Statistic Question?

Question

Another important difference is that while t-test can be one tailed or two tailed, ANOVA does not have that distinction. Why do you think that is the case?

Answer 1

Both methods are decided by the α-risk that is stated before the tests are analyzed. The calculated values of the z-test or t-test can either be positive or negative and, therefore, the left tail (negative values) or the right tail (positive values) or both contain the critical region(s). But the calculated ANOVA F-value is the ratio of 2 variances and variances can only be positive, so the critical region is always in right tail (positive values) of the F distribution.

Answer 2

The t -test is used for determining a difference between two groups. this give use the freedom to look at a hypothesis test for equality or an inequality.

the ANOVA is used to test for the equality of multiple means, i.e.,

H0: μ1 = μ2 = μ3 = ... = μn
the alternate hypothesis is that at least one of the means is different from the others.

testing to see if one or more of the means is larger than the others would be overly complicated.

if you want to test for the pairs of mean being larger than others then you can conduct multiple tests.

let's assume that you have four hospitals. Let's also assume you have parametric data. if you do a t/z - test or the Mann-Whitney then you have to conduct 6 different hypothesis tests.

H0: μ1 = μ2
H0: μ1 = μ3
H0: μ1 = μ4
H0: μ2 = μ3
H0: μ2 = μ4
H0: μ3 = μ4

if you are testing at the α level then the probability of a Type I error is bounded between 0 and α.

The probability of not committing a Type I error is (1 - α)

for 6 tests the probability of not committing a Type I error is (1 - α) ^6. For example, testing at the 5% level means the probability of not committing a Type I error is 0.7350919. The probability of committing a Type I error is 0.2649081. This shows that for finding a difference between the means of four groups you have a probability of being wrong going from 5% to over 26%.


testing the inequalities have the same set up. using the Bonferroni correction or Fisher's LSD will help to have a multiple tests with the correct significance level.

Answer 3

I don't know how important this is. Either you do an ANOVA analysis or you don't. t-tests are limited to 2 samples or populations.

With ANOVAs, you can deal with multiple samples. However, you can't apportion the variation due to samples to do one-tailed or two-tailed tests, since you don't know a priori that each sample of combination of samples represent statistically different populations.