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Some Basic Stat Concepts from my notes

CorrelationPearson: Product CorelationSpearman: Rank CorelationWhenever we have Ordinal Data (Rank Column, Marks of Students ).If two ranks are same, then (next rank+(next+1))/2 rank, i:e (2+3)/2If Three ranks are same, then rank is: (next rank+(next+1)+(next+2))/2. i: 1+3+4/2And so on.Hypothesis TestFind t valueFind p valueCheck it p value less than  or more than alphaIf p value is less than alpha then T Test: Sample Size is Small: Say 30.One Tail TestWhen Hypothesis talks about mean as > or < of Certail valueWhen Fail to reject Null Hypothesis?If the P from test is > than Alpha(0.05, generally), then Null Hypothesis is Accepted(Failed to reject)If T from experiment is < than T found out for Degree of Freedom(Number of Observations) and Critical Value (Alpha say 0.05) from T Table of One Tail, then Null Hypothesis is Accepted.Two Tail TestWhen Null Hypothesis talks about = of certain value.When Fail to reject Null Hypothesis?If the P from test is > than Alpha(0.05, gen…
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ORE-OBIEE Integration

Let us say I want to have prediction and linear model plots of data table in Oracle having data of height and weight using ORE, to use server memory execution rather than client memory execution and show the result in OBIEE reporting tool.

Prerequisite:

1. I have ORE in 12c pdb configured with a user, say RUSER2 with RQROLE and RQADMIN roles.
2. OBIEE 11.1.1.7.0 is configured.
3. Table PRED_TBL is created with height, weight and ID columns.
4. I know R,ORE and OBIEE

Let us start with scripts:

#olm is a ORE script which returns predicted table.
begin
sys.rqScriptDrop('olm');
sys.rqScriptCreate('olm', 'function(dat) {
raw<-dat
library(ORE)
lmdl<-lm(WEIGHT~.,raw)
lpred<-predict(lmdl,raw)
pred_tbl<-cbind(raw,lpred)
pred_tbl
}');
end;

#Creating view which calls olm script with PRED_TBL as input
create or replace view R_PREDICT as
select HEIGHT, WEIGHT, ID,LPRED from table( rqTableEval(
        cursor(select * from PRED_TBL),cursor(select 1 as "ore.con…

Chi Square Distribution and Difference from T

To which question Chi square distribution answers?

Suppose you have a data population. It is

1. Normally distributed
2. You know the standard deviation of population.

Now you want to conduct an experiment out of the sample of data. You will get standard deviation of that sample too.

Now what is the probability that next sample of same size you pick will have less than or equal to standard deviation of earlier sample standard deviation computed? (Greater than case is nothing but 1- (less than case). By default it is left tail test.)

First Step: Find chisquare critical value.

Χ2 = [ ( n - 1 ) * s2 ] / σ2

R function:

chisqcv <- function(samplesize,samplestandarddeviation,populationstandarddeviation){
result<-((samplesize-1)*(samplestandarddeviation*samplestandarddeviation)/(populationstandarddeviation*populationstandarddeviation))
return(result)
}

example:

chisquarecriticalvalue<-chisqcv(7,6,4)

Once you find critical value, find out cumulative probability distribution:
pchisq(chis…