F-TEST (FRIEDMAN’S TEST)
 Friedman's Test is a nonparametric hypothesis test, meaning that it is used to
study populations with rankings but no clear numerical interpretation.
 It is used when no distributional assumptions are necessary.
 This is used for three or more groups of data and the variables measured are
ordinally scaled.
 It is used to test the null hypothesis that the samples have been drawn from the
same population.
To use the F-test or Friedman’s test:
 State the null hypothesis (samples come from populations with equal
distributions) and the alternative hypothesis (samples come from populations
with different distributions).
 State the level of significance. This is the error probability.
 Set up a table based on the data. The rows represent the number of samples
(the blocks), while the columns represent the sample treatments.
 In the table, rank the data within the blocks, giving the lowest value rank 1 and
tied values the average of the two ranks.
 Get the sum of the ranks per treatment.
 Compute the Friedman test statistic, which is obtained through either formula:
S = ∑ Ri2
– R2
/n
where Ri2 is the sum of the squares of the rank sums, R2 is the square of the
total rank sum and n is the number of treatments.
Fr = 12/bt(t+1) * ∑Tj
2
– 3b(t + 1)
 where b is the number of blocks, t is the number of treatments, and ∑Tj
2
is the
sum of the squares of the rank sums.
 Test the null hypothesis by comparing the test statistic with the chi-square
critical value from the χ2
distribution table (if you used the first formula, find the
equivalent in the Friedman statistics table).
 To find the chi-square critical value mentioned above, subtract 1 from the
number of treatments to get the number of degrees of freedom. Look up the chi-
square value equivalent to the degrees of freedom and the significance level. Do
this is you calculated the test statistic with the second formula.
 If Fr is greater than or equal to the critical chi-square value/Friedman statistics
table value found, the null hypothesis is dropped.
EXAMPLE:
A nutritionist compared three treatments of squash-malunggay-banana baby food
preparations with the ingredients in different proportions. The three treatments
were subjected to a sensory evaluation using four evaluators. The treatments were
rated 1 to 4, 4 being the highest and best rating. Assume significance is 5%.
 State the hypotheses: Ho = samples are from same-distribution populations, Ha
= samples come from different-distribution populations.
 Create a table first, then list down all the data. Then rank the data across blocks
(in black) and find the sum (in red).
BlockTreatment Treatment
1
Treatment
2
Treatment
3
Block 1 3 (1.5) 4 (3) 3 (1.5)
Block 2 4 (3) 2 (1.5) 2 (1.5)
Block 3 3 (1.5) 4 (3) 3 (1.5)
Block 4 4 (3) 3 (2) 2 (1)
∑Rank 9 9.5 5.5
 Compute for the Friedman test statistic.
Fr = 12/bt(t+1) * ∑Tj
2
– 3b(t + 1)
Fr = 12/(4)(3)(3+1) * ∑Tj
2
– 12(3 + 1)
Fr = 12/(4)(3)(3+1) * ∑Tj
2
– 12(3 + 1)
Fr = 12/(4)(3)(3+1) * (92
+ 9.52
+ 5.52
) – 12(3 + 1)
Fr = 2.375
 Next, find the degrees of freedom.
DOF = treatments – 1 = 3 – 1 = 2
 Look for the chi-square value with DOF 2 and significance level 0.05 (5%). This is
5.991.
 Compare the Fr and the chi-square value.
Since Fr < 5.991, the null hypothesis is then true.
OR
 After constructing the table, use the formula S = ∑ Ri2
– R2
/n.
S = ∑ Ri2
– R2
/n
S = (92
+ 9.52
+ 5.52
) – (24) 2
/3
S = 9.5
 Here, use n instead of degrees of freedom. , and significance is 0.05.
 Look up the value of S5,3 and compare its value to S.
 Since the calculated S-value is less than S5,3 (S5,3 = 37), then the null hypothesis is
correct.
HOW TO PERFORM A FRIEDMAN TEST IN EXCEL
Note: You will need to install the Analyse-it toolbar for Microsoft Excel 2003/2007. The
instructions shown here are for Microsoft Excel 2003. For help on using the toolbar on
Excel 2007, refer to the documentation at: http://www.analyse-
it.com/support/documentation/220/Welcome.htm
Note: The Analyse-it toolbar has a limited trial of 30 days.
Given this data set:
 To perform the Friedman test, select the Analyse-it toolbar and drag the cursor
to “Compare Pairs”. Then select “Friedman”. (NOTE: There must be a variable
name across the blocks.)
 Select everything when the window pops out by clicking on the green figures.
 It will display a set of results for Friedman test.

