91影视

When a set of data is sorted in ascending (more common) or descending order, the median is the middle number.

The data in the example is,

\(\begin{array}{l}{\rm{24}}{\rm{.46,25}}{\rm{.61,26}}{\rm{.25,26}}{\rm{.42,26}}{\rm{.66,27}}{\rm{.15,27}}{\rm{.31,27}}{\rm{.54,27}}{\rm{.74,27}}{\rm{.94,}}\\{\rm{27}}{\rm{.98,28}}{\rm{.04,28}}{\rm{.28,28}}{\rm{.49,28}}{\rm{.50,28}}{\rm{.87,29}}{\rm{.11,29}}{\rm{.13,29}}{\rm{.50,30}}{\rm{.88}}{\rm{.}}\end{array}\)

The Sample Variance\({{\rm{s}}^{\rm{2}}}\)is,

\({{\rm{s}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{n - 1}}}}{\rm{ \times }}{{\rm{S}}_{{\rm{xx}}}}\)

Where,

\(\begin{array}{c}{{\rm{S}}_{{\rm{xx}}}}{\rm{ = }}\sum {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - \bar x}}} \right)}^{\rm{2}}}} \\{\rm{ = }}\sum {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ - }}\frac{{\rm{1}}}{{\rm{n}}}{\rm{ \times }}{\left( {\sum {{{\rm{x}}_{\rm{i}}}} } \right)^{\rm{2}}}\end{array}\)

The standard deviation of the sample s is,

\({\rm{s = }}\sqrt {{{\rm{s}}^{\rm{2}}}} {\rm{ = }}\sqrt {\frac{{\rm{1}}}{{{\rm{n - 1}}}}{\rm{ \times }}{{\rm{S}}_{{\rm{xx}}}}} \)

The sum of squared data is required to calculate\({{\rm{S}}_{{\rm{xx}}}}\). The data has been squared.

\(\begin{array}{l}{\rm{598}}{\rm{.2916,655}}{\rm{.8721,689}}{\rm{.0625,698}}{\rm{.0164,710}}{\rm{.7556,737}}{\rm{.1225,745}}{\rm{.8361,}}\\{\rm{758}}{\rm{.4516,769}}{\rm{.5076,780}}{\rm{.6436,782}}{\rm{.8804,786}}{\rm{.2416,799}}{\rm{.7584,811}}{\rm{.6801,}}\\{\rm{812}}{\rm{.25,833}}{\rm{.4769,847}}{\rm{.3921,848}}{\rm{.5569,870}}{\rm{.25,953}}{\rm{.5744}}\end{array}\)

and the total is,

\(\begin{array}{c}\sum {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ = 598}}{\rm{.2916 + 655}}{\rm{.8721 + \ldots + 953}}{\rm{.5744}}\\{\rm{ = }}{\rm{.15489}}{\rm{.6204}}\end{array}\)

The total of the information is,

\(\sum {{{\rm{x}}_{\rm{i}}}} {\rm{ = 24}}{\rm{.46 + 25}}{\rm{.61 + \ldots + 30}}{\rm{.88 = 555}}{\rm{.86}}\)

As a result,\({{\rm{S}}_{{\rm{xx}}}}\),

\(\begin{array}{c}{{\rm{S}}_{{\rm{xx}}}}{\rm{ = }}\sum {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ - }}\frac{{\rm{1}}}{{\rm{n}}}{\rm{ \times }}{\left( {\sum {{{\rm{x}}_{\rm{i}}}} } \right)^{\rm{2}}}\\{\rm{ = 15489}}{\rm{.6204 - }}\frac{{\rm{1}}}{{{\rm{20}}}}{\rm{ \times (555}}{\rm{.86}}{{\rm{)}}^{\rm{2}}}\\{\rm{ = 40}}{\rm{.6034}}{\rm{.}}\end{array}\)

The varianceis now,

\(\begin{array}{c}{{\rm{s}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{19}}}}{\rm{ \times 40}}{\rm{.6034}}\\{\rm{ = 2}}{\rm{.137}}\end{array}\)

And the standard deviation s is,

\(\begin{array}{c}{\rm{s = }}\sqrt {{{\rm{s}}^{\rm{2}}}} \\{\rm{ = }}\sqrt {{\rm{2}}{\rm{.137}}} \\{\rm{ = 1}}{\rm{.462}}{\rm{.}}\end{array}\)

To calculate the specified estimate of standard deviation, first compute the sample median, then calculate the absolute difference of data points, and divide the median by \({\rm{0}}{\rm{.6745}}\).

The Sample Median is calculated with n observations ordered from least to greatest, including repeated values. As a result, sample median is,

\(\begin{array}{l}{\rm{\tilde x = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{ The only middle value if n is odd }}}\\{{\rm{ The average of the two middle values if n is even }}}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{l}}{{{\left( {\frac{{{\rm{n + 1}}}}{{\rm{2}}}} \right)}^{{\rm{th}}}}}&{{\rm{, if n = odd }}}\\{{\rm{ average of }}{{\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}^{{\rm{th}}}}{\rm{ and }}{{\left( {\frac{{\rm{n}}}{{\rm{2}}}{\rm{ + 1}}} \right)}^{{\rm{th}}}}{\rm{ ordered values }}}&{{\rm{, if n = even}}}\end{array}} \right.\end{array}\)

The data has been sorted.

