91影视

The mean of the sample mean X that we just calculated is identical to the population mean. We just calculated the standard deviation of the sample mean X, which is the population standard deviation divided by the square root of the sample size: 10=20/2.

(a)

The following holds,

\(\begin{array}{l}{\rm{E(\bar X - \bar Y) = E(\bar X) - E(\bar Y)}}\\{\rm{ = E}}\left( {\frac{{\rm{1}}}{{\rm{m}}}\sum\limits_{{\rm{i = 1}}}^{\rm{m}} {{{\rm{X}}_{\rm{i}}}} } \right){\rm{ - E}}\left( {\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{n = 1}}}^{\rm{m}} {{{\rm{Y}}_{\rm{i}}}} } \right)\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{m}}}\sum\limits_{{\rm{i = 1}}}^{\rm{m}} {\rm{E}} \left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ - }}\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{E}} \left( {{{\rm{Y}}_{\rm{i}}}} \right){\rm{n}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{\rm{1}}}{{\rm{m}}}{\rm{ \times m \times E}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ - }}\frac{{\rm{1}}}{{\rm{n}}}{\rm{ \times n \times E}}\left( {{{\rm{Y}}_{\rm{1}}}} \right)\\{\rm{ = }}{{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}\end{array}\)

(1): The distributions of \({X_i}\)and \({Y_j}\)are identical.

This indicates that \(\bar X - \bar Y\)is an unbiased estimate of \({\mu _1} - {\mu _2}\).

The Sample Mean \(\bar x\)of observations \({x_1},{x_2}, \ldots ,{x_n}\)is calculated as follows:

\(\begin{array}{c}{\rm{\bar x = }}\frac{{{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + \ldots + }}{{\rm{x}}_{\rm{n}}}}}{{\rm{n}}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} \end{array}\)

The sample mean for the beam strengths data was calculated in exercise 1 and is barx=8.141. The sample mean for cylinder strengths is

\(\begin{array}{c}{\rm{\bar y = }}\frac{{\rm{1}}}{{{\rm{20}}}}{\rm{(6}}{\rm{.1 + 5}}{\rm{.8 + \ldots + 11}}{\rm{.12)}}\\{\rm{ = 8}}{\rm{.575}}\end{array}\)

Therefore, the estimate value is \(\begin{array}{c}{\rm{\bar x - \bar y = 8}}{\rm{.141 - 8}}{\rm{.575}}\\{\rm{ = 0}}{\rm{.434}}\end{array}\)

(b)

The following is true for the variance due to independence:

\(\begin{array}{c}{\rm{V(\bar X - \bar Y)}}\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{V(\bar X) + ( - 1}}{{\rm{)}}^{\rm{2}}}{\rm{V(\bar Y)}}\\{\rm{ = V}}\left( {\frac{{\rm{1}}}{{\rm{m}}}\sum\limits_{{\rm{i = 1}}}^{\rm{m}} {{{\rm{X}}_{\rm{i}}}} } \right){\rm{ + V}}