Given the following sample information, test the hypothesis that the treatment means are equal at the .05 significance level.
Treatments 1, 2, 3:
3, 9, 6
2, 6, 3
5, 5, 5
1, 6, 5
3, 8, 5
1, 5, 4
-, 4, -
-, 7, -
-, 6, -
-, 4, -
a. State the null hypothesis and the alternate hypothesis.
b. What is the decision rule?
c. Compute SST, SSE, and SS total.
d. Complete an ANOVA table.
e. State your decision regarding the null hypothesis.
f. If H0 is rejected, can we conclude that treatment 1 and treatment 2 differ? Use the 95 percent level of confidence.
The solution gives step by step procedure for analysis of variance problem with calculation attached in Excel.
When we want to test two samples to determine if it is likely that the population means (estimated by the sample means) are different, we typically use a t-test. If the samples are large, we can also use a z-test. (Note that the formulas for computing s, t and/or z in the case of a two-sample test are different than the formulas for computing the same values in a one-sample test. Use Excel data analysis to conduct tests comparing two sample means.)
Using ANOVA (short for Analysis of Variance), however, we can test 3 or more sample means to determine if at least one of the sample means comes from a population with a mean that is significantly different from all of the others in the test. We actually do this by estimating a combined population variance two different ways and comparing the two estimates (the ratio of these two variance estimates follows the so-called "F distribution").
Why do we need a new test method to compare the means of 3 or more populations? Why can't we just use a series of z-tests or t-tests to compare all of the possible pairs of population means to see if one (or more) is different?
Most of the testing is to determine one or two things:
1. Is there a statistically significant difference between two or more population means? (based on comparison of 2 or more sample means)
2. Is there a statistically significant relationship between two or more variables? We can use regression analysis or chi-square tests to answer this second question.)View Full Posting Details