91影视

The size of the sample is $n=22$.

The sample standard deviation is $s=14.3$.

The 95% confidence interval estimate of $蟽$is $11.0<蟽<20.4$.

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

$CI=\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_R^2}}<蟽<\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_L^2}}$

where

n is the sample size

s is the sample standard deviation

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

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

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

It is incorrect to express the confidence interval as role="math" localid="1648040974197" $\left(15.7卤4.7\right)$because the confidence interval of role="math" localid="1648040987212" $蟽$ 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;

$\left(s卤\text{Margin}\text{of}\text{Error}\right)=\left(15.7卤4.7\right)$

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, $\left(15.7卤4.7\right)$is an absolutely incorrect way of expressing the confidence interval.