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Find the p-value for a one-way ANOVA with a test statistic f = 6.89, where the degrees of freedom for the error is 76 and the degrees of freedom for the groups is 5?

Answer :

The p-value is calculated using the F-distribution cumulative distribution function: p-value = 1 - F(6.89; 5, 76) ≈ 0.0000395. This small p-value indicates strong evidence against the null hypothesis, suggesting significant differences between the group means.

To calculate the p-value, we can use the F-distribution cumulative distribution function (CDF) denoted as F(x; ν1, ν2), where:

- x is the test statistic (F = 6.89)

- ν1 is the degrees of freedom for the groups (5)

- ν2 is the degrees of freedom for the error (76)

The p-value is the probability of observing a test statistic at least as extreme as the one we have, assuming that the null hypothesis is true. This is represented by the tail probability:

p-value = 1 - F(6.89; 5, 76)

Using the F-distribution CDF formula:

F(x; ν1, ν2) = I(x; ν1, ν2) / B(ν1/2, ν2/2)

where:

- I(x; ν1, ν2) is the incomplete beta function

- B(ν1/2, ν2/2) is the beta function

We can calculate the p-value as:

p-value = 1 - I(6.89; 5, 76) / B(5/2, 76/2)

Using numerical integration or approximation methods, we get:

p-value ≈ 0.0000395

This is the probability of observing an F-statistic at least as extreme as 6.89, assuming that the group means are equal (null hypothesis). Since this probability is very small, we reject the null hypothesis and conclude that there is a significant difference between the group means.

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