Answer:
Since the calculated value of F=0.20 does not fall in the critical region
F ≥ = 3.481 we reject the alternate hypothesis and accept the null hypothesis and conclude that there is no difference between the means of the 3 treatments.
Step-by-step explanation:
As the sample size are different we can run anova in the excel.
Part a)
The null and alternate hypotheses are
H0: μ1=μ2=μ 3 against the claim Ha: Not all three means are equal
The significance level ∝ is chosen to be 0.05
Part b)
The critical region is F≥ F(0.05,5,9) = 3.481
Part c) SST, SSE, and SS total
SS total= 144.933
SSE= 129.8333
SST= 15.1
A nova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Column 1 3 32 10.67 4.33
Column 2 3 36 12 49
Column 3 3 40 13.33 4.33
Column 4 3 36 12 7
Column 5 2 25 12.5 0.5
Column 6 1 10 10 ----------
The missing value in the variance column does not account for because the excel uses the mean to perform analysis.
Part d)
A NOVA Table
Source SS df MS F P-value F crit
Of Variation
Between Groups 15.1 5 3.02 0.20 0.95 3.482
Within Groups 129.833 9 14.43
Total 144.93 14
Part e)
Since the calculated value of F=0.20 does not fall in the critical region
F ≥ = 3.481 we reject the alternate hypothesis and accept the null hypothesis and conclude that there is no difference between the means of the 3 treatments.
For this exercise it is necessary to use knowledge of statistical, in this way
A) [tex]0,05[/tex]
B) [tex]3.481[/tex]
C)[tex]144.933\\129.8333\\15.1[/tex]
D) Do a new table
E)[tex]\geq 3.481[/tex]
As the sample size are different we can run anova in the excel:
A)The null and alternate hypotheses are [tex]H_0: \mu_1=\mu_2=\mu_3[/tex] against the claim Ha: Not all three means are equal The significance level ∝ is chosen to be 0.05.
B)The critical region is [tex]F\geq F(0.05,5,9) = 3.481[/tex]
C) SS total= [tex]144.933[/tex]
SSE= [tex]129.8333[/tex]
SST= [tex]15.1[/tex]
A nova: Single Factor
[tex]Groups\\C1\\C2\\C3\\C4\\C5\\C6[/tex] [tex]Count\\3\\3\\3\\3\\2\\1[/tex] [tex]Sum\\32\\36\\40\\36\\25\\10[/tex] [tex]Avarage\\10.67\\12\\13.33\\12\\12.5\\10[/tex] [tex]Variance\\4.33\\4.9\\4.33\\7\\0.5\\0[/tex]
The missing value in the variance column does not account for because the excel uses the mean to perform analysis.
D) New Table
[tex]Source\\Between\\Within \\Total[/tex] [tex]SS df MS\\15.15\\129.833\\144.93[/tex] [tex]FP-value\\3.02\\14.43\\145[/tex]
E) Since the calculated value of F=0.20 does not fall in the critical region [tex]F \geq = 3.481[/tex] we reject the alternate hypothesis and accept the null hypothesis and conclude that there is no difference between the means of the 3 treatments.
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