61. Willows in Yellowstone Writers in some fields summarize data by giving $\overline{x}$and its standard error rather than $\overline{x}$and $s_x$. Biologists studying willow trees in Yellowstone National Park reported their results in a table with columns labeled $\overline{x}+SE$. The table entry for the heights of willow trees (in centimeters) in one region of the park was $61.55\pm 19.03.$The researchers measured a total of$23$ trees.
(a) Find the sample standard deviation $s_x$ for these measurements. Show your work.
(b) Explain why the given interval is not a confidence interval for the mean height of willow trees in this region of the park.

A summarize data by giving $\overline{x}$and its standard error rather than $\overline{x}$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,

$s=SE\times \sqrt{n}$

$=19.03\times \sqrt{23}$

$=91.2647$

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 $\overline{x}\pm \left(t^{*}\times \overline{E}\right)$.

The interval in the problem is $\overline{x}\pm \overline{E}$.

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

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