What is the most appropriate test to determine if the three feed additives resulted to different mean cortisol level of chickens at 5% level of significance? Justify your answer. T.
What is the most appropriate test to determine if the three feed additives resulted to different mean cortisol level of chickens at 5% level of significance? Justify your answer. T.
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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Question
Answer letter a, use image two as a guide to determine the test to be use and provide justification
Transcribed Image Text:1. CHICKEN FEED ADDITIVE STUDY
Instruction: For all test of hypotheses, write Ho and Ha in words, specify test
procedure, show computed test statistic with gf (if applicable), p-value, decision
and conclusion.
A study was conducted to determine the effect of different types of feed additives
(A1, A2, A3) in the cortisol level (ng/ml) among chickens. Fifteen chickens were
randomly assigned to the three feed additives. The cortisol level (ng/ml) of the
chickens was then measured after four weeks.
a. What is the most appropriate test to determine if the three feed additives
resulted to different mean cortisol level of chickens at 5% level of significance?
Justify your answer.
Test to be used:
Justification (provide test of hypothesis if you must):
Transcribed Image Text:1. CHICKEN FEED ADDITIVE STUDY
Wilk-Shapiro Test for Normality
data:
A1
data:
A2
data:
АЗ
W = 0.98999
W = 0.87485
p-value = 0.2866
W = 0.97079
p-value = 0.9797
p-value = 0.8803
Bartlett test of homogeneity of variances
cortisol by additive
= 0.39956, df = 2, p-value
data:
Bartlett's K-squared
= 0.8189
Analysis of Variance
Pr(>F)
Df Sum Sq Mean Sq F value
2.700
additive
1.3501
14.24 0.000678 ***
Residuals
12
1.137
0.0948
Signif. codes:
0 '*** ! 0.001 '** !
0.01 '*' 0.05 '.' 0.1 ' ' 1
Kruskal-Wallis Rank Sum Test
data:
cortisol by additive
Kruskal-Wallis chi-squared
9.42, df = 2, p-value = 0.009005
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