91Ó°ÊÓ

A summarize data by giving $\stackrel{¯}{x}$and its standard error rather than $\stackrel{¯}{x}$and $s_x$. To find the sample standard deviation $s_x$.

The standard deviation is divided by the square root of the sample size to obtain the standard error of the mean: $SE=\frac{s_x}{\sqrt{n}}$

The Sample size $n$is $23$. The Standard error of mean $SE$is $19.03$Multiply the two sides by $\sqrt{n}$in the equation $SE=\frac{s_x}{\sqrt{n}}$.

Hence,

$s=SE×\sqrt{n}$

$=19.03×\sqrt{23}$

$=91.2647$

Therefore, the standard deviation is $91.2647$.

To explain why the supplied interval for the mean height of willow trees in this park region is not a confidence interval.

The confidence interval is calculated using the formula $\stackrel{¯}{x}Â\pm \left(t^{*}×\stackrel{¯}{E}\right)$.

The interval in the problem is $\stackrel{¯}{x}Â\pm \stackrel{¯}{E}$.

And here, $t^{*}$is unknown.

As a result, the provided interval is not a confidence interval.