of factor A and B words) Σ ij (aß)ij = 0: There is no interaction between s method of application. words) Σ ij (aß)ij ‡ 0 : There is an interaction between th s method of application. MSAB MSE istic: 1.000 659 > 0.05, we fail to reject Ho. vel of significance, there is no sufficient evidence to say th the type of pesticide used and its method of application. ecessary) words) [i ai = 0: There is no difference in the mean su erent types of pesticides used. words) Σi ai #0 : At least one type of pesticide used has
of factor A and B words) Σ ij (aß)ij = 0: There is no interaction between s method of application. words) Σ ij (aß)ij ‡ 0 : There is an interaction between th s method of application. MSAB MSE istic: 1.000 659 > 0.05, we fail to reject Ho. vel of significance, there is no sufficient evidence to say th the type of pesticide used and its method of application. ecessary) words) [i ai = 0: There is no difference in the mean su erent types of pesticides used. words) Σi ai #0 : At least one type of pesticide used has
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Explain what the owner should do and how you should proceed given the result of recent hypothesis tests.
Transcribed Image Text:5) Suppose all assumption of parametric test were satisfied, perform necessary
parametric tests of hypothesis at 5% level of significance.
Test for Interaction of factor A and B
Ho: (in symbols and words) Σ ij (aß)ij = 0: There is no interaction between the type of
pesticide used and its method of application.
Ha: (in symbols and words) Σ ij (aß)ij = 0: There is an interaction between the type of
pesticide used and its method of application.
MSAB
Test statistic formula: MSE
Value of the test statistic: 1.000
p-value: 0.34659
Decision: 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 an
interaction between the type of pesticide used and its method of application.
Test for factor A: (if necessary)
Ho: (in symbols and words) [i ai = 0: There is no difference in the mean survival time (in
mins) among the different types of pesticides used.
Ha: (in symbols and words) Ei ai = 0: At least one type of pesticide used has a different
survival time (in mins).
Test statistic formula: MSA
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
data: Pesticide 1 Method A data:
Pesticide 1 Method B
Shapiro-Wilk normality test
Shapiro-Wilk normality test
W 1.00000,
W 0.96429,
p-value 1.0000
p-value 0.6369
data: Pesticide 2 Method A data:
data: Pesticide 2 Method B
Shapiro-Wilk normality test Shapiro-Wilk normality test
W 0.98684
0.96429
p-value 0.7804
p-value 0.6369
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")
DE F value Pr (>F)
group 3 0.8718 0.4947
8
Anova Table (Type II testa)
Response: survival time
method
pesticide
Sum Sq Df value
Pr (>F)
21.33 1 7.1111 0.02851.
341.33 1 113.7778 0.000005232 ***
0.34659
methodipesticide 3.00 1 1.0000
Residuals
24.00 8
。
---
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
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