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.

Math

Statistics

To assess the reliability of timber structures and related building design codes, many researchers have studied strength factors of structural lumber. In one such study, three species of Canadian softwood were analyzed for bending strength. Varying numbers of observations were made for each species. Assume all three populations are normally distributed with equal variances. The results are given here: Bending Strength (in MPa) Mean (in MPa)

To assess the reliability of timber structures and related building design codes, many researchers have studied strength factors of structural lumber. In one such study, three species of Canadian softwood were analyzed for bending strength. Varying numbers of observations were made for each species. Assume all three populations are normally distributed with equal variances. The results are given here: Bending Strength (in MPa) Mean (in MPa)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.

Question

Solve

Douglas-Fir
Hem-Fir
Spruce pine-Fir
75
45
42
57
48
39
51
33
30
33
Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
Douglas-Fir
3
183
61
156
Hem-Fir
3
126
42
63
Spruce pine-Fir
4
144
36 30
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
1118.1
2
559.05 7.411648 0.018691 4.737414
Within Groups
528
7 75.42857
Total
1646.1
9

To assess the reliability of timber structures and related building design codes, many
researchers have studied strength factors of structural lumber. In one such study, three
species of Canadian softwood were analyzed for bending strength. Varying numbers of
observations were made for each species. Assume all three populations are normally
distributed with equal variances. The results are given here:
Bending Strength (in MPa)
Mean (in MPa)
Douglas-Fir
75, 57, 51
61
Hem-Fir
45, 48, 33
42
Species
Spruce Pine-
42, 39, 30, 33
Fir
a) State the null and alternative hypotheses.
Null
Ho: Hi =
Hz = M3
Alternative Ha Not all the population means are equal
b) Construct the appropriate ANOVA table.
Anova: Single Factor
SUMMARY
Douglas-Fir
Hem-Fir
Spruce pine-Fir
Groups
Count
Sum
Average
Variance
75
45
42
57
48
39
Douglas-Fir
3
183
61
156
51
33
30
Hem-Fir
126
42
63
33
Spruce pine-Fir
144
36
30
ANOVA
Source of Variation
SS
df
MS
P-value
E crit
Test statistic value :
7.4|16
Between Groups
1118.1
2
559.05 7.411648 0.018691 4.737414
Within Groups
528
7 75.42857
P-value is : 0.0187
Total
1646.1
9
c) Use the ANOVA table from part b) to test whether there is a difference in the mean
bending 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 level
of 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 why
not?
e) Determine which pair(s) of means are significantly different using the Tukey-Kramer
method.
37

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