Haemoglobin and sickle-cell type

Topic: Analysis of variance examples · more questions on this topic
Five treatments, ten pairs ›

Three groups have sizes 16, 15 and 10. The treatment sum of squares is 99.89 and the error mean square is 1.00.

1. State the null and alternative hypotheses for a one-factor analysis of variance.

2. Complete the degrees of freedom and the F statistic.

3. The p-value is about 2.3\times 10^{-11}. What is the conclusion at level 0.05, and which assumptions does the F test use?

Worked answer (click to show)

H_0 is equal mean haemoglobin in the three disease types. H_1 is that the means are not all equal. The degrees of freedom are 3-1=2 and 41-3=38. The F statistic is

F=\frac{99.89/2}{1.00}=49.95.

The p-value is about 2.3\times 10^{-11}, so the means differ at level 0.05. The F test assumes independent normal observations with a common variance.