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Author Question: A researcher wishes to test H0: 1 = 2 = 3 against H1:at least one of the means is different. Based ... (Read 51 times)

mikaylakyoung

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on: Jan 16, 2020
A researcher wishes to test H0: μ1 = μ2 = μ3 against H1:at least one of the means is different. Based on the ANOVA table, the null hypothesis is rejected and the following output represents the results of Tukey's test. What should the researcher conclude?

Tukey Simultaneous Tests
Response Variable Response
All Pairwise Comparisons among Levels of Treatment
Treatment = 1 subtracted from:
Treatment    Difference            SE of            T- Value      Adjusted
                        of Means        Difference                               P-Value
2                       7.1887              1.982                 3.627              0.0165
3                       0.7488              1.982                 0.3778            0.9251    

Treatment = 2 subtracted from:
Treatment    Difference            SE of            T- Value      Adjusted
                        of Means        Difference                               P-Value
3                      -6.4393               1.982               -3.2489         0.0282
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Statistics: Informed Decisions Using Data ¦ Sullivan ¦ 5th Edition
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Marked as best answer by mikaylakyoung on Jan 16, 2020

Madisongo23

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Reply #1 on: Jan 16, 2020
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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mikaylakyoung

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Reply 2 on: Jan 16, 2020
Excellent

xiaomengxian

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Reply 3 on: Yesterday
Gracias!

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