[R] p-values calculation

jd6688 jdsignature at gmail.com
Mon Jul 26 07:32:59 CEST 2010


Id	cat1	location	item_values	p-values	sequence
a111	1	3002737	0.196504377	0.01	1
a112	1	3017821	0.196504377	0.05	2
a113	1	3027730	0.196504377	0.02	3
a114	1	3036220	0.196504377	0.04	4
a115	1	3053984	0.196504377	0.03	5
a116	1	3063892	0.196504377	0.07	6
a117	1	3076333	0.196504377	0.08	7
a118	1	3090500	0.196504377	0.02	8
a119	1	3103304	0.196504377	0.03	9
a120	1	3119350	0.196504377	0.05	10
a121	1	3129884	0.196504377	0.01	11
a122	1	3154598	0.196504377	0.03	12
a123	1	3170910	0.196504377	0.05	13
a124	1	3180712	0.196504377	0.06	14
a125	1	3186519	0.196504377	0.07	15
a126	1	3192256	0.196504377	0.09	16
a127	1	3198441	0.196504377	0.01	17
a128	1	3205784	0.196504377	0.02	18
a129	1	3210685	0.196504377	0.03	19
a130	1	3218542	0.196504377	0.04	20
a131	1	3234318	0.196504377	0.05	21
a132	1	3239972	0.196504377	0.09	22
a133	1	3245663	0.196504377	0.05	23
a134	1	3257997	0.196504377	0.02	24
a135	1	3273226	0.196504377	0.03	26
a136	1	3285404	0.196504377	0.04	27
a137	1	3290332	0.196504377	0.05	28
a138	1	3300679	0.196504377	0.03	29
a139	1	3310164	0.196504377	0.09	30


I would like to calculate the p values for the item_values above.

option 1: z score based
> wData <- within(wData, { 
+ z <- (values - mean(values))/sd(values) 
+ "p-value" <- 2*pnorm(-abs(z))  ## 2-sided 
+ rm(z) }) 
> wData 

option 2: t distibution based
   
p_value<-2*pt(-abs(t),df=n-1)


my questions:
what R function can assist the distribution analysis to the item_values
column above? for instance: normal distribution analysis?

how to calculate the p values if the distribution of the item_values is
no-parameter distribution?

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