Data analysis Part 10: Analysis of variance (ANOVA) test 

ANOVA test i.e. Analysis of variance test is used to compare the means among three or more groups.

In this test inferences about means are made by analyzing variance.

Let us take three sets of data as follows (Actual weight of paracetamol tablets that are marketed as 500mg)

Batch1	Batch2	Batch3
499.643	500.240	500.124
499.475	501.055	499.845
500.459	500.120	500.295
499.692	500.680	498.655
497.946	500.478	500.087
500.280	500.157	499.872
500.072	499.961	501.363
500.054	500.833	500.048
499.619	501.052	499.567
500.761	499.996	500.311
499.251	499.264	499.168
500.808	499.221	500.523
500.224	500.384	499.283
500.083	500.999	500.882
500.747	499.507	499.234
498.547	500.690	497.703
501.282	499.731	500.901
499.672	499.934	500.452
499.086	497.957	499.746
499.600	499.131	499.894
500.828	499.257	500.476
501.772	502.237	503.315
499.615	498.184	497.129
500.491	500.381	501.061
499.105	501.112	498.413

Results of ANOVA test is as follows

Anova: Single Factor

Summary

Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	0.403	2	0.202	0.192	0.825	3.124
Within Groups	75.413	72	1.047
Total	75.817	74

Groups	Count	Sum	Average	Variance
Batch1	25	12499.113	499.965	0.725
Batch2	25	12502.561	500.102	0.904
Batch3	25	12498.347	499.934	1.513

ANOVA

Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	0.403	2	0.202	0.192	0.825	3.124
Within Groups	75.413	72	1.047
Total	75.817	74

 

As per Fig.1 (ANOVA test, Single factor), it can be seen that F falls within the specified critical value (F critical).  

• Acceptance of Null hypothesis means wt. of tablet (population mean) from three batches (population mean) are equal (Batch 1 = Batch 2 = Batch 3).
  

• Rejection of Null hypothesis/ acceptance of alternative hypothesis means  wt. of tablets from three batches (population mean) are different. (Batch 1 ≠ Batch 2 ≠ Batch 3). So means of three population is not equal. At least one of the means is different. However, the ANOVA does not tell you where the difference lies. A t-Test has to be performed to test each pair of means.    
  

 In this case as per Fig.1., Null hypothesis is accepted. Hence it is conclusively proved that there is no difference between the weight of batch 1, batch 2 and batch 3 and mean weight of batch 1, batch 2 and 3 matches with each other.

Population mean of batch 1 = batch 2 = batch 3 (Statistically proven without individual's bias)

Reference

Hyperlinked in the text.