Question

2. There are four fertilizers that can be applied to a plot of land. A farmer wants to check which one of them boosts the fertility of a a crop most. He has a land which he divides into 16 1meter by 1 meter plots and applied four fertilizers to four plots each at random. At the end of the crop season he notes the yield from the 16 plots, a summary of which is given below. Fertilizer replications sample mean sample variance 570.75 860.50 433.00 496.25 1360.250 3619.000 2064.667 2426.917 4 4 4 (a) Construct an ANOVA table from the above information and test whether there was a difference in yield using the four fertilizers. State the model, hypotheses, test statistic, decision rule and decision clearly. (b) Compare all treatments through a pairwise comparison procedure. Which method among Bonferroni, Scheffe and Tukey will be the most suitable here? Use the best among these three for the pairwise comparisons and rank the fertilizers from best to worst according to these tests

Answer 1

Solution Given N 16 k 4 (a) To detemine the differences between the means yield of 4 different fertilizers use the One way ANOVA model The hypothesis is: Ho: There is difference in the mean yields of 4 different fertilizers, i.e. = , = 4 = 44 no H2 There is a difference in the mean yields of 4 different fertilizers, i.e. at least on yields is different of the mean Decision: rule: If the calculated value of the F statistic is more than the critical value F d, the null hypothesis is rejected The ANOVA table is: of Sum of Squares Degrees of Mean square freedom Source F Variation 2 k- 1 SSB Between MSB F = MSE SSB nXX MSB i treatments k-1 N-k Error SSE -ΣΣ (x-x)' SSE MSE N-k Total 2 N-1 -ΣΣx-X SST Compute the sum of squares between (SSB) as follows:

4 Σ 41 i-l (570.75 +860.50+433.00 496.25 =590.125 The value of SSB is- 2 SSB nX-X _ -(4x(570.75-590.125))+ +4x(496.25-590.125)) (4x(860.50-590.125)+(4x(433.00-590.125)) 427915.25 Compute the sum of squares of error (SSE) as follows: SSE ΣΣ (x - 1 -Σ(1-1) Here s is the sample variance offertilizer i= 1, 2, 3 and 4 The value of SSE is, SSE Σ (η -1)s ((4-1)x1360.250)+((4-1)x3619.000) +((4-1)x2064.667) +((4-1)x2426.917) 28412.502 Compute the sum of squares of total (SST) as follows: SSTΣΣ|Χ-x = SSB SSE = 427915.25 2412.502 = 456327.752

The complete ANOVA table is: of Degrees Source of Mean square Sum F of Variation Squares freedom 4 1 3 142638.417 Between 427915.25 427915.25 MSB .=142638.417 F = 60.243 treatment 2367.7085 28412.502 2367.7085 28412.502 Error 16 4 MSE 12 12 Total 456327.75 16 1 = 2 15 The critical value of Fis, Use the F table: TABLE E F critical values (continued) Degrees of freedom in the merato 273 247 259 144 7.57 5.42 50 482 465 637 12.86 4.53 6.15 12404 441 4.36 791 010 001 11.26 7.59 6.63 1205 2541 1849 15.83 14.39 13.48 1.77 261 3.45 J00 336 S12 7.21 1056 301 426 281 249 2.5 2.51 329 247 321 244 025 010 001 571 5,08 472 44 412 S80 11.13 4.20 410 547 4.03 535 699 642 6.06 561 22 860 1639 13.90 12.56- 11.71 10.70- 10.37 10.11 292 24 238 100 329 2.73 261 2.52 246 4.96 4.10 3.71 148 331 122 314 дет 302 694 483 385 025 546- 4.47 424 407 3.95 010 10.04 7.56- 655 599 564 539 5204 50- 4.94 1491 001 21.04 12.55 11.28 10.48 993 952 920 8.96 2.54 J00 256- 3982 526- 7.21 13.81 200 245 239 309 3.88 2.34 3.0m 3.76 489 2.30 295 2.27 290 3.59 463 8.12 484 672 3.59 3.20 025 010 001 306- 4.6 622 11.56 428 547 10.35 404 5.32 9.58 965 19.09 s07 9.05 474 835 8.66 J00 18. 281 3.89 5.30 261 349 2.48 126 2.39 233 100 2.28 2.91 3.61 224 285 351 221 os0 475 3.11 280 6.55 12 025 447 4.12 3.89 373 3.44 18.64 12 97 1080- 961 745 For degrees of freedom (3, 12) all the critical F values are less than the calculated value of the test statistic

Since calculated F< critical F the null hypothesis is rejected Conclusion There is a difference in yield on using the four different fertilizers (b) The Tukey's method considers the differences among all the pairs of means divided by the estimated standard deviation of mean (standard error) and compares the value with the critical value of Tukey's Studentizied statistic S VMSE n The Tukey method is better than the Bonferroni and Scheffe's method. Because the Tukey method is exact in case of equal sample sizes whereas Bonferroni only approximates significantly different or not is: The decision rule to determine whether the means are If the calculated q value is more than the critical value ofqathe two means are significantly different Compute the different Tukey statistic values and compare them with the critical Tukey value as follows

Decision Means q0.05, 4, 12 Not significantly different Significantly different Significantly different Significantly different Significantly different Not significantly different -x, -289.75 - 137.75 - 74.50 - 127.50 1, 2 4.20 1, 3 4.20 1, 4 4.20 2, 3 4.20 4.20 2, 4 64.25 3, 4 4.20 - -63.25 Ranks of the fertilizers: Fertilizers Rank 1 2 2 1 3 4 4 3