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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(chisquarecriticalvalue,degreesoffreedom)

Good Source:

http://stattrek.com/probability-distributions/chi-square.aspx

Value you will get is the answer for our question.

In T: You have sample mean and population mean. You are giving a probability that next sample mean is equal to earlier sample mean. It will not talk about standard deviation.

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