Question

The following data are from a completely randomized design. Treatment Treatment Treatment A B C 32 44 34 30 43 37 30 44 36 26 46 37 32 48 41 Sample mean 30 45 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) For the CRD, we have various sum of squares defined as following

Sum of Squares, Treatment
Sum of Squares, Error
Mean Squares, Treatment
Mean Squares, Error

where is the jth observation of ith treatment, is the ith treatment mean and is the grand mean.

k= number of treatments, n= number of observation of each treatment and k-1 is degree of freedom(d.f.) of treatment sum of square and k(n-1) is d.f. of error sum of square.

using the values given we can compute,

Sum of Squares, Treatment	563.3
Sum of Squares, Error	66
Mean Squares, Treatment	281.7
Mean Squares, Error	5.5

The test statistic for testing the hypothesis that mean of three treatments is equal is given by,

Using the calculated values we have, F= 51.21, the tabulated F at 0.05 level of significance is = 3.89.

As tabulated F < F, we reject the null hypothesis.

Also the p value for above statistic is <0.05, which also suggests no strong evidence against the null hypothesis. Thus we can conclude that Mean of three treatments are different.

b) When there are equal no. of observations in each treatment Fisher's LSD between any two treatment is defined as

, here d if d.f. of Error, a is level of significance and MSE is Sum of Square, Error.

Here d= 12 and a=0.05 then two tailed t (12,0.05) = 2.18

Using above data LSD = 11.201

Difference (mean)	Absolute Value	Conclusion
(A-B) = -15	15	As AbsVal > LSD, we conclude there is significant diff between treatment A & B
(A-C) = -7	7	As AbsVal < LSD, we conclude there is no significant diff between treatment A & B
(B-C) = 8	8	As AbsVal < LSD, we conclude there is no significant diff between treatment A & B

c)