Question

Use technology to construct the confidence intervals for the population variance sigmasquared and the population standard deviation sigma. Assume the sample is taken from a normally distributed population.

cequals0.98​, ssquaredequals23.04​, nequals25

Answer 1

Given

n = 25

So the degrees of freedom i s (n-1) = 25 - 1 = 24

Sample Variance = = 23.04

Confidence Interval = 98% = 0.98

We use the

Since the Confidence Interval is 98% i.e 0.98

Therefore = 2% = 0.02

To get the ; we have to search the value which is intersected between the degrees of freedom 24 and 0.99 i.e 0.98+0.01 = 0.99; and we get the as 10.856

Similary to get the we have to search the value which is intersected between the degrees of freedom 24 and 0.01; and we get the   as 42.980

NOTE: We have to search this values from CRITICAL VALUES UNDER CHI - SQUARE DISTRIBUTION.

Now the 98% Confidence Interval of Population Variance and Standard Deviation is

Is the 98% confidence Interval of

is the 98% confidence interval of