Chapter 14.5, Problem 28E

a.

To determine

Find the standard error of the mean.

Answer to Problem 28E

  $0.20$

Explanation of Solution

Given information:

Given: A random samples of $50acorns$ reveals a mean diameter of $16.2millimeters$ with a standard deviation of $1.4millimeters$ .

Calculation:

Here, we have to find the standard error of the mean:

  $\begin{array}{l}{\sigma }_{\overline{X}}=\frac{Standarddeviation}{\sqrt{N}}\\ {\sigma }_{\overline{X}}=\frac{1.4}{\sqrt{50}}=0.20\end{array}$

Hence, the standard error of the mean is $0.20$ .

b.

To determine

Find the interval about the sample mean having $5\%$ confidence level.

Answer to Problem 28E

  $\left[15.81,16.59\right]$

Explanation of Solution

Given information:

Given: A random samples of $50acorns$ reveals a mean diameter of $16.2millimeters$ with a standard deviation of $1.4millimeters$ .

Calculation:

Here, we have to find the interval:

  $\begin{array}{l}5\%confidencelevelmeansP=95\%\\ P=95\%correspondstot=1.96\\ \left[\overline{X}-t{\sigma }_X,\overline{X}+t{\sigma }_X\right]\\ \left[15.81,16.59\right]\end{array}$

Hence, the interval is $\left[15.81,16.59\right]$ .

c.

To determine

Find the interval about the sample mean having $99\%$ chances that true mean lies within the interval.

Answer to Problem 28E

  $\left[15.68,16.71\right]$

Explanation of Solution

Given information:

Given: A random samples of $50acorns$ reveals a mean diameter of $16.2millimeters$ with a standard deviation of $1.4millimeters$ .

Calculation:

Here, we have to find the interval:

  $\begin{array}{l}P=99\%correspondstot=2.58\\ \left[\overline{X}-t{\sigma }_{\overline{X}},\overline{X}+t{\sigma }_{\overline{X}}\right]\\ \left[15.68,16.71\right]\end{array}$

Hence, the interval is $\left[15.68,16.71\right]$ .

d.

To determine

Find the interval about the sample mean having $0.80$ chances that true mean lies within the interval.

Answer to Problem 28E

  $\left[15.94,16.46\right]$

Explanation of Solution

Given information:

Given: A random samples of $50acorns$ reveals a mean diameter of $16.2millimeters$ with a standard deviation of $1.4millimeters$ .

Calculation:

Here, we have to find the interval:

  $\begin{array}{l}P=80\%correspondstot=1.3\\ \left[\overline{X}-t{\sigma }_{\overline{X}},\overline{X}+t{\sigma }_{\overline{X}}\right]\\ \left[15.94,16.46\right]\end{array}$

Hence, the interval is $\left[15.94,16.46\right]$ .