Question

Try the following experiment

a. pick a number for mean and standard deviation

b. generate 5 random numbers using a normal distribution.

repeat steps (b) 1000 times. you had 1000 sets of 5 normally distributed random variables. Try the following

Convert each one to a z-score, (value - mean) / (std dev). Square each one of the 5 and add them.

you should know have a list of 1000 sums of 5 squared standard normals. Confirm you have the Chi squared table for 5 degrees of freedom

Answer 1

we have run this in R, The R code is

#===========================================================================

#selecting mean and sd
m=0
s=1

chi=rep(0,1000)
for(i in 1:1000)
{
x=rnorm(5,m,s)
chi[i]=sum(((x-m)/s)^2)
}
chisq.test(chi,rchisq(1000,5))

#=====================================================================

the test result is

Pearson's Chi-squared test

data: chi and rchisq(1000, 5)
X-squared = 999000, df = 998000, p-value = 0.2397

since p-value>0.05 we can say that the data will have chi-sq distributed at 95% confidence.