91影视

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

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

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,

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

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.

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 \(p\).

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,\((0.1667,0.4109)\).