More Related Content

PDF
Geometric Series and Finding the Sum of Finite Geometric Sequence
PPTX
Computing the Variance of a Discrete Probability Distribution.pptx
PPTX
RANDOM SAMPLING, PARAMETER AND STATISTICS, SAMPLING DISTRIBUTION OF STATISTIC...
PPTX
grade 11 Statistics and probability Week 4.pptx
PPTX
Mean of grouped data
PDF
Anthropology of disaster
PPSX
Measures of central tendency ict integration
PPTX
4-Computing-the-Variance-of-Discrete-Probability-Distribution.pptx
Geometric Series and Finding the Sum of Finite Geometric Sequence
Computing the Variance of a Discrete Probability Distribution.pptx
RANDOM SAMPLING, PARAMETER AND STATISTICS, SAMPLING DISTRIBUTION OF STATISTIC...
grade 11 Statistics and probability Week 4.pptx
Mean of grouped data
Anthropology of disaster
Measures of central tendency ict integration
4-Computing-the-Variance-of-Discrete-Probability-Distribution.pptx

Viewers also liked (13)

PPT
F Distribution
PPTX
Anova (f test) and mean differentiation
PPTX
Z-Test and Standard error
PPT
F test Analysis of Variance (ANOVA)
PPTX
Anova ppt
PPT
Data Analysis Using Spss T Test
PDF
Analysis of Variance (ANOVA)
PPTX
Analysis of variance (ANOVA)
PDF
Student's T-test, Paired T-Test, ANOVA & Proportionate Test
PPTX
Student t-test
PPTX
Chi square test
PDF
Hypothesis testing; z test, t-test. f-test
PPTX
Reporting Chi Square Test of Independence in APA
F Distribution
Anova (f test) and mean differentiation
Z-Test and Standard error
F test Analysis of Variance (ANOVA)
Anova ppt
Data Analysis Using Spss T Test
Analysis of Variance (ANOVA)
Analysis of variance (ANOVA)
Student's T-test, Paired T-Test, ANOVA & Proportionate Test
Student t-test
Chi square test
Hypothesis testing; z test, t-test. f-test
Reporting Chi Square Test of Independence in APA
Ad

Similar to Str f-test (20)

PDF
THE FRIEDMAN TEST ( Biostatics B. Pharm)
PPTX
Friedman's 2way ANOVA.pptx
PPTX
Friedman FR TEST FOR RANDOMIZED BLOCK DESIGN.pptx
PPTX
Non parametric test- Muskan (M.Pharm-3rd semester)
PPTX
Presentation non parametric
PDF
Non Parametric Test by Vikramjit Singh
PPTX
Statistics-Non parametric test
PPTX
biostat__final_ppt_unit_3.pptx
PPSX
Dalut ppt. of factorial analysis of variance-b
PPSX
Dalut ppt. of factorial analysis of variance-b
PPSX
Dalut ppt. of factorial analysis of variance-b
PPTX
Chi square test
PPTX
PPT
9-NON PARAMETRIC TEST in public health .ppt
PPTX
All non parametric test
PPTX
All non parametric test
PPTX
Chisquare Test
PPTX
Tests of Significance.pptx powerpoint presentation
PPTX
chi-Square. test-
THE FRIEDMAN TEST ( Biostatics B. Pharm)
Friedman's 2way ANOVA.pptx
Friedman FR TEST FOR RANDOMIZED BLOCK DESIGN.pptx
Non parametric test- Muskan (M.Pharm-3rd semester)
Presentation non parametric
Non Parametric Test by Vikramjit Singh
Statistics-Non parametric test
biostat__final_ppt_unit_3.pptx
Dalut ppt. of factorial analysis of variance-b
Dalut ppt. of factorial analysis of variance-b
Dalut ppt. of factorial analysis of variance-b
Chi square test
9-NON PARAMETRIC TEST in public health .ppt
All non parametric test
All non parametric test
Chisquare Test
Tests of Significance.pptx powerpoint presentation
chi-Square. test-
Ad

More from iamkim (20)

PDF
Nat sci minerals part1
PDF
Nat Sci - Minerals
DOCX
Batch 2012 schedule of exit interview
DOCX
College test results b2012
PDF
Chem cations
PDF
Chem anions
PDF
Grad ball collections per section(01 28-12)
DOCX
Congratulations to batch 2012 star scholar candidates
DOC
Retreat consent form
DOC
Retreat agreements
PDF
Fil la loba negra
PDF
Fil fray botod
PDF
Fil 3 q readings
DOC
Dasalan at tocsohan
PDF
Chem ps electrolysis
PDF
Physics waves
PDF
Math 3 hw ps2
DOCX
Memo circular # 4 dtd nov 4, 2011
DOC
Final creative shots hair & makeup evaluation (110211)
DOC
Creative shots hair & makeup evaluation (110211)
Nat sci minerals part1
Nat Sci - Minerals
Batch 2012 schedule of exit interview
College test results b2012
Chem cations
Chem anions
Grad ball collections per section(01 28-12)
Congratulations to batch 2012 star scholar candidates
Retreat consent form
Retreat agreements
Fil la loba negra
Fil fray botod
Fil 3 q readings
Dasalan at tocsohan
Chem ps electrolysis
Physics waves
Math 3 hw ps2
Memo circular # 4 dtd nov 4, 2011
Final creative shots hair & makeup evaluation (110211)
Creative shots hair & makeup evaluation (110211)

