  The following are six observations collected from treatment 1, four observations collected from treatment 2, and five observationscollected from treatment 3. Test the hypothesis at the 0.05 significance level that the treatment means are equalTreatment 1 Treatment 2 Treatment 3913102091114151314121510a. State the null and the alternate hypothesis.Ho u1р2рзH1 Treatment means are notall the same.b. What is the decision rule? (Round your answer to 2 decimal places.)Reject Ho if F>3.89 b. What is the decision rule? (Round your answer to 2 decimal places.)Reject Ho if F3.89Rectangular Snipc. Compute SST, SSE, and SS total. (Round your answers to 2 decimal places.)70.40 SSESST=SS total =152.93d. Complete the ANOVA table. (Round your SS, MS and Fvalues to 2 decimal places.)SourceSSdfMSFTreatments70.4025.12Error126.88Total152.9314

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Step 2

Explanation:

Part (a):

The test hypotheses are:

Null hypothesis: H0: μ1 = μ2 = μ3; vs.

Alternate hypothesis: H1: Treatment means are not all the same.

Part (b):

The F-test is used for testing the above hypotheses.

Here, number of treatments, k = 3.

Total number of observations, n = 6 + 4 + 5 = 15.

Thus, the numerator degrees of freedom for the F-distribution is, k – 1 = 2; the denominator degrees of freedom, nk = 12.

The significance level is, α = 0.05.

Thus, the critical value is, ; k – 1, nk = F0.05; 2, 12.

Using Excel formula, =F.INV.RT(0.05,2,12), the critical value is 3.89.

The decision rule is: Reject H0 if F > 3.89.

Part (c... help_outlineImage TranscriptioncloseAnova: Single Factor SUMMARY Groups Count Variance Sum Average Treatment 1 6 58 9.666666667 3.066666667 4 Treatment 2 60 15 11.33333333 Treatment 3 5 63 12.6 8.3 ANOVA Source of Variation SS df F P-value F crit MS Between Groups 70.4 35.2 5.117932149 0.024703779 3.885293835 Within Groups 82.53333333 12 6.877777778 Total 152.9333333 14 fullscreen

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