[R] anova analysis on factors...

[Sarah Goslee]

And a further problem with both approaches: is 11pm (23) really that
different from 1am?

On Thu, Dec 8, 2011 at 3:28 PM, Michael <comtech.usa at gmail.com> wrote:
> Hi all,
>
> If we wanted to study the effect on the mean of the hourly data based 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
>
> Residuals 14860 2.20072 1.4810e-04
>
>        [[alternative HTML version deleted]]
>
-- 
Sarah Goslee
http://www.functionaldiversity.org