In an experiment designed to test the output levels of three different treatments, the following results were obtained: SST=390, SSTR=140 ,nt =18 . Set up the ANOVA table and test for any significant difference between the mean output levels of the three treatments. Use 0.05alpha. Do not round your intermediate calculations
In an experiment designed to test the output levels of three different treatments, the following results were obtained: SST=390, SSTR=140 ,nt =18 . Set up the ANOVA table and test for any significant difference between the mean output levels of the three treatments. Use 0.05alpha. Do not round your intermediate calculations
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
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In an experiment designed to test the output levels of three different treatments, the following results were obtained: SST=390, SSTR=140 ,nt =18 . Set up the ANOVA table and test for any significant difference between the mean output levels of the three treatments. Use 0.05alpha. Do not round your intermediate calculations.

Transcribed Image Text:In an experiment designed to test the output levels of three different treatments, the following results were obtained: SST = 390, SSTR =
140, nT = 18. Set up the
ANOVA table and test for any significant difference between the mean output levels of the three treatments. Use a = 0.05. Do not round your intermediate calculations.
Source
Sum
Degrees
Mean Square
F
p-value
(to 2 decimals)
of Variation
of Squares of Freedom
(to 2 decimals)
(to 4 decimals)
Treatments
140
Error
250
Total
390
The p-value is
Select your answer -
What is your conclusion?
Select your answer -
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