Question

In a completely randomized design, seven experimental units were used for each of the five levels of the factor. Complete the following ANOVA table (to 2 decimals, if necessary). If your answer is zero enter "0".

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
Treatments
Error
Total

a. What hypotheses are implied in this problem

: - Select your answer -Not all five treatment means are equalAll five treatment means are equalItem 9

: - Select your answer -Not all five treatment means are equalAll five treatment means are equalItem 10

b. At the  level of significance, can we reject the null hypothesis in part (a)?

Calculate the value of the test statistic (to 2 decimals).

The -value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 12

What is your conclusion?

Answer 1

As no data is given in the question except the sample size, so the "Sum of Squares" taken in this question is for reference purpose only.

Part A:

The filled table is given below:

SSE = SST - SSTR = 1006 - 345 = 661

Degree of Freedom (error) = n -1 = 35 - 1 = 34 (total members per sample - 1)

Degree of Freedom (treatment) = k -1 = 5 - 1 = 4 (total groups - 1)

MSE = SSE/df

MSTR = SSTR/df

F = MSTR/MSE

The F table is given below:

So, the P value is between Less than 0.01

Exact P value from Excel is 0.00541

As the P vale is less than 0.05, so the null hypothesis is not accepted which is explained in Part B.

Part B:

Here, the hypothesis are:

H₀: U₁ = U₂ = U₃ (Null Hypothesis)

H_a: All the means are not equal (Alternate Hypothesis)

U is the mean...

As the P value is less than 0.05, it can be stated that there occurs a significant difference in the means of all the five groups. In this case the null hypothesis is not accepted that means the means of all the five groups are not equal.

End of the Solution...