Question

In: Statistics and Probability

The following is sample information. Test the hypothesis that the treatment means are equal. Use the...

The following is sample information. Test the hypothesis that the treatment means are equal. Use the 0.01 significance level.

Treatment 1 Treatment 2 Treatment 3
9            9            3           
8            10            5           
4            10            4           
9            7            4           

a. State the null hypothesis and the alternate hypothesis.

H0: click to select (The treatment means are not all the same, the treatment means are the same) *pick which answer is correct from the options in the brackets*

H1:   click to select (The treatment means are not all the same, the treatment means are the same) *pick which answer is correct from the options in the brackets*          

b. What is the decision rule? (Round the final answer to 2 decimal places.)

Reject H0 if the test statistic is greater than _______    *fill in _______*

c. Compute SST, SSE, and SS total. (Round the final answers to 3 decimal places.)

SST = ________ *fill in ______*             

SSE = ________ *fill in ______*             

SS total = ________ *fill in ______*

d. Complete the ANOVA table. (Round the SS, MS, and F values to 3 decimal places.)

Source SS DF MS F
Treatment       ______ ______ ______ _____
Error    _______ _______ _____    n / a
Total _______ _____ n / a n / a
*fill in _____, ignore N/A)

e. State your decision regarding the null hypothesis.

click to select (Do not reject, reject )H0.*pick which answer is correct from the options in the brackets*

Solutions

Expert Solution

a.

Here we define the null and alternate hypothesis as follows:

H0: The treatment means are the same

H1: The treatment means are not all the same.

b.

We reject the null hypothesis if the test statistic is greater than F statistic with level of significance =0.01 and degrees of freedom 2,9

c.yij is the jth treatment corresponding to the ith level i=1,2,3 j=1,2,3,4

=52.670 where Ti0 is the treatment means corresponding to the ith treatment i=1,2,3

and CF is the correction factor = and

=77.670

and SSE= TSS-SST=25.00

d.

Source DF SS   MS F-Value
treatments 2 52.67 26.333 9.48   
Error 9 25.00 2.778
Total 11 77.67

Here MS= SS/DF

e.

here =8.0215

since F value >

we reject the null hypothesis at 99% level of significance.


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