Q. 61 - The Practice of Statistics for AP

Question

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.

Step-by-Step Solution

Answer

(a) The sample standard deviation $s_x$ for these measurements is $91.2647$

(b) The given interval is not a confidence interval because the factor $t^{*}$is unknown.

Part (a) Step 1 : Given information

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

Part (a) Step 2 : Explanation

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 $91.2647$.

Part (b) Step 1 : Given Information

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

Part (b) step 2 : Explanation

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.