Chapter 6, Problem 6.1P

For each set of sample outcomes below, use Formula 6.2 to construct the 95% confidence interval for estimating $\mu$, the population mean.

a.	$\begin{array}{c}\bar{X}=5.2\\ s=0.7\\ N=157\end{array}$	b.	$\begin{array}{c}\bar{X}=100\\ s=9\\ N=620\end{array}$	c.	$\begin{array}{c}\bar{X}=20\\ s=3\\ N=220\end{array}$
d.	$\begin{array}{c}\bar{X}=1020\\ s=50\\ N=329\end{array}$	e.	$\begin{array}{c}\bar{X}=7.3\\ s=1.2\\ N=105\end{array}$	f.	$\begin{array}{c}\bar{X}=33\\ s=6\\ N=220\end{array}$

To determine

a.

The 95% confidence interval for estimating the population mean.

Answer to Problem 6.1P

Solution:

The 95% confidence interval for estimating $\mu$ is $5.2\pm 0.11$.

Explanation of Solution

Given Information:

$\begin{array}{rll}\overline{X}& =& 5.2\\ s& =& 0.7\\ N& =& 157\end{array}$

Formula used:

Confidence interval for a sample mean (large sample, population standard deviation unknown) is given by,

$c.i.=\overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)$

Calculation:

The value of statistic at 95% level of significance is given as,

$Z_{0.5}=1.96$

Substitute $5.2$ for $\overline{X}$, $1.96$ for $Z$, $0.7$ for $s$ and 157 for $N$ in the above mentioned formula,

$\begin{array}{rll}c.i.& =& \overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)\\ & =& 5.2\pm Z\left(\frac{0.7}{\sqrt{157-1}}\right)\\ & =& 5.2\pm 1.96\left(\frac{0.7}{12.49}\right)\\ & =& 5.2\pm 0.11\end{array}$

Conclusion:

Thus, the 95% confidence interval is $5.2\pm 0.11$ for estimating population mean $\mu$.

To determine

b.

The 95% confidence interval for estimating the population mean.

Answer to Problem 6.1P

Solution:

The 95% confidence interval for estimating $\mu$ is $100\pm 0.71$.

Explanation of Solution

Given Information:

$\begin{array}{rll}\overline{X}& =& 100\\ s& =& 9\\ N& =& 620\end{array}$

Formula used:

Confidence interval for a sample mean (large sample, population standard deviation unknown) is given by,

$c.i.=\overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)$

Calculation:

The value of statistic at 95% level of significance is given as,

$Z_{0.5}=1.96$

Substitute 100 for $\overline{X}$, $1.96$ for $Z$, 9 for $s$ and 620 for $N$ in the above mentioned formula,

$\begin{array}{rll}c.i.& =& \overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)\\ & =& 100\pm Z\left(\frac{9}{\sqrt{620-1}}\right)\\ & =& 100\pm 1.96\left(\frac{9}{24.87}\right)\\ & =& 100\pm 0.71\end{array}$

Conclusion:

Thus, the 95% confidence interval is $100\pm 0.71$ for estimating population mean $\mu$.

To determine

c.

The 95% confidence interval for estimating the population mean.

Answer to Problem 6.1P

Solution:

The 95% confidence interval for estimating $\mu$ is $20\pm 0.40$.

Explanation of Solution

Given Information:

$\begin{array}{rll}\overline{X}& =& 20\\ s& =& 3\\ N& =& 220\end{array}$

Formula used:

Confidence interval for a sample mean (large sample, population standard deviation unknown) is given by,

$c.i.=\overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)$

Calculation:

The value of statistic at 95% level of significance is given as,

$Z_{0.5}=1.96$

Substitute 20 for $\overline{X}$, $1.96$ for $Z$, 3 for $s$ and 220 for $N$ in the above mentioned formula,

$\begin{array}{rll}c.i.& =& \overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)\\ & =& 20\pm Z\left(\frac{3}{\sqrt{220-1}}\right)\\ & =& 20\pm 1.96\left(\frac{3}{14.80}\right)\\ & =& 20\pm 0.40\end{array}$

Conclusion:

Thus, the 95% confidence interval is $20\pm 0.40$ for estimating population mean $\mu$.

To determine

d.

The 95% confidence interval for estimating the population mean.

Answer to Problem 6.1P

Solution:

The 95% confidence interval for estimating $\mu$ is $1020\pm 5.41$.

Explanation of Solution

Given Information:

$\begin{array}{rll}\overline{X}& =& 1020\\ s& =& 50\\ N& =& 329\end{array}$

Formula used:

Confidence interval for a sample mean (large sample, population standard deviation unknown) is given by,

$c.i.=\overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)$

Calculation:

The value of statistic at 95% level of significance is given as,

$Z_{0.5}=1.96$

Substitute 1020 for $\overline{X}$, $1.96$ for $Z$, 50 for $s$ and 329 for $N$ in the above mentioned formula,

$\begin{array}{rll}c.i.& =& \overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)\\ & =& 1020\pm Z\left(\frac{50}{\sqrt{329-1}}\right)\\ & =& 1020\pm 1.96\left(\frac{0.7}{18.11}\right)\\ & =& 1020\pm 5.41\end{array}$

Conclusion:

Thus, the 95% confidence interval is $1020\pm 5.41$ for estimating population mean $\mu$.

To determine

e.

The 95% confidence interval for estimating the population mean.

Answer to Problem 6.1P

Solution:

The 95% confidence interval for estimating $\mu$ is $7.3\pm 0.23$.

Explanation of Solution

Given Information:

$\begin{array}{rll}\overline{X}& =& 7.3\\ s& =& 1.2\\ N& =& 105\end{array}$

Formula used:

Confidence interval for a sample mean (large sample, population standard deviation unknown) is given by,

$c.i.=\overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)$

Calculation:

The value of statistic at 95% level of significance is given as,

$Z_{0.5}=1.96$

Substitute $7.3$ for $\overline{X}$, $1.96$ for $Z$, $1.2$ for $s$ and 105 for $N$ in the above mentioned formula,

$\begin{array}{rll}c.i.& =& \overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)\\ & =& 7.3\pm Z\left(\frac{1.2}{\sqrt{105-1}}\right)\\ & =& 7.3\pm 1.96\left(\frac{1.2}{10.20}\right)\\ & =& 7.3\pm 0.23\end{array}$

Conclusion:

Thus, the 95% confidence interval is $7.3\pm 0.23$ for estimating population mean $\mu$.

To determine

f.

The 95% confidence interval for estimating the population mean.

Answer to Problem 6.1P

Solution:

The 95% confidence interval for estimating $\mu$ is $33\pm 0.79$.

Explanation of Solution

Given Information:

$\begin{array}{rll}\overline{X}& =& 33\\ s& =& 6\\ N& =& 220\end{array}$

Formula used:

Confidence interval for a sample mean (large sample, population standard deviation unknown) is given by,

$c.i.=\overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)$

Calculation:

The value of statistic at 95% level of significance is given as,

$Z_{0.5}=1.96$

Substitute 33 for $\overline{X}$, $1.96$ for $Z$, 6 for $s$ and 220 for $N$ in the above mentioned formula,

$\begin{array}{rll}c.i.& =& \overline{X}\pm Z\left(\frac{s}{\sqrt{N-1}}\right)\\ & =& 33\pm Z\left(\frac{6}{\sqrt{220-1}}\right)\\ & =& 33\pm 1.96\left(\frac{6}{14.80}\right)\\ & =& 33\pm 0.79\end{array}$

Conclusion:

Thus, the 95% confidence interval is $33\pm 0.79$ for estimating population mean $\mu$.