In: Statistics and Probability
A one-way analysis of variance experiment produced the following ANOVA table. (You may find it useful to reference the t table and the q table.)
SUMMARY | ||||||
Groups | Count | Average | ||||
Column 1 | 10 | 349 | ||||
Column 2 | 10 | 348 | ||||
Column 3 | 10 | 366 | ||||
Column 4 | 10 | 365 | ||||
Source of Variation | SS | df | MS | F | p-value | |
Between Groups | 2,997.11 | 3 | 999.04 | 15.54 | 1.2E–06 | |
Within Groups | 2,314.71 | 36 | 64.30 | |||
Total | 5,311.82 | 39 | ||||
a. Use Fisher’s LSD method to determine which means differ at the 5% level of significance. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and Round your answers to 2 decimal places.)
b. Use Tukey’s HSD method to determine which means differ at the 5% level of significance. (If the exact value for nT – c is not found in the table, then round down. Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)
c. Given your responses to parts (a) and (b), do the population means differ at the 5% significance level?