Question

The following data are from a completely randomized design. Treatment Treatment Treatment A B C 32 47 34 30 46 37 30 47 36 26 49 37 32 51 41 Sample mean 30 48 37 Sample variance 6 4 6.5 At the = .05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean Squares, Treatment Mean Squares, Error Calculate the value of the test statistic (to 2 decimals). The p-value is Selectless than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 6 What is your conclusion? SelectConclude that not all treatment means are equalDo not reject the assumption that the treatment means are equalItem 7 Calculate the value of Fisher's LSD (to 2 decimals). Use Fisher's LSD procedure to test whether there is a significant difference between the means for treatments A and B, treatments A and C, and treatments B and C. Use = .05. Difference Absolute Value Conclusion A - B SelectSignificant differenceNo significant differenceItem 10 A - C SelectSignificant differenceNo significant differenceItem 12 B - C SelectSignificant differenceNo significant differenceItem 14 Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the difference between the means of treatments A and B (to 2 decimals). Since treatment B has the larger mean, the confidence interval for the difference between the means of treatments A and B ( A - B) should be reported with negative values. ( , )

Answer 1

a. The below table shows all the formula that are required to solve the question. Source of Variation Analysis of Variance(ANOVA) Sum of Squares Degrees of Mean Squares (MS) Freedom df = k-1 SSW = 2(X-X) Within MS: ss. Écr-.) 4-2-1 145 - MS MSW Between Total df = n-1 ss.=, -23° The null and alternative hypotheses are: Null hypothesis: All the treatment means are same. Alternative hypothesis: At least one of the treatment mean is different. The analysis is done using Excel. Step by step procedure is provided below: 1. Enter the data in the spreadsheet. 2. Go to Data>Data Analysis>select Anova: Single factor and click ok. 3. Fill the options as required and click ok. 4. The output is obtained as:

Anova: Single Factor SUMMARY Groups Treatment A Treatment B Treatment C Count 5 5 5 Sum Average Variance 150 30 240 48 4 185 37 6.5 df ANOVA Source of Variation Between Groups Within Groups SS 823.3333 66 MS F P-value F crit 2 411.6667 74.84848 1.67E-07 3.885294 12 5.5 Total 889.3333 14 The required values are: Sum of Squares. Treatment = 823.33 Sum of Squares. Error = 66 Mean Squares. Treatment = 411.66 Mean Squares, Error = 5.5 The value of the test statistic is Fcalculated = 74.84. The p-value is almost zero which is less than 0.01 and also less than the level of significance, 0.05. Therefore, the null hypothesis is rejected. So, it can be concluded that not all treatment means are equal.

6. The value of Fischer's LSD is. Fischer's LSD = 24. fasean = 3.23 Difference Absolute Value A-B L 18 A- C7 Conclusion Difference is significant Difference is significant Difference is significant c. The required 95% confidence interval estimate is, Confidence Interval = Difference in Means + Fischer's LSD value = (30 - 48)+ 3.23 = - 18 = 3.23 =(-21.23, -14.77)