Complete the following ANOVA table. Then determine if there are any differences using alpha = 0.01
Df. Sum Sq. Mean Sq. F value
Tube 4. 0.6596
Error. 25. 2077.23
Total
Here we want to test the
null hypothesis H0: All tubes are sam
alternate hypothesis Ha: atleast one tube is different from others
since the critical F(0.01,4,25)=4.1774 is more than calculate F=0.0020, so we fail to reject (or accept H0) and conclude that there are no differences among tubes.
following information has been generated from the given information
Mean_Sq= Sum_Sq/Df
Mean_Sq_Tube=Mean_Sq_Tube/Df_Tube=0.6596/4=0.1649
F=Mean_Sq_Tube/Mean_Sq_Error=0.1649/0.83.0892=0.0020
source | Df | Sum Sq | Mean Sq | F value |
Tube | 4 | 0.6596 | 0.1649 | 0.0020 |
Error | 25 | 2077.23 | 83.0892 | |
Total | 29 | 2077.89 |
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