Question

Example 1.11 introduced the accompanying observations on bond strength.

11.5 12.1 9.9 9.3 7.8 6.2 6.6 7.0

13.4 17.1 9.3 5.6 5.7 5.4 5.2 5.1

4.9 10.7 15.2 8.5 4.2 4.0 3.9 3.8

3.6 3.4 20.6 25.5 13.8 12.6 13.1 8.9

8.2 10.7 14.2 7.6 5.2 5.5 5.1 5.0

5.2 4.8 4.1 3.8 3.7 3.6 3.6 3.6

a.Estimate true average bond strength in a way that conveys information about precision and reliability.

(Hint: \(\sum {{{\bf{x}}_{\bf{i}}}} {\bf{ = 387}}{\bf{.8}}\) and \(\sum {{{\bf{x}}^{\bf{2}}}_{\bf{i}}} {\bf{ = 4247}}{\bf{.08}}\).)

b. Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10.

Short answer

a)\((6.702,9.456)\)

b)The 95% confidence interval for the proportion p is,\((0.1667,0.4109)\).

Step-by-step solution

Step 1:The sample mean.

The sample mean \(\overline x \) of observations \({x_1},{x_2},.......{x_n}\) is given by.

\(\begin{array}{l}\overline x = \frac{{{x_1} + {x_2} + ....... + {x_n}}}{n}\\\overline x = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \end{array}\)

Step 2:Solution for part a).

For large \(n\), the standardized random variable,\(Z = \frac{{\overline X - \mu }}{{S/\sqrt n }}\) has approximately a normal distribution with expectation \(0\) and standard deviation \(1\).

Therefore, a large sample confidence interval for \(\mu \) is,

\(\overline x \pm {Z_{\alpha /2}} \cdot \frac{s}{{\sqrt n }}\)with confidence level of approximately \(100(1 - \alpha )\% \). This stands regardless the population distribution.

The sample mean \(\overline x \) of observations \({x_1},{x_2},.......{x_n}\) is given by.

\(\begin{array}{l}\overline x = \frac{{{x_1} + {x_2} + ....... + {x_n}}}{n}\\\overline x = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \end{array}\)

The sample mean is,

\(\begin{array}{l}\overline x = \frac{1}{{48}}(387.8)\\\overline x = 8.079\end{array}\)

The sample variance \({s^2}\) is

\({s^2} = \frac{1}{{n - 1}} \cdot {S_{xx}}\)

Where

\(\begin{array}{c}{S_{xx}} = \sum {({x_i} - \overline x } {)^2}\\ = \sum {{x_i}^2} - \frac{1}{n} \cdot {\left( {\sum {{x^i}} } \right)^2}\end{array}\)

The sample standard deviation \(s\)is,

\(\begin{array}{l}s = \sqrt {{s^2}} \\s = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \end{array}\)

The sample variance is,

\(\begin{array}{l}{s^2} = \frac{1}{{48 - 1}} \cdot \left( {4247.08 - \frac{1}{{48}}{{(387.8)}^2}} \right)\\{s^2} = 23.7017\end{array}\)

The sample standard deviation is,

\(\begin{array}{l}s = \sqrt {23.7017} \\s = 4.868\end{array}\)

Step 3:95% confidence interval for the mean.

Finally, the \(95\% \)confidence interval for the mean is,

\(\begin{array}{l}\left( {\overline x - {Z_{\alpha /2}} \cdot \frac{s}{{\sqrt n }},\overline x + {Z_{\alpha /2}} \cdot \frac{s}{{\sqrt n }}} \right)\\ = \left( {8.079 - 1.96\left( {\frac{{4.868}}{{\sqrt {48} }}} \right),8.079 + 1.96\left( {\frac{{4.868}}{{\sqrt {48} }}} \right)} \right)\\ = (6.702,9.456)\end{array}\)

Where,

\(\begin{array}{c}100(1 - \alpha ) = 95\\\alpha = 0.05\end{array}\)

And,

\(\begin{array}{c}{Z_{\alpha /2}} = {Z_{0.05/2}}\\ = {Z_{0.025}}\\ = 1.96\end{array}\)

The obtained form is,

\(P(Z > {Z_{0.025}}) = 0.025\)and from the normal probability table in the appendix. The probability can also be computed by a software.

Step 4:Solution for part b).

Denote with,

\(\begin{array}{l}\widetilde p = \frac{{\left( {\frac{{\widehat p + {Z^2}_{\alpha /2}}}{{2n}}} \right)}}{{\left( {\frac{{1 + {Z^2}_{\alpha /2}}}{n}} \right)}}\\\widehat q = 1 - \widehat p\end{array}\)

A confidence interval for a population proportion \(p\) with confidence level approximately \(100(1 - \alpha )\) is,

\(\left( {\widetilde p - {Z_{\alpha /2}} \cdot \frac{{\sqrt {\widehat p\widehat q/n + {Z^2}_{\alpha /2}/4{n^2}} }}{{1 + {Z^2}_{\alpha /2}/n}},\widetilde p + {Z_{\alpha /2}} \cdot \frac{{\sqrt {\widehat p\widehat q/n + {Z^2}_{\alpha /2}/4{n^2}} }}{{1 + {Z^2}_{\alpha /2}/n}}} \right)\)

This is also called a score CI for \(p\).

Step 5:95% confidence interval for the mean.

The number of all such bond which exceeds \(10\) is \(13\)out of \(48\), therefore,

\(\begin{array}{l}\widehat p = \frac{{13}}{{48}}\\ = 0.2708\end{array}\)

Also, \(\widetilde p\) is,

\(\begin{array}{c}\widetilde p = \frac{{\left( {\frac{{\widehat p + {Z^2}_{\alpha /2}}}{{2n}}} \right)}}{{\left( {\frac{{1 + {Z^2}_{\alpha /2}}}{n}} \right)}}\\ = \frac{{\left( {\frac{{0.2708 + {{(1.96)}^2}}}{{2(48)}}} \right)}}{{\left( {1 + \frac{{{{(1.96)}^2}}}{{48}}} \right)}}\\ = 0.2888\end{array}\)

Therefore, the \(95\% \) confidence interval for the proportion \(p\) is,

\(\begin{array}{l}\left( {\widetilde p - {Z_{\alpha /2}} \cdot \frac{{\sqrt {\widehat p\widehat q/n + {Z^2}_{\alpha /2}/4{n^2}} }}{{1 + {Z^2}_{\alpha /2}/n}},\widetilde p + {Z_{\alpha /2}} \cdot \frac{{\sqrt {\widehat p\widehat q/n + {Z^2}_{\alpha /2}/4{n^2}} }}{{1 + {Z^2}_{\alpha /2}/n}}} \right)\\ = \left( \begin{array}{l}0.2888 - 1.96\left( {\frac{{\sqrt {0.2708(0.7292/48) + {{1.96}^2}/4{{(48)}^2}} }}{{1 + {{(1.96)}^2}/48}}} \right)\\,0.2888 + 1.96\left( {\frac{{\sqrt {0.2708(0.7292/48) + {{1.96}^2}/4{{(48)}^2}} }}{{1 + {{(1.96)}^2}/48}}} \right)\end{array} \right)\\ = (0.1667,0.4109)\end{array}\)

Hence, The \(95\% \) confidence interval for the proportion p is,\((0.1667,0.4109)\).