Question

# The following data were obtained for a randomized block design involving five treatments and three blocks:...

The following data were obtained for a randomized block design involving five treatments and three blocks: SST = 510, SSTR = 370, SSBL = 95. Set up the ANOVA table. (Round your value for F to two decimal places, and your p-value to three decimal places.)

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

Test for any significant differences. Use α = 0.05.

State the null and alternative hypotheses.

H0: Not all the population means are equal.
Ha: μ1 = μ2 = μ3 = μ4 = μ5H0: μ1 = μ2 = μ3 = μ4 = μ5
Ha: μ1μ2μ3μ4μ5    H0: At least two of the population means are equal.
Ha: At least two of the population means are different.H0: μ1μ2μ3μ4μ5
Ha: μ1 = μ2 = μ3 = μ4 = μ5H0: μ1 = μ2 = μ3 = μ4 = μ5
Ha: Not all the population means are equal.

Find the value of the test statistic. (Round your answer to two decimal places.)

p-value =

Do not reject H0. There is sufficient evidence to conclude that the means of the treatments are not all equal.

Reject H0. There is not sufficient evidence to conclude that the means of the treatments are not all equal.

Do not reject H0. There is not sufficient evidence to conclude that the means of the treatments are not all equal.

Reject H0. There is sufficient evidence to conclude that the means of the treatments are not all equal.

Solution;-

 Source of Variation Sum of squares Degree of Freedom Mean Square F p-value Treatments 370 4 92.5 16.444 0.001 Blocks 95 2 47.5 8.4444 0.011 Error 45 8 5.625 Total 510 14

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Ha: μ1 = μ2 = μ3 = μ4 = μ5
Ha: At least two of the population means are different.

Formulate an analysis plan. For this analysis, the significance level is 0.05.

Analyze sample data.

F statistics is given by:- F = 16.444

The P-value = 0.001

Interpret results. Since the P-value (0.001) is less than the significance level (0.05), we have to reject the null hypothesis.

Conclusion:-

Reject H0. There is sufficient evidence to conclude that the means of the treatments are not all equal.

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