Question

Willows in Yellowstone Writers in some fields summarize data by giving x ̄ and its

standard error rather than $\overline{x}\mathrm{and}{S}_X$.Biologists studying willow plants in Yellowstone National Park reported their results in a table with columns labeled $\overline{x}\mathrm{and}{S}_X$ The table entry for the heights of willow plants (in centimeters) in one region of the park was $61.55\pm 19.03$. The researchers measured a total of $23$plants.

a. Find the sample standard deviation $s_x$for these measurements.

b. A hasty reader believes that the interval given in the table is a $95\%$ confidence interval for the mean height of willow plants in this region of the park. Find the actual confidence level for the given interval.

Short answer

a. Standard deviation is $91.2547\mathrm{mm}$

b. Actual confidence level is $66\%$

Step-by-step solution

Given Information

It is given that:

$SE=19.03$

$\overline{x}=61.55$

$n=23$

Standard Deviation

We know that $SE=\frac{s_x}{\sqrt{n}}$

$19.03=\frac{s_x}{\sqrt{23}}$

$s_x=19.03\times \sqrt{23}$

$s_x=91.2647$

Actual Confidence Level

The degree of freedom is $=\mathrm{df}=n-1=23-1=22$

Using excel formula, confidence level is found as $=2^{*}TDIST(1,22,2)$

Confidence Level is$66\%$