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? Hint: use principles of probability to help your calculations: P(accurate AND accurate AND accurate))  (Write your answer accurate without rounding.)

Why would we use ANOVA instead of three separate tests?

Why would we want to use three separate tests instead of ANOVA?

Answer 1

Solution:

Dimension of centrality is the likelihood of making type 1 mistake in the test.

So, alpha=10% it implies that there will make a mistake in the test with likelihood of 0.10

In the mean time, it implies that there will get the precise outcome with a likelihood of 0.90.

(a)

On the off chance that we perform ANOVA test to think about the three methods.

At that point we are leading one single test for all the three methods and along these lines the likelihood of exact outcome is 0.90.

(b)

Now, instead of ANOVA we utilize three distinctive test for three pair wise correlation of mean then precise outcome will be delivered if all the three test are exact.

So, the likelihood for precision of the three test will turn into

Likelihood of exact result=0.90*0.90*0.90=0.729

P(accurate result)=0.729

(c)

Here we use ANOVA rather than numerous test when we simply need to see that if the methods for the gathering contrast or not:

Why ANOVA rather than 3 distinctive test.

we need to perform more test in the event that we don't utilize ANOVA.

Using the different test will expand the likelihood of making type I blunder.