Chapter 8.1, Problem 68E

a.

To determine

To fill: The blanks in thegiven statement with appropriate words.

Answer to Problem 68E

98%, 0.1233 and 5.1007.

Explanation of Solution

Given information:

  $\begin{array}{l}\text{98\%}\text{Confidence}\text{interval}\text{=}\left(0.1233,5.1007\right)\\ \text{Sample}\text{mean}\left(\text{<mover accent="true"><mo>x</mo><mo>¯</mo></mover>}\right)=2.657\\ \text{Sample}\text{size}\left(N\right)=58\\ \text{Population}\text{standard}\text{deviation}\left(\sigma \right)=8.0000\\ {\text{SE}}_{Mean}\left(\frac{\sigma }{\sqrt{N}}\right)=1.0505\end{array}$

Confidence interval is the interval estimate of unknown population parameter. It gives the probability that population parameter will fall in this interval.It means that there is 98% surety that population number of years employed lie between 0.1233 and 5.1007.

Thus, the complete statement is, “We are $\underset{\_}{98\%}$ confident that the population number of years employed is between $\underset{\_}{0.1233}$ and $\underset{\_}{5.1007}$ ”.

b.

To determine

To find:The 95% confidence interval.

Answer to Problem 68E

The confidence interval is $\left(0.598,4.716\right)$ .

Explanation of Solution

Given:

Confidence level = 95%

Formula used:

  $\overline{x}-z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}<\mu <\overline{x}+z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}$

Calculation:

The value of $z_{\frac{\alpha }{2}}$ corresponding to $95\%$ confidence level is 1.96, which is obtained from the standard normal table.

The 95% confidence interval using the provided output can be calculated as:

  $\begin{array}{c}\overline{x}-z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}<\mu <\overline{x}+z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}\\ 2.657-1.96\times 1.0505<\mu <2.657+1.96\times 1.0505\\ 2.657-2.059<\mu <2.657+2.059\\ 0.598<\mu <4.716\end{array}$

Hence, the confidence interval is $\left(0.598,4.716\right)$ .

c.

To determine

To find: The sample size to have the margin of error of 1 at $98\%$ confidence level.

Answer to Problem 68E

The required sample size is 347.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Margin of error }\left(m\right)=1\\ \text{Confidence level = 98\%}\end{array}$

Formula used:

  $n={\left(\frac{z_{\frac{\alpha }{2}}\times \sigma }{m}\right)}^2$

Calculation:

The value of $z_{\frac{\alpha }{2}}$ corresponding to $98\%$ confidence level is 2.33.

The sample size at $98\%$ confidence level can be calculated as:

  $\begin{array}{c}n={\left(\frac{z_{\frac{\alpha }{2}}\times \sigma }{m}\right)}^2\\ ={\left(\frac{2.33\times 8}{1}\right)}^2\\ =347.450\\ \approx 347\end{array}$

Hence, the sample size is 347.

d.

To determine

To find: The sample size to have the margin of error of 1 at $95\%$ confidence level.

Answer to Problem 68E

The required sample size is 246.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Confidence level = 95\%}\\ \text{Margin of error }=1\end{array}$

Formula used:

  $n={\left(\frac{z_{\frac{\alpha }{2}}\times \sigma }{m}\right)}^2$

Calculation:

The value of $z_{\frac{\alpha }{2}}$ corresponding to $95\%$ confidence level is 1.96.

The sample size at $95\%$ confidence level can be calculated as:

  $\begin{array}{c}n={\left(\frac{z_{\frac{\alpha }{2}}\times \sigma }{m}\right)}^2\\ ={\left(\frac{1.96\times 8}{1}\right)}^2\\ =245.862\\ \approx 246\end{array}$

Hence, the sample size is 246.