# How to do a one-way analysis of variance (ANOVA) (in Julia)

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If we have multiple independent samples of the same quantity (such as students’ SAT scores from several different schools), we may want to test whether the means of each of the samples are the same. Analysis of Variance (ANOVA) can determine whether any two of the sample means differ significantly. How can we do an ANOVA?

## Solution

Let’s assume we have our samples in several different Julia arrays. Here I’ll construct some made-up data about SAT scores at four different schools.

 1 2 3 4 school1_SATs = [ 1100, 1250, 1390, 970, 1510 ]; school2_SATs = [ 1010, 1050, 1090, 1110 ]; school3_SATs = [ 900, 1550, 1300, 1270, 1210 ]; school4_SATs = [ 900, 850, 1110, 1070, 910, 920 ];

ANOVA tests the null hypothesis that all group means are equal. You choose $\alpha$, the probability of Type I error (false positive, finding we should reject $H_0$ when it’s actually true). I will use $\alpha=0.05$ in this example.

 1 2 3 4 5 using HypothesisTests alpha = 0.05 p_value = pvalue( OneWayANOVATest( school1_SATs, school2_SATs, school3_SATs, school4_SATs ) ) reject_H0 = p_value < alpha alpha, p_value, reject_H0
 1 (0.05, 0.03405326535040251, true)

The result we see above is to reject $H_0$, and therefore conclude that at least one pair of means is statistically significantly different.

If you are using the most common $\alpha$ value of $0.05$, you can save a few lines of code and get a more detailed printout by just printing out the test itself:

 1 OneWayANOVATest( school1_SATs, school2_SATs, school3_SATs, school4_SATs )
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 One-way analysis of variance (ANOVA) test ----------------------------------------- Population details: parameter of interest: Means value under h_0: "all equal" point estimate: NaN Test summary: outcome with 95% confidence: reject h_0 p-value: 0.0341 Details: number of observations: [5, 4, 5, 6] F statistic: 3.69513 degrees of freedom: (3, 16)

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Contributed by Nathan Carter (ncarter@bentley.edu)