Question

In a completely randomized design,12 experimental units were used for the first treatment, 15 for thesecond treatment, and 20 for the third treatment. Complete thefollowing analysis of variance (to 2 decimals, if necessary).

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatments	1200
Error
Total	1600

At a .05 level of significance, isthere a significant difference between the treatments?

P -value is?

Less than 0.1

Between 0.1 and 0.25

Between 0.25 and 0.05

Between 0.05 and 0.10

Greater than 0.10

What is yourconclusion?

Conclude not all treatmentmeans are equal

OR

Cannot reject the assumptionall treatment means are equal

Answer 1

Concepts and reason

Analysis of variance is a technique to test the means of the more than two groups. It is a collection of statistical models used to analyse the differences among group means and their associated procedures developed by statistician and evolutionary biologist Ronald Fisher.

Fundamentals

The formula for the error sum of squares is as follows:

The formula of the degrees of freedom for treatment sum of squares is as follows:

The formula of the degrees of freedom for error sum of squares is as follows:

$df.=n-k $

The formula of degrees of freedom for total sum of squares is as follows:

The formula of the mean sum of squares for the treatments is as follows:

The formula for the mean sum of squares for the error is as follows:

The formula of the F statistic is as follows:

Calculate the error sum of squares.

Calculate the degrees of freedom for treatment sum of squares.

Calculate the degrees of freedom for error sum of squares.

Calculate the degrees of freedom for total sum of squares.

Calculate the mean sum of squares for treatments.

Calculate the mean sum of squares for error.

Calculate the F statistic as follows:

Calculate the p-value.

The p-value is 0.05

Consider Null and Alternative hypothesis.

Null hypothesis, all treatment means are equal.

Alternative hypothesis, all treatment means are not equal.

Hence, conclude that all treatment means are not equal

Ans:

The ANOVA table is as follows:

The p-value is less than 0.10

All treatment means are not equal.