Question

Suppose we want to test whether or not three means are equal. We want to perform this test with a 10% significance level.

If we perform an ANOVA test, what is the probability of the test producing accurate results (avoiding a Type I error)?

Suppose we, instead, run three separate hypothesis tests (t-tests), each with 10% significance level.

• Mean 1 = Mean 2
• Mean 1 = Mean 3
• Mean 2 = Mean 3

What is the probability that all three tests would be accurate? P(accurate AND accurate AND accurate)  (Write your answer accurate without rounding.)

Why would we use ANOVA instead of three separate tests?

Answer 1

(a)

If we perform an ANOVA test, what is the probability of the test producing accurate results (avoiding a Type I error) =

1 - = 1 - 0.10 =.0.90

Answer is:

0.90

(b)

the probability that all three tests would be accurate? P(accurate AND accurate AND accurate) =

(1- 0.1)³ = 0.9³ = 0.729

Answer is:

0.729

(c)

We would use ANOVA instead of three separate tests because in the case of three separate tests, the probability of Type I Error is additive and thus, the probability that all three tests would be accurate is considerbly reduced, whereas in the case of ANOVA controls the Type I Error sothat the Type I Error remains at 10%.