91Ó°ÊÓ

The size of the sample is\(n = 22\).

The sample standard deviation is\(s = 14.3\).

The 95% confidence interval estimate of \(\sigma \)is \(11.0 < \sigma < 20.4\).

The confidence interval to estimate the population standard deviation\(\left( \sigma \right)\)has the following expression:

\(CI = \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _R^2}}} < \sigma < \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _L^2}}} \)where

n is the sample size

s is the sample standard deviation

\(\chi _R^2\)is the right-tailed critical value of\({\chi ^2}\)with\(\left( {n - 1} \right)\)degrees of freedom

\(\chi _L^2\)is the left-tailed critical value of\({\chi ^2}\)with\(\left( {n - 1} \right)\)degrees of freedom

Here, the value of the confidence interval is equal to (11.0, 20.4).

It is incorrect to express the confidence interval as\(\left( {15.7 \pm 4.7} \right)\)because the confidence interval of\(\sigma \)is not symmetric about the sample standard deviation.

Also, it makes the reader believe that the value of the sample standard deviation is equal to 15.7 by comparing the value as follows;

\(\left( {s \pm {\rm{Margin}}\;{\rm{of}}\;{\rm{Error}}} \right) = \left( {15.7 \pm 4.7} \right)\)

It can be assumed that s takes the value equal to 15.7 and the margin of error takes a value equal to 4.7 when in reality, the value of the sample standard deviation is equal to 14.3.

Therefore, \(\left( {15.7 \pm 4.7} \right)\) is an absolutely incorrect way of expressing the confidence interval.