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

Answers

Answer 1

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.

Answer 2

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.

See more about statistical at : brainly.com/question/15692038


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Answers

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Answers

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Answer:

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-----------------

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Answers

Answer:

12

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x + 4z + z = 25

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Answer:

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