\(\begin{array}{l}{\rm{24}}{\rm{.46,25}}{\rm{.61,26}}{\rm{.25,26}}{\rm{.42,26}}{\rm{.66,27}}{\rm{.15,27}}{\rm{.31,27}}{\rm{.54,27}}{\rm{.74,27}}{\rm{.94,27}}{\rm{.98,28}}{\rm{.04,}}\\{\rm{28}}{\rm{.28,28}}{\rm{.49,28}}{\rm{.5,28}}{\rm{.87,29}}{\rm{.11,29}}{\rm{.13,29}}{\rm{.5,30}}{\rm{.88}}{\rm{.}}\end{array}\)

Since this sample has an even number of observations (\({\rm{n = 20}}\)), the values of interest are,

\(\begin{array}{c}{\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)^{{\rm{th}}}}{\rm{ = }}{\left( {\frac{{{\rm{20}}}}{{\rm{2}}}} \right)^{{\rm{th}}}}\\{\rm{ = 1}}{{\rm{0}}^{{\rm{th}}}}\\{\left( {\frac{{\rm{n}}}{{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{th}}}}{\rm{ = }}{\left( {\frac{{{\rm{20}}}}{{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{th}}}}\\{\rm{ = 1}}{{\rm{1}}^{{\rm{th}}}}\end{array}\)

In the ordered sample data, the tenth and eleventh values are highlighted in red. The sample median is the average of the two figures previously given.

\(\begin{array}{c}{\rm{\tilde x = }}\frac{{{\rm{27}}{\rm{.94 + 27}}{\rm{.98}}}}{{\rm{2}}}\\{\rm{ = 27}}{\rm{.96}}\end{array}\)

The median is subtracted from the data.

\(\begin{array}{l}{\rm{ - 3}}{\rm{.5, - 2}}{\rm{.35, - 1}}{\rm{.71, - 1}}{\rm{.54, - 1}}{\rm{.3, - 0}}{\rm{.81, - 0}}{\rm{.65, - 0}}{\rm{.42, - 0}}{\rm{.22,}}\\{\rm{ - 0}}{\rm{.02,0}}{\rm{.02,0}}{\rm{.08,0}}{\rm{.32,0}}{\rm{.53,0}}{\rm{.54,0}}{\rm{.91,1}}{\rm{.15,1}}{\rm{.17,1}}{\rm{.54,2}}{\rm{.92}}\end{array}\)

This data's absolute value is,

\(\begin{array}{l}{\rm{3}}{\rm{.5,2}}{\rm{.35,1}}{\rm{.71,1}}{\rm{.54,1}}{\rm{.3,0}}{\rm{.81,0}}{\rm{.65,0}}{\rm{.42,0}}{\rm{.22,0}}{\rm{.02,0}}{\rm{.02,0}}{\rm{.08,}}\\{\rm{0}}{\rm{.32,0}}{\rm{.53,0}}{\rm{.54,0}}{\rm{.91,1}}{\rm{.15,1}}{\rm{.17,1}}{\rm{.54,2}}{\rm{.92}}\end{array}\)

Finally, the data that is utilised to calculate the median estimations\({\rm{\sigma }}\)(absolute value divided by\({\rm{0}}{\rm{.6745}}\))

\(\begin{array}{l}{\rm{5}}{\rm{.19,3}}{\rm{.48,2}}{\rm{.54,2}}{\rm{.28,1}}{\rm{.93,1}}{\rm{.20,0}}{\rm{.96,0}}{\rm{.62,0}}{\rm{.33,0}}{\rm{.03,0}}{\rm{.03,}}\\{\rm{0}}{\rm{.12,0}}{\rm{.47,0}}{\rm{.79,0}}{\rm{.80,1}}{\rm{.35,1}}{\rm{.70,1}}{\rm{.73,2}}{\rm{.28,4}}{\rm{.33}}{\rm{.}}\end{array}\)

The data has been sorted.

\(\begin{array}{l}{\rm{0}}{\rm{.03,0}}{\rm{.03,0}}{\rm{.12,0}}{\rm{.33,0}}{\rm{.47,0}}{\rm{.62,0}}{\rm{.79,0}}{\rm{.80,0}}{\rm{.96,}}\\{\rm{1}}{\rm{.20,1}}{\rm{.35,1}}{\rm{.70,1}}{\rm{.73,1}}{\rm{.93,2}}{\rm{.28,2}}{\rm{.28,2}}{\rm{.54,3}}{\rm{.48,4}}{\rm{.33,5}}{\rm{.19}}{\rm{.}}\end{array}\)

Since this sample has an even number of observations (\({\rm{n = 20}}\)), the values of interest are,

\(\begin{array}{c}{\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)^{{\rm{th}}}}{\rm{ = }}{\left( {\frac{{{\rm{20}}}}{{\rm{2}}}} \right)^{{\rm{th}}}}\\{\rm{ = 1}}{{\rm{0}}^{{\rm{th}}}}\\{\left( {\frac{{\rm{n}}}{{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{th}}}}{\rm{ = }}{\left( {\frac{{{\rm{20}}}}{{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{th}}}}\\{\rm{ = 1}}{{\rm{1}}^{{\rm{th}}}}\end{array}\)

In the ordered sample data, the tenth and eleventh values are highlighted in red. The sample median is the average of the two figures previously given.

\({\rm{ \tilde x = }}\frac{{{\rm{1}}{\rm{.2 + 1}}{\rm{.35}}}}{{\rm{2}}}{\rm{ = 1}}{\rm{.275}}{\rm{.}}\)

As a result, using the statistic median provided in the exercise, the estimate\({\rm{\sigma }}\)is,

\({\rm{\hat \sigma = 1}}{\rm{.275}}\)

It's worth noting that the sample standard deviation exceeds the estimate\({\rm{\hat \sigma }}\).