The following are six observations collected from treatment 1, four observations collected from treatment 2, and five observations collected from treatment 3. Test the hypothesis at the 0.05 significance level that the tre atment means are equal. a) state the null and the alternate hypotheses
b) What is the decision rulec).
c) Compute SST, SSE, and SS total
d) Complete an ANOVA table
e) State your decision regarding the null hypothesis
Treatment 1
9, 7, 11, 9, 12, 10
Treatment 2
13, 20, 14, 13
Treatment 3
10, 9, 15, 14, 15

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      

GroupsCountSumAverage    Variance  

Column 1332          10.67            4.33

Column 2336          12               49  

Column 3340        13.33     4.33

Column 4336        12               7  

Column 5225       12.5               0.5  

Column 6110         10            

The missing value in the variance column does not account for because the excel uses the mean to perform analysis.      

Part d)      

A NOVATable    

Source              SSdfMS               FP-valueF crit

Of Variation

Between Groups       15.1   53.020.20        0.95        3.482

Within Groups  129.833   914.43                                          

Total       144.9314                                                          

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.

answer: all the values are within because it has a pattern that repeats from -6 to 4.

good luck!