A oneway analysis of variance experiment produced the following ANOVA table. (You may find it useful to reference the q table).
SUMMARY  
Groups  Count  Average  
Column 1  6  0.71  
Column 2  6  1.43  
Column 3  6  2.15  
Source of Variation  SS  df  MS  F  pvalue  
Between Groups  10.85  2  5.43  20.88  0.0000  
Within Groups  3.86  15  0.26  
Total  14.71  17  
SUMMARY  
Groups  Count  Average  
Column 1  6  0.71  
Column 2  6  1.43  
Column 3  6  2.15  
ANOVA  
Source of Variation  SS  df  MS  F  pvalue 
Between Groups  10.85  2  5.43  20.88  0.0000 
Within Groups  3.86  15  0.26  
Total  14.71  17 
a. Conduct an ANOVA test at the 1% significance level to determine if some population means differ.
Reject H_{0}; we can conclude that some population means differ.
Reject H_{0}; we cannot conclude that some population means differ.
Do not reject H_{0}; we can conclude that some population means differ.
Do not reject H_{0}; we cannot conclude that some population means differ.
b. Calculate 99% confidence interval estimates of μ_{1} − μ_{2,}μ_{1} − μ_{3,} and μ_{2} − μ_{3} with Tukey’s HSD approach. (If the exact value for n_{T} – c is not found in the table, then round down. Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)

c. Given your response to part b, which means significantly differ?

a)since p value <0.01
Reject H_{0}; we can conclude that some population means differ.
b)
MSE=  0.2600  
df(error)=  15  
number of treatments =  3  
pooled standard deviation=Sp =√MSE=  0.510 
critical q with 0.01 level and k=3, Nk=15 df=  4.84  
Tukey's (HSD) =(q/√2)*(sp*√(1/ni+1/nj) =  1.01 
Lower bound  Upper bound  
(x_{i}x_{j} )  ME  (x_{i}x_{j})ME  (x_{i}x_{j})+ME  
μ1μ2  0.72  1.01  1.73  0.29 
μ1μ3  1.44  1.01  2.45  0.43 
μ2μ3  0.72  1.01  1.73  0.29 
c)
μ1μ2  no  
μ1μ3  YEs  
μ2μ3  No 
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