative form. i.e. there is no significant correlation between the variables) and the alternative hypothesis (H 1 ). Method H 0 - There is no significant difference between the variable at Site 1 and Site 2 H 1 - There is a significant difference between the variable at Site 1 and Site 2
2. Copy your data into the table below as variable x and variable y and label the data sets Rank 1 R 1 Data Set 1 Beech Hill (m) Data Set 2 Rushey Plain (m) Rank 2 R 2
Rank 1 R 1 Data Set 1 Beech Hill (m) Data Set 2 Rushey Plain (m) Rank 2 R 2 Start from the lowest and put the numbers in order: 16, 17, 18, 19, 20, 20, 21, 21, 22, 23, 23, 24, Treat both sets of data as one data set and rank them in increasing order (the lowest data value gets the lowest rank)
When you have data values of the same value, they must have the same rank. Take the ranks you would normally assign (5 and 6) and add them together (11) and divide the ranks between the data values(5.5) The lowest data value gets a rank of The same thing is done for all data values that are the same
When you have data values of the same value, they must have the same rank. Take the ranks you would normally assign (5 and 6) and add them together (11) and divide the ranks between the data values(5.5) The lowest data value gets a rank of The assigned ranks can then be put into the table
Rank 1 R Data Set 1 Beech Hill (m) Data Set 2 Rushey Plain (m) Rank 2 R Sum the ranks for each set of data ( R) R 1 = = 68 R 2 = = 23
5. Calculate the number of samples in each data set (n) Count the number of samples in each of the data sets Rank 1 R Data Set 1 Beech Hill (m) Data Set 2 Rushey Plain (m) Rank 2 R n 1 = 7 n 2 = 6
U1U1 = n 1 x n 2 + ½ n 2 (n 2 + 1) - R 2 U2U2 = n 1 x n 2 + ½ n 1 (n 1 + 1) - R 1 It is a good idea to break the equations down into three bite size chunks that will then give you a very easy three figure sum U1U1 =n 1 x n 2 +½ n 2 (n 2 + 1)- R 2 U2U2 =n 1 x n 2 +½ n 1 (n 1 + 1)- R 1 6. Calculate the Values for U 1 and U 2 using the equations
U1U1 = n 1 x n 2 + ½ n 2 (n 2 + 1) - R 2 (7x6) +3(6+1) -23 (7x6) +3(7) -23 U 1 = = 40 U 2 = = 2 (7x6) +3.5(8) -68 (7x6) +3.5(7+1) -68 U2U2 = n 1 x n 2 + ½ n 1 (n 1 + 1) - R 1
The smallest U value isU 2 = 2 6. Compare the smallest U value against the table of critical values
Value of n n1n (at the 0.05 or 95% confidence level i.e. we are 95% confident our data was not due to chance) We us the values of n 1 and n 2 to find our critical value
Is 2 (our smallest U value) smaller or larger than 6 (our critical value from the Mann Whitney Table)? Smaller The smallest U value is less than the critical value; therefore the null hypothesis is rejected The alternative Hypothesis can be accepted – There is a significant difference between the tree heights of Beech Hill and Rushey Plain
Use the following data to calculate U values independently Rank 1 R 1 Velocity cm.s -1 Pools Velocity cm.s -1 Riffles Rank 2 R 2 Rank 1 R 1 Abundance of Gammarus pulex Pools Abundance of Gammarus pulex Riffles Rank 2 R 2 Abundance of Gammarus pulex in pools and riffles of an Exmoor stream
Key questions Is there a significant relationship? Which data value/s would you consider to be anomalous and why? What graph would you use to present this data?