Question

3. Please consider Problem 4.12 given in the additional document Text Scan 2 (a) Carry out an appropriate analysis of variance and cut and paste the ANOVA table below (you may use SAS). (b) Test to see if the means associated with each stirring rate are all equal or if at least one mean is different from the rest (use 0.05 significance level). Please write down the statistics you used from the SAS output. State your conclusion (c) Is the blocking that was done helped reduce the error significantly? [Hint look at the significance of the blocking factor] Please write down the statistics you used from the SAS output. State your conclusions

answer this using SAS program and this example??

Answer 1

Based on the data set provided in the question we ran the one way ANOVA and obtained the following result as attached.

ANOVA TABLE Furnace 1 Furnace 3 Furnace 4 Furnace 2 4 14 14 17 6 9 6 9 6 13.250 3.775 7.688 6.000 2.160 5.750 2.500 5.750 2.872 MS 55.062 8.354 P-value source treatments error total 165.188 100.250 265.438 6.5910 0.0070 12 19 17 15 13 Furnace1 F ace 2 Furnace 3 Furnace 4 urn

From the ANOVA table above we can observe that the P value =0.007 <0.05 . Hence the result is significant and we can say that the grain size is dependent on stirring rate.

Treatment 8.0 4.0 5.0 6.0 14.0 5.0 6.0 9.0 14.0 6.0 9.0 2.0 17.0 9.0 3.0 6.0 Input Data Descriptive statistics of your k=4 independent treatments: Treatment → observations N sum Xi Pooled Total 16 53.0000 24.0000 23.0000 23.0000 123.0000 7.6875 745.0000 | 158.0000 | 151.0000 | 157.0000 | 1,211.0000 13.2500 6.0000 5.7500 5.7500 sum of squares ΣΖ sample variance s2 sample std. dev.s 4.6667 6.2500 8.2500 3.7749 2.1602 2.5000 2.8723 std. dev. of mean SE 1.8875 1.08011.2500 1.4361 17.6958 4.2066 1.0517 One-way ANOVA of your k-4 independent treatments: sum oT deorees OT mean square source F statistic p-value squares SS freedom V MS treatment 165.1875 55,0625 6.5910 0.0070 100.2500 12 8.3542 total 265.4375 15 Conclusion from Anova: The p-value corresponing to the F-statistic of one-way ANOVA is lower than 0.05, suggesting that the one or more treatments are significantly different. The Tukey HSD test, Scheffé, Bonferroni and Holm multiple comparison tests follo. These post-hoc tests would likely identify which of the pairs of treatments are significantly differerent from each other.

The p-value corrresponing to the F-statistic of one-way ANOVA is lower than 0.01 which strongly suggests that one or more pairs of treatments are significa ntly different. You have k= 4 treatments, for which we shall apply Tukey's HSD test to each of the 6 pairs to pinpoint which of them exhibits statistially significant difference. We first establish the critical value of the Tukey-Kramer HSD Q statistic based on the k = 4 treatments and v= 12 degrees of freedom for the error term, for significance level α= 0.01 and 0.05 (p-values) in the Studentized Range distribution. We obtain these ctitical values for Q for α of 0.01 and 0.05, 4 12 critical

post-hoc Tukey HSD Test Calculator results: に4 treatments degrees of freedom for the error term v = 12 Critical values of the Studentized Range Q statistic: -0.06,k-Ay-12 = 4.1992 Qcritical,A-4 12-5.50 20 Qaitioalk-4-12-4.1992

Tukey HSD results treatments Tukey HSD Tukey HSD Tukey HSD pair Q statistic p-value inferfence A vs B 5.01670.0182223 p<0.05 5.1897 0.0147080 p<0.05 5.1897 0.0147080p<0.05 0.1730 0.8999947 insignificant 0.1730 0.8999947 insignificant 0.0000 0.8999947 insignificant A vs C A vs D B vs C B vs D C vs D