Chapter 7.1, Problem 66E

Interpret computer output: The following MINITAB output presents a 98% confidence interval.

1. a. Fill in the blanks: We are______ confident that the population mean is between____ and _____
2. b. Use the appropriate critical value along with the information in the computer output to construct a 95% confidence interval.
3. c. Find the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.
4. d. Find the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

To determine

Identify for what percent it is confident that the population mean will lie between.

Answer to Problem 66E

We are 98% confident that the population mean is between 0.2133 and 5.1007.

Explanation of Solution

From the display of MINITAB output, the 98% confidence interval for the population mean is $\left(0.2133,5.1007\right)$.

Hence, there is 98% confidence that the population mean will lie between 0.2133 and 5.1007.

To determine

Construct a 95% confidence interval using the appropriate critical value and the information in the output.

Answer to Problem 66E

The 95% confidence interval using the appropriate critical value and the information in the output is $\left(0.598,4.716\right)$

Explanation of Solution

Calculation:

Critical value:

If $\alpha$ is a number between 0 and 1 or $0<\alpha <1$, then

• The notation $z_{\alpha }$ represents the z-score with an area $\alpha$ to the right.
• The notation $z_{\frac{\alpha }{2}}$ represents the z-score with an area $\frac{\alpha }{2}$ to the right.

Here the confidence level is provided as 95% that is, the significance level is 0.05.

$\begin{array}{c}{z}_{\frac{\alpha }{2}}=z_{\frac{0.05}{2}}\\ =z_{0.025}\end{array}$

For critical value $z_{0.025}$:

Software procedure:

Step-by-step software procedure to obtain the critical value using MINITAB software is as follows,

• Choose Stat > Graphs > Probability Distribution Plot.
• Choose View Probability.
• From Distribution, choose ‘Normal’ distribution.
• Enter the Mean as 0 and Standard deviation as 1.
• Click Shaded Area tab.
• Choose Probability and right tail for the region of the curve to shade.
• Enter the Probability as 0.025.
• Click OK in all the dialogue boxes.

Output using MINITAB software is as follows,

Thus, the critical value is 1.96.

Confidence interval:

The interval which is used to estimate the value of a parameter is termed as confidence interval.

$\begin{array}{c}\text{Confidence interval}=\text{Point estimate}\pm \text{Margin of error}\\ =\overline{x}\pm z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\end{array}$

Substitute, $\overline{x}$ as 2.657, $z_{\frac{\alpha }{2}}$ as 1.96, $\sigma$ as 8 and n as 58.

$\begin{array}{c}\text{Confidence interval}=2.657\pm 1.96\left(\frac{8}{\sqrt{58}}\right)\\ =2.657\pm 1.96\left(1.0505\right)\\ =2.657\pm 2.0589\\ =\left(0.598,4.716\right)\end{array}$

Thus, the 95% confidence interval is $\left(0.598,4.716\right)$.

To determine

Obtain the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 98% confidence interval will have a margin of error of 1.0 is 345.

Explanation of Solution

Calculation:

Sample size:

Assume m as the desired margin of error, $\sigma$ is the population standard deviation and $z_{\frac{\alpha }{2}}$ as the critical value of a confidence interval. The sample size n needed for the confidence interval will have margin of error approximately equal to m is,

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

If the value of n obtained from the formula is not a whole number, round it up to the nearest whole number. By rounding up, it can be sure that the margin of error is no greater than the desired value m.

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $z_{\frac{\alpha }{2}}$ for 98% confidence level is observed as 2.326.

Substitute $z_{\frac{\alpha }{2}}$ as 2.326, $\sigma$ as 8 and m as 1.0 in the formula,

$\begin{array}{c}n={\left(\frac{2.326\cdot 8}{1.0}\right)}^2\\ ={\left(18.608\right)}^2\\ =346.2577\\ \approx 347\end{array}$

Thus, the sample size is 347.

To determine

Obtain the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 95% confidence interval will have a margin of error of 1.0 is 246.

Explanation of Solution

Calculation:

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $z_{\frac{\alpha }{2}}$ for 95% confidence level is observed as 1.96.

Substitute $z_{\frac{\alpha }{2}}$ as 1.96, $\sigma$ as 8 and m as 1.0 in the formula,

$\begin{array}{c}n={\left(\frac{1.96\cdot 8}{1.0}\right)}^2\\ ={\left(15.68\right)}^2\\ =245.8624\\ \approx 246\end{array}$

Thus, the sample size is 246.