Question

The statistics of nequals22 and sequals14.3 result in this​ 95% confidence interval estimate of sigma​: 11.0less thansigmaless than20.4. that confidence interval can also be expressed as​ (11.0, 20.4). given that 15.7plus or minus4.7 results in values of 11.0 and​ 20.4, can the confidence interval be expressed as 15.7plus or minus4.7 as​ well?

Answer 1

Answer: no, the confidence interval for the standard deviation σ cannot be expressed as 15.7 $\pm$ 4.7

There are three ways in which you can possibly express a confidence interval:

1) inequality
The two extremities of the interval will be each on one side of the "less then" symbol (the smallest on the left, the biggest on the right) and the symbol for the standard deviation (σ) will be in the middle:
11.0 < σ < 20.4
This is the first interval given in the question and it means that the standard deviation can vary between 11.0 and 20.4

2) interval
The two extremities will be inside a couple of round parenthesis, separated by a comma, always the smallest on the left and the biggest on the right:
(11.0, 20.4)
This is the second interval given in the question.

3) point estimate $\pm$ margin of error
This is the most common way to write a confidence interval because it shows straightforwardly some important information. 
However, this way can be used only for the confidence interval of the mean or of the popuation, not for he confidence interval of the variance or of the standard deviation.

This is due to the fact that in order to calculate the confidence interval of the standard variation (and similarly of the variance), you need to apply the formula:
$\sqrt{ \frac{(n-1) s^{2} }{\chi^{2}_{\alpha / 2} } } \leq \sigma \leq \sqrt{ \frac{(n-1) s^{2} }{\chi^{2}_{1 - \alpha / 2} } }$

which involves a χ² distribution, which is not a symmetric function. For this reason, the confidence interval we obtain is not symmetric around the point estimate and the third option cannot be used to express it.