Treatment Treatment Treatment A B с 32 45 35 30 44 38 30 45 37 26 47 38 32 49 42 Sample mean 30 46 38 Sample variance 6.00 4.00 6.50 a. At the a = 0.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean Squares, Treatment 640 66 320 5.5 Mean Squares, Error Calculate the value of the test statistic (to 2 decimals). 58.18
Treatment Treatment Treatment A B с 32 45 35 30 44 38 30 45 37 26 47 38 32 49 42 Sample mean 30 46 38 Sample variance 6.00 4.00 6.50 a. At the a = 0.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean Squares, Treatment 640 66 320 5.5 Mean Squares, Error Calculate the value of the test statistic (to 2 decimals). 58.18
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
Please help with (C).
Transcribed Image Text:Calculate the value of the test statistic (to 2 decimals).
58.18
The p-value is less than 0.01
What is your conclusion?
Conclude that not all treatment means are equal
b. Calculate the value of Fisher's LSD (to 2 decimals).
3.23
Use Fisher's LSD procedure to test whether there is a significant difference between the means for treatments A and B, treatments A and C, and
treatments B and C. Use a = 0.05.
Difference Absolute Value
TA-TB
TA-EC
16
8
8
Conclusion
Significant difference
Significant difference
TB - IC
Significant difference
c. Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the difference between the means of treatments A and B (to 2
decimals). Enter negative values as negative numbers.
Transcribed Image Text:Treatment Treatment Treatment
A
B
с
32
45
35
30
44
38
30
45
37
26
47
38
32
49
42
Sample mean
30
46
38
Sample variance
6.00
4.00
6.50
a. At the a= 0.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal?
Compute the values below (to 1 decimal, if necessary).
Sum of Squares, Treatment
Sum of Squares, Error
Mean Squares, Treatment
640
The n-value is less than 0.01
66
320
Mean Squares, Error
Calculate the value of the test statistic (to 2 decimals).
58.18
5.5
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