search for: hours_factors

...sed on the hours within a day... and we wanted to do Anova analysis... We have two choices: Please see below: Why are these two approaches giving very different p-values? And which one shall I use? Thanks a lot! 1. treating the hours as double/floating numbers: anova(lm(hourlydata~as.double(hours_factors))) Df Sum Sq Mean Sq F value Pr(>F) as.double(hours_factors) 1 0.0002 0.00019876 1.3425 0.2466 Residuals 14868 2.2013 0.00014806 2. treating the hours as factors: anova(lm(hourlydata~hours_factors)) Df Sum Sq Mean Sq F value Pr(>F) hours_factors 9 0.00077 8.5979e-05 0.5806 0.8142 Re...