Please answer question 28. F table included if needed. CHAPTER 10 Comparisons In

Question

Please answer question 28. F table included if needed. CHAPTER 10 Comparisons Involving Means, Experimental Design, and Analysis of Variance 10.4 AN INTRODUCTION TO CONTENTS EXPERIMENTAL STATISTICS IN PRACTICE DESIGN AND ANALYSIS U.S. FOOD AND DRUG OF VARIANCE ADMINISTRATION Data Collection 10.1 INFERENCES ABOUT THE Assumptions for Analysis of DIFFERENCE BETWEEN Variance TWO POPULATION MEANS: Analysis of Variance: AConcep. AND O, KNOWN ual Overview Interval Estimation of u u2 10.5 ANALYSIS OF VARIANCE Hypothesis Tests About Au2 AND THE COMPLETELY Practical Advice RANDOMIZED DESIGN Between-Treatments Estimate of 10.2 INFERENCES ABOUT THE DIFFERENCE BETWEEN Population Variance TWO POPULATION MEANS: Within-Treatments Estimate of 0, AND o, UNKNOWN Population Variance Comparing the Variance Interval Estimation of Au1-u2 Estimates: The FTest Hypothesis Tests About u u2 Table Computer Practical Advice Results for Analysis of 10.3 INFERENCES ABOUT THE Variance DIFFERENCE BETWEEN Testing for the Equality of k TWO POPULATION MEANS: Population Means: An MATCHED SAMPLES Observational Study

Explanation / Answer

Given,

SSTR = 300

SST = 460

So,

SSE = SST - SSTR

SSE = 460 - 300

SSE = 160

There are 5 level of factors (k) and 7 experimental units for each factor. So,

Total number of elements (N) = 7x5 = 35

Treatment degrees of freedom = k - 1 = 5 - 1 = 4

Error degrees of freedom = N - k = 35 - 5 = 30

Total degrees of freedom = N - 1 = 35 - 1 = 34

Also,

Mean square for treatment (MSTR) = SSTR / (k - 1) = 300 / 4 = 7.5

Mean square for error (MSE) = SSE / (N - k) = 160 / 30 = 5.33

F = MSTR / MSE = 75 / 5.33 = 14.06

p - value = 0.000

Hence,

ANOVA table will be:

Sum of variations Sum of squares Degrees of freedom Mean Square F p - value Treatment 300 4 75 14.06 0.000 Error 160 30 5.33 - - Total 460 34 - - -

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