Explain briefly how you should proceed given the result of recent hypothesis tests. 5) Suppose all assumption of parametric test were satisfied, perform necessaryparametric tests of hypothesis at 5% level of significance.Test for Interaction of factor A and BHo: (in symbols and words) Σ ij (aß)ij = 0: There is no interaction between the type ofpesticide used and its method of application.Ha: (in symbols and words) Σ ij (aß)ij = 0: There is an interaction between the type ofpesticide used and its method of application.Test statistic formula: MSABMSEValue of the test statistic: 1.000p-value: 0.34659Decision: Since 0.34659 > 0.05, we fail to reject Ho.Conclusion: At 5% level of significance, there is no sufficient evidence to say that there is aninteraction between the type of pesticide used and its method of application.Test for factor A: (if necessary)Ho: (in symbols and words) Σ i ai = 0: There is no difference in the mean survival time (inmins) among the different types of pesticides used.Ha: (in symbols and words) Σi ai = 0: At least one type of pesticide used has a differentsurvival time (in mins).Test statistic formula: MSEMSA Problem: Mr. Kinalabaw, an owner of a mango plantation, is worried about theincreasing number of fruit piercing moths present in most of the mango trees in hisfarm. He then asked for Concon's help, his trusted statistician, in resolving theproblem. They identified two types of pesticide (1, 2) and two methods ofapplication (A, B). They applied each treatment combination to 3 fruit piercingmoths. Their survival time (in minutes) was recorded.PesticideMethod ofApplicationSurvival Time (in minutes)A3129301B283027A2318202B171518185data: Pesticide 1 Method AShapiro-Wilk normality testW = 1.00000,p-value = 1.0000data: Pesticide 2 Method AShapiro-Wilk normality testW 0.98684p-value 0.7804=Bartlett test of homogeneity of variancesdata: survival time by trmcombinBartlett's K-squared 1.4136, df = 3, p-value = 0.7024Levene's Test for Homogeneity of Variance (center "mean")Df F value Pr (>F)group 3 0.8718 0.49478Anova Table (Type II tests)Response: survival timeSum Sq Df F value21.33 1 7.1111methodpesticidemethod:pesticide 3.00 1 1.0000Residuals24.00 8Pr (>F)0.02851 *341.33 1 113.7778 0.000005232 ***0.34659Signif. codes: 0 ***** 0.001 **** 0.01 * 0.05. 0.1''Kruskal-Wallis rank sum test.data: survival time by trmcombinKruskal-Wallis chi-squared= 9.6573, df = 3, p-value = 0.02172data: Pesticide 1 Method BShapiro-Wilk normality testW 0.96429,p-value 0.6369data: Pesticide 2 Method BShapiro-Wilk normality testW= 0.96429p-value 0.6369

Testing for Interaction term: H0: There is no effect due to interaction between application method…

Answered: 5) Suppose all assumption of parametric…

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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