Chapter 7, Problem 47SE

Example 1.11 introduced the accompanying observations on bond strength.

1. a. Estimate true average bond strength in a way that conveys information about precision and reliability. [Hint: Σx_i = 387.8 and $\sum x_i^2=4247.08$.]
2. b. Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10.

To determine

Find the estimate of true average bond strength of observations.

Answer to Problem 47SE

The 95% confidence interval for population mean bond strength of observations is $\underset{\_}{\left(6.702<\mu <9.456\right)}$

Explanation of Solution

Given info:

The data represents the bond strength of observations.

Calculation:

Here, the estimate of the true average bond strength of observations is obtained by finding the confidence interval about mean for the bond strength of observations.

The prior confidence level 95% is used to estimate the true average bond strength.

Sample standard deviation:

Step by step procedure to obtain sample standard deviation using the MINITAB software:

• Choose Stat > Basic Statistics > Display descriptive statistics.
• In Sample from columns, enter the column of bond strength.
• In Options, select Standard deviation.
• Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

From MINITAB output, the sample standard deviation is 4.868.

Confidence interval:

Software Procedure:

Step by step procedure to obtain the 95% confidence interval to estimate the population mean bond strength of observations using the MINITAB software:

• Choose Stat > Basic Statistics > 1-Sample z.
• In Sample from columns, enter the column of bond strength.
• Enter sample standard deviation as 4.868.
• In Options, enter Confidence level as 95.0.
• Choose not equal in alternative.
• Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

Thus, the 95% confidence interval for population mean bond strength of observations is $\underset{\_}{\left(6.702<\mu <9.456\right)}$.

Interpretation:

There is 95% confident that the population mean bond strength of observations lies between 6.702 and 9.456.

To determine

Find the 95% confidence interval for the proportion of bond strengths greater than 10.

Answer to Problem 47SE

The 95% confidence interval for population proportion of bond strengths greater than 10 is $\underset{\_}{\left(0.166<p<0.410\right)}$

Explanation of Solution

Calculation:

Point estimate:

Here, the total number of bond strengths surveyed is $n=48$.

The number of bond strength values greater than 10 is 13.

That is, the number of specified characteristics is $x=13$.

The population proportion of bond strengths greater than 10 is obtained as follows:

$\begin{array}{c}\widehat{p}=\frac{x}{n}\\ =\frac{13}{48}\\ =0.2708\end{array}$

Thus, the population proportion of bond strengths greater than 10 is $\widehat{p}=0.2708$

Confidence interval:

Critical value:

For 95% level of significance,

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Hence, the cumulative area to the left is,

$\begin{array}{c}\text{Area to the left}=1-\text{Area to the right}\\ =1-0.025\\ =0.975\end{array}$

From Table A.3 of the standard normal distribution in Appendix , the critical value is 1.96.

Thus, the critical value is $\left(z_{\frac{\alpha }{2}}\right)=1.96$.

Here, the sample size is $n=48$ which is large in size.

Hence, the population proportion is,

$\begin{array}{c}\tilde{p}=\frac{\left(\widehat{p}+\frac{z_{\frac{\alpha }{2}}{}^2}{2\times n}\right)}{\left(1+\frac{z_{\frac{\alpha }{2}}{}^2}{n}\right)}\\ =\frac{0.2708+\frac{{1.96}^2}{2\times 48}}{1+\frac{{1.96}^2}{48}}\\ =\frac{0.3108}{1.0800}\\ =0.2878\end{array}$

Thus, the adjusted population proportion is $\tilde{p}=0.2878$

Confidence interval:

The upper bound of the confidence interval is,

$\begin{array}{c}CI=\tilde{p}\pm z_{\frac{\alpha }{2}}\times \frac{\sqrt{\left(\frac{\widehat{p}\times \left(1-\widehat{p}\right)}{n}+\frac{z_{\frac{\alpha }{2}}{}^2}{4\times n^2}\right)}}{\left(1+\frac{z_{\alpha }{}^2}{n}\right)}\\ =0.2878\pm 1.96\times \frac{\sqrt{\frac{0.2708\times \left(1-0.2708\right)}{48}+\frac{{1.96}^2}{4\times {48}^2}}}{1+\frac{{1.96}^2}{48}}\\ =0.2878\pm 0.1221\\ \simeq \left(0.166,0.410\right)\end{array}$

Thus, the 95% confidence interval for the population proportion of bond strengths greater than 10 is $\underset{\_}{\left(0.166<p<0.410\right)}$.

Interpretation:

There is 95% confident that the population proportion of bond strengths greater than 10 lies between 0.166 and 0.410.