Solved)e) Douglas-FirHem-FirSpruce pine-Fir75454257483951333033Anova: Single FactorSUMMARYGroupsCountSumAverageVarianceDouglas-Fir318361156Hem-Fir31264263Spruce pine-Fir414436 30ANOVASource of VariationSSdfMSFP-valueF critBetween Groups1118.12559.05 7.411648 0.018691 4.737414Within Groups5287 75.42857Total1646.19 To assess the reliability of timber structures and related building design codes, manyresearchers have studied strength factors of structural lumber. In one such study, threespecies of Canadian softwood were analyzed for bending strength. Varying numbers ofobservations were made for each species. Assume all three populations are normallydistributed with equal variances. The results are given here:Bending Strength (in MPa)Mean (in MPa)Douglas-Fir75, 57, 5161Hem-Fir45, 48, 3342SpeciesSpruce Pine-42, 39, 30, 33Fira) State the null and alternative hypotheses.NullHo: Hi =Hz = M3Alternative Ha Not all the population means are equalb) Construct the appropriate ANOVA table.Anova: Single FactorSUMMARYDouglas-FirHem-FirSpruce pine-FirGroupsCountSumAverageVariance754542574839Douglas-Fir318361156513330Hem-Fir126426333Spruce pine-Fir1443630ANOVASource of VariationSSdfMSP-valueE critTest statistic value :7.4|16Between Groups1118.12559.05 7.411648 0.018691 4.737414Within Groups5287 75.42857P-value is : 0.0187Total1646.19c) Use the ANOVA table from part b) to test whether there is a difference in the meanbending strengths among the three types of wood at the .05 level.Given that the level of significance is 0.05. Here , the p-value (0.0187) is less than the levelof significance (0.05). Therefore, reject the null hypothesis.Thus, there is difference in mean bending strengths among the 3 types of wood.d) Is it applicable to follow up with Tukey-Kramer pairwise comparisons? Why or whynot?e) Determine which pair(s) of means are significantly different using the Tukey-Kramermethod.37

d) From part c, there exists significant difference among three trending strengths. Hence, we use Tukey's Kramer post-hoc test for finding the pair-wise comparisons. e. Critical Value: From the given information, there are 3 groups and the total number of observations is 10. From Tukey’s table, the value QU corresponding with k=3, error degrees of freedom as 7 and 0.05 level of significance is 4.16. Rejection Rule: If the test statistic is greater than the critical value, then the pairs of means are significant. If the test statistic is less than or equal to the critical value, then the pairs of means are not significant. Pair Test statistic,HSD=Mi-MjMSwnh Critical value Decision Conclusion Doulas-Fir VS Hem-Fir Mi-MjMSwnh=61-4275.428573=3.789 4.16 Test statistic is less than critical value Not significant Doulas-Fir VS Spruce pine-Fir Mi-MjMSwnh=61-3675.428573=4.986 4.16 Test statistic is greater than critical value Significant Hem-Fir VS Spruce pine-Fir Mi-MjMSwnh=42-3675.428574=1.382 4.16 Test statistic is less than critical value Not significant In the given situation, the significant pair is Doulas-Fir and Spruce pine-Fir.