Chapter 14, Problem 14.31E

(a)

To determine

To find out what is the standard deviation of the sampling distribution of mean body length $\overline{x}$ .

Answer to Problem 14.31E

The standard deviation is $2.1381$ .

Explanation of Solution

In the question, it is given that the deer mice adult body lengths are known to be normally distributed with,

  $\begin{array}{l}\overline{x}=91.1\text{mm}\\ \sigma =8\text{mm}\\ \mu =86\text{mm}\\ n=14\\ c=95\%\end{array}$

And also from the graphing calculator for a $95\%$ confidence interval is as follows:

  $\text{Confidence interval}=(86.909,95.291)$

Thus, the standard deviation of the sampling distribution of mean body length $\overline{x}$ can be calculated as:

  $\begin{array}{l}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{8}{\sqrt{14}}\\ =2.1381\end{array}$

(b)

To determine

To find out what critical value was used to compute this $95\%$ confidence interval.

Answer to Problem 14.31E

The critical value is $1.96$ .

Explanation of Solution

In the question, it is given that the deer mice adult body lengths are known to be normally distributed with,

  $\begin{array}{l}\overline{x}=91.1\text{mm}\\ \sigma =8\text{mm}\\ \mu =86\text{mm}\\ n=14\\ c=95\%\end{array}$

And also from the graphing calculator for a $95\%$ confidence interval is as follows:

  $\text{Confidence interval}=(86.909,95.291)$

Thus, the critical value used to compute this $95\%$ confidence interval can be obtained from the z-table at $\alpha =0.05$ is:

  $z_{\alpha /2}=z_{0.025}=1.96$

(c)

To determine

To show the step-by-step computations required to arrive at the interval provided by the graphing calculator.

Explanation of Solution

In the question, it is given that the deer mice adult body lengths are known to be normally distributed with,

  $\begin{array}{l}\overline{x}=91.1\text{mm}\\ \sigma =8\text{mm}\\ \mu =86\text{mm}\\ n=14\\ c=95\%\end{array}$

And also from the graphing calculator for a $95\%$ confidence interval is as follows:

  $\text{Confidence interval}=(86.909,95.291)$

Thus, the step-by-step computations required to arrive at the interval provided by the graphing calculator are as follows:

  $\begin{array}{l}\overline{x}\pm z^{*}\times \frac{\sigma }{\sqrt{n}}\\ =91.9\pm 1.96\times \frac{8}{\sqrt{14}}\\ =91.9\pm 4.191\\ =(86.909,95.291)\end{array}$

Hence the result is as above.

(d)

To determine

To explain would a $90\%$ confidence interval based on the same data be larger or smaller.

Answer to Problem 14.31E

The $90\%$ confidence interval based on the same data be smaller.

Explanation of Solution

In the question, it is given that the deer mice adult body lengths are known to be normally distributed with,

  $\begin{array}{l}\overline{x}=91.1\text{mm}\\ \sigma =8\text{mm}\\ \mu =86\text{mm}\\ n=14\\ c=95\%\end{array}$

And also from the graphing calculator for a $95\%$ confidence interval is as follows:

  $\text{Confidence interval}=(86.909,95.291)$

Thus, the confidence interval at $90\%$ can be calculated as:

  $\begin{array}{l}\overline{x}\pm z^{*}\times \frac{\sigma }{\sqrt{n}}\\ =91.9\pm 1.645\times \frac{8}{\sqrt{14}}\\ =91.9\pm 3.52\\ =(88.38,95.42)\end{array}$

As the critical value of at $\alpha =0.10$ is $1.645$ . Thus, we conclude that a $90\%$ confidence interval based on the same data be smaller.