13.3

Q1

The following data are from a completely randomized design. Treatment Treatment Treatment A B C 32 44 33 30 43 36 30 44 35 26 46 36 32 48 40 Sample mean 30 45 36 Sample variance 6.00 4.00 6.50 a. At the o = 0.05 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 66.0 Mean Squares, Treatment Mean Squares, Error 5.5Calculate the value of the test statistic [to 2 decimals}. Mama9 What is your conclusion? Conclude that not all treatment means are equal V 9 h. Calculate the value of Fisher's LSD {to 2 decimals). 3.23 9 Use Fisher's LSD procedure to test I.vhether there is a signicant difference between the means for treatments A and B, treatments A and C, and treatments B and C. Use a = 0.05. Difference Absolute Value Conclusion its 9 9 are. 9 9 are. 9 9 c. Use Fisher's LSD procedure to develop a 95% condence interval estimate of the difference between the means of treatments A and B {to 2 decimals). Enter negative values as negative numbers. .: 9. 9: The following data are from a completely randomized design. Treatment Treatment Treatment A B C 32 46 34 30 45 37 30 46 36 26 48 37 32 50 41 Sample mean 30 47 37 Sample variance 6.00 4.00 6.50 a. At the o = 0.05 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 730 Sum of Squares, Error 66.0 Mean Squares, Treatment 365 Mean Squares, Error 5.5Calculate the value of the test statistic [to 2 decimals). 66.36 (g WWWco What is your conclusion? Conclude that not all h-eatrnenl means are equal V 9 I3. Calculate the value of Fisher's LSD {to 2 decimals]. 3.23 9 Use Fisher's LSD procedure to test Iwhether there is a signicant difference between the means for treatments A and BI treatments A and C, and treatments B and C. Use a: = 0.05. Difference Absolute Value Conclusion - u 9 9 15. 1 ea 9 1954:: m 9 9 c. Use Fisher's LSD procedure to develop a 95% condence interval estimate of the difference between the means of treatments A and B (to 2 decimals). Enter negative values as negative numbers. [I 13.?3' 9, 20.23 9) Consider the experimental results for the following randomized block design. Make the calculations necessaryr to set up the analysis of variance table. Treatment A B C 1 10 9 9 2 12 7' 6 Blocks 3 13 1 5 15 4 20 19 18 5 B B 8 Use a: = 0.05 to test for any signicant differences. ShowI entries to 2 decimals but p-value to 4 decimals, if necessary. Round all intermediate values to two decimal places in your calculations. If your answer is zero enter "".l] Source of Variaon Sum of Squares Degrees of Freedom Mean Square F pvalue Treatments 9 a 8 a 6 Blocks E6 9 9 Error 6 8 9 1.61] g Total 33333 Q 111 9 The pvalue is 9 What is your conclusion? Conclude not all treatment means are equal V 9 Consider the experimental results for the following randomized block design. Make the calculations necessary to set up the analysis of variance table. Treatment A B C 1 10 9 B 2 12 6 5 Blocks 3 18 1 5 14 4 20 18 18 5 B 7" 8 Use a = 0.05 to test for any signicant differences {to 2 decimals but Ismralue to 4 decimals, if necessary). If your answer is zero, enter \"1]". Source Di Varialion Sum of Squares Degrees of Freedom Mean Square F pvalue Treatments 26.53 g 2 @ 13.2]r @ 9 9 Blocks 312.2? @ 4 9 19.0? 9 Error 9 8 9 2.04 9 Total C 9 14 9 They-value ._e What is your conclusion? Conclude not all treatment me: n5 are equal V 9