91影视

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

Where

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

The sample standard deviation \(s\)is,\(s = \sqrt {{s^2}} \),

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

The sample median \(\widetilde x\) is obtained from ordered \(n\) observations from smallest to largest when we include the repeated values. So, sample median is

\(\begin{aligned}{}\widetilde {\rm{x}}{\rm{ = }}\left\{ \begin{aligned}{l}{\rm{The only middle value if n is odd}}\\{\rm{The average of the two middle value if n is even}}\end{aligned} \right.\\\widetilde {\rm{x}}{\rm{ = }}\left\{ \begin{aligned}{l}{\left( {\frac{{{\rm{n + 1}}}}{{\rm{2}}}} \right)^{{\rm{th}}}}{\rm{ , if n is odd}}\\{\rm{average of }}{\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)^{{\rm{th}}}}{\rm{and}}{\left( {\frac{{\rm{n}}}{{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{th}}}}{\rm{ordered values,if n is even}}\end{aligned} \right.\end{aligned}\)

Assume that we have \(n\) observations. By ordering the observation from smallest to largest and dividing he two halves (smallest half/largest half, if \(n\) is odd the median \(\widetilde x\) is included in both halves) we define the lower fourth as the median of the smallest half and the upper fourth as the median of the largest half. Fourth spread \({f_s}\) is defined as

\({f_s}\)= upper fourth 鈥� lower fourth.

Given data in the exercise 45 is, \(21,16,29,35,42,24,24,25\).

First order the given data,

\(16,21,24,24,25,29,35,42,\)

This sample consists of \(n = 8\) observations, which is an even number, therefore the values of the interest are

\(\begin{aligned}{}{\left( {\frac{n}{2}} \right)^{th}} & = {\left( {\frac{8}{2}} \right)^{th}}\\{\left( {\frac{n}{2}} \right)^{th}} & = {4^{th}}\\{\left( {\frac{n}{2} + 1} \right)^{th}} & = {5^{th}}\end{aligned}\)

Where the \({4^{th}}\) and \({5^{th}}\) values in the ordered sample date. The sample median is average of the two mentioned values, hence

\(\begin{aligned}{}\widetilde x & = \frac{{24 + 25}}{2}\\\widetilde x & = 24.5\end{aligned}\)

For the lower fourth, the ordered smallest half is

\(16,21,24,24\)

The smallest half consists of \(n = 4\) observations, which is an even number, therefore the values of the interest are

\(\begin{aligned}{}{\left( {\frac{n}{2}} \right)^{th}} & = {\left( {\frac{4}{2}} \right)^{th}}\\{\left( {\frac{n}{2}} \right)^{th}} & = {2^{th}}\\{\left( {\frac{n}{2} + 1} \right)^{th}} & = {3^{th}}\end{aligned}\)

Where the \({2^{th}}\) and \({3^{th}}\) values in the ordered sample date. The lower fourth is average of the two mentioned values, hence

\(\begin{aligned}{}\widetilde {{x_l}} & = \frac{{21 + 24}}{2}\\{\widetilde x_l} & = 22.5\end{aligned}\)

For the upper fourth, the ordered largest half is

\(25,29,35,42\)

The largest half consists of \(n = 4\) observations, which is an even number, therefore the values of the interest are

\(\begin{aligned}{}{\left( {\frac{n}{2}} \right)^{th}} & = {\left( {\frac{4}{2}} \right)^{th}}\\{\left( {\frac{n}{2}} \right)^{th}} & = {2^{th}}\\{\left( {\frac{n}{2} + 1} \right)^{th}} & = {3^{th}}\end{aligned}\)

Where the \({2^{th}}\) and \({3^{th}}\) values in the ordered sample date. The upper fourth is average of the two mentioned values, hence

\(\begin{aligned}{}\widetilde {{x_u}} & = \frac{{29 + 35}}{2}\\{\widetilde x_u} & = 32\end{aligned}\)

Finally, the fourth spread is difference between upper and lower fourth,

\(\begin{aligned}{}{f_s} & = {\widetilde x_u} - {\widetilde x_l}\\{f_s} & = 32 - 22.5\\{f_s} & = 9.5\end{aligned}\)

The quantity in the parentheses is \(2.11\) and the \(95\% \) confidence interval is

\(\left( {24.5 - 2.11 \times \frac{{9.5}}{{\sqrt 8 }},24.5 + 2.11 \times \frac{{9.5}}{{\sqrt 8 }}} \right) = (17.413,31.587)\)

Upper confidence bound for \(\mu \) and lower confidence bound for \(\mu \) are given by,

\(\overline x + {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-upper bound,

\(\overline x - {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-lower bound.

With confidence level of \(100(1 - \alpha )\% \). Then distribution the random sample is taken from is normal.

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

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

Where

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

The sample standard deviation \(s\)is,

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

Using this above two formulas it is easy to obtain the sample standard deviation, and it is,

\(s = 8.2115\)

The sample mean is,

\(\begin{aligned}{}\widetilde x & = \frac{1}{8}(16 + 21 + ... + 42)\\\widetilde x & = 27\end{aligned}\)

And the \(t\) value for \(\alpha = 0.5\), from the appendix of the book is,

\({t_{\alpha /2,7}} = 2.365\)

The t interval is no,

\(\left( {27 - 2.365 \times \frac{{8.2115}}{{\sqrt 8 }},27 + 2.365 \times \frac{{8.2115}}{{\sqrt 8 }}} \right) = (20.134,33.866)\)

The widths of the intervals are similar. The \(t\) interval is centred at the mean and the mean is bigger than the median because of observation with value \(42\) (mild outlier by the definition of the outliers).