Str f-test

  • 1. F-TEST (FRIEDMAN’S TEST)  Friedman's Test is a nonparametric hypothesis test, meaning that it is used to study populations with rankings but no clear numerical interpretation.  It is used when no distributional assumptions are necessary.  This is used for three or more groups of data and the variables measured are ordinally scaled.  It is used to test the null hypothesis that the samples have been drawn from the same population. To use the F-test or Friedman’s test:  State the null hypothesis (samples come from populations with equal distributions) and the alternative hypothesis (samples come from populations with different distributions).  State the level of significance. This is the error probability.  Set up a table based on the data. The rows represent the number of samples (the blocks), while the columns represent the sample treatments.  In the table, rank the data within the blocks, giving the lowest value rank 1 and tied values the average of the two ranks.  Get the sum of the ranks per treatment.  Compute the Friedman test statistic, which is obtained through either formula: S = ∑ Ri2 – R2 /n where Ri2 is the sum of the squares of the rank sums, R2 is the square of the total rank sum and n is the number of treatments. Fr = 12/bt(t+1) * ∑Tj 2 – 3b(t + 1)  where b is the number of blocks, t is the number of treatments, and ∑Tj 2 is the sum of the squares of the rank sums.  Test the null hypothesis by comparing the test statistic with the chi-square critical value from the χ2 distribution table (if you used the first formula, find the equivalent in the Friedman statistics table).  To find the chi-square critical value mentioned above, subtract 1 from the number of treatments to get the number of degrees of freedom. Look up the chi- square value equivalent to the degrees of freedom and the significance level. Do this is you calculated the test statistic with the second formula.
  • 2.  If Fr is greater than or equal to the critical chi-square value/Friedman statistics table value found, the null hypothesis is dropped. EXAMPLE: A nutritionist compared three treatments of squash-malunggay-banana baby food preparations with the ingredients in different proportions. The three treatments were subjected to a sensory evaluation using four evaluators. The treatments were rated 1 to 4, 4 being the highest and best rating. Assume significance is 5%.  State the hypotheses: Ho = samples are from same-distribution populations, Ha = samples come from different-distribution populations.  Create a table first, then list down all the data. Then rank the data across blocks (in black) and find the sum (in red). BlockTreatment Treatment 1 Treatment 2 Treatment 3 Block 1 3 (1.5) 4 (3) 3 (1.5) Block 2 4 (3) 2 (1.5) 2 (1.5) Block 3 3 (1.5) 4 (3) 3 (1.5) Block 4 4 (3) 3 (2) 2 (1) ∑Rank 9 9.5 5.5  Compute for the Friedman test statistic. Fr = 12/bt(t+1) * ∑Tj 2 – 3b(t + 1) Fr = 12/(4)(3)(3+1) * ∑Tj 2 – 12(3 + 1) Fr = 12/(4)(3)(3+1) * ∑Tj 2 – 12(3 + 1) Fr = 12/(4)(3)(3+1) * (92 + 9.52 + 5.52 ) – 12(3 + 1) Fr = 2.375  Next, find the degrees of freedom. DOF = treatments – 1 = 3 – 1 = 2  Look for the chi-square value with DOF 2 and significance level 0.05 (5%). This is 5.991.  Compare the Fr and the chi-square value. Since Fr < 5.991, the null hypothesis is then true. OR  After constructing the table, use the formula S = ∑ Ri2 – R2 /n.
  • 3. S = ∑ Ri2 – R2 /n S = (92 + 9.52 + 5.52 ) – (24) 2 /3 S = 9.5  Here, use n instead of degrees of freedom. , and significance is 0.05.  Look up the value of S5,3 and compare its value to S.  Since the calculated S-value is less than S5,3 (S5,3 = 37), then the null hypothesis is correct. HOW TO PERFORM A FRIEDMAN TEST IN EXCEL Note: You will need to install the Analyse-it toolbar for Microsoft Excel 2003/2007. The instructions shown here are for Microsoft Excel 2003. For help on using the toolbar on Excel 2007, refer to the documentation at: http://www.analyse- it.com/support/documentation/220/Welcome.htm Note: The Analyse-it toolbar has a limited trial of 30 days. Given this data set:  To perform the Friedman test, select the Analyse-it toolbar and drag the cursor to “Compare Pairs”. Then select “Friedman”. (NOTE: There must be a variable name across the blocks.)  Select everything when the window pops out by clicking on the green figures.
  • 4.  It will display a set of results for Friedman test.