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2) Specify the appropriate linear model and define each component in the context with the problem.. Yijk = μ + α₁ + B₁ + (aß)¡j + Eįjk; i = 1,2 ; j = 1,2 ; k = 1,2,3 Yijk = μl = a¡ = B₁ = (aß) ¡j = Eijk ijk =

2) Specify the appropriate linear model and define each component in the context with the problem.. Yijk = μ + α₁ + B₁ + (aß)¡j + Eįjk; i = 1,2 ; j = 1,2 ; k = 1,2,3 Yijk = μl = a¡ = B₁ = (aß) ¡j = Eijk ijk =

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.3: Spanning Sets And Linear Independence
Problem 28EQ: Use the method of Example 2.23 and Theorem 2.6 to determine if the sets of vectors in Exercises...
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Problem: Mr. Kinalabaw, an owner of a mango plantation, is worried about the increasing number of fruit piercing moths present in most of the mango trees in his farm. He then asked for Concon's help, his trusted statistician, in resolving the problem. They identified two types of pesticide (1, 2) and two methods of application (A, B). They applied each treatment combination to 3 fruit piercing moths. Their survival time (in minutes) was recorded. Pesticide Method of Application Survival Time (in minutes) A 31 29 30 1 B 28 30 27 A 23 18 20 2 B 17 15 18 185 data: Pesticide 1 Method A Shapiro-Wilk normality test W = 1.00000, p-value = 1.0000 data: Pesticide 2 Method A Shapiro-Wilk normality test W 0.98684 p-value 0.7804 = Bartlett test of homogeneity of variances data: survival time by trmcombin Bartlett's K-squared 1.4136, df = 3, p-value = 0.7024 Levene's Test for Homogeneity of Variance (center "mean") Df F value Pr (>F) group 3 0.8718 0.4947 8 Anova Table (Type II tests) Response: survival time Sum Sq Df F value 21.33 1 7.1111 method pesticide method:pesticide 3.00 1 1.0000 Residuals 24.00 8 Pr (>F) 0.02851 * 341.33 1 113.7778 0.000005232 *** 0.34659 Signif. codes: 0 ***** 0.001 **** 0.01 * 0.05. 0.1'' Kruskal-Wallis rank sum test. data: survival time by trmcombin Kruskal-Wallis chi-squared= 9.6573, df = 3, p-value = 0.02172 data: Pesticide 1 Method B Shapiro-Wilk normality test W 0.96429, p-value 0.6369 data: Pesticide 2 Method B Shapiro-Wilk normality test W= 0.96429 p-value 0.6369
Transcribed Image Text:Problem: Mr. Kinalabaw, an owner of a mango plantation, is worried about the increasing number of fruit piercing moths present in most of the mango trees in his farm. He then asked for Concon's help, his trusted statistician, in resolving the problem. They identified two types of pesticide (1, 2) and two methods of application (A, B). They applied each treatment combination to 3 fruit piercing moths. Their survival time (in minutes) was recorded. Pesticide Method of Application Survival Time (in minutes) A 31 29 30 1 B 28 30 27 A 23 18 20 2 B 17 15 18 185 data: Pesticide 1 Method A Shapiro-Wilk normality test W = 1.00000, p-value = 1.0000 data: Pesticide 2 Method A Shapiro-Wilk normality test W 0.98684 p-value 0.7804 = Bartlett test of homogeneity of variances data: survival time by trmcombin Bartlett's K-squared 1.4136, df = 3, p-value = 0.7024 Levene's Test for Homogeneity of Variance (center "mean") Df F value Pr (>F) group 3 0.8718 0.4947 8 Anova Table (Type II tests) Response: survival time Sum Sq Df F value 21.33 1 7.1111 method pesticide method:pesticide 3.00 1 1.0000 Residuals 24.00 8 Pr (>F) 0.02851 * 341.33 1 113.7778 0.000005232 *** 0.34659 Signif. codes: 0 ***** 0.001 **** 0.01 * 0.05. 0.1'' Kruskal-Wallis rank sum test. data: survival time by trmcombin Kruskal-Wallis chi-squared= 9.6573, df = 3, p-value = 0.02172 data: Pesticide 1 Method B Shapiro-Wilk normality test W 0.96429, p-value 0.6369 data: Pesticide 2 Method B Shapiro-Wilk normality test W= 0.96429 p-value 0.6369
2) Specify the appropriate linear model and define each component in the context with the problem.. Yijk = μ + α₁ + B₁ + (aß)¡j + Eijk; i = 1,2 ; j = 1,2; k = 1,2,3 Y μl = α; = B₁ = (ap) = & ijk =
Transcribed Image Text:2) Specify the appropriate linear model and define each component in the context with the problem.. Yijk = μ + α₁ + B₁ + (aß)¡j + Eijk; i = 1,2 ; j = 1,2; k = 1,2,3 Y μl = α; = B₁ = (ap) = & ijk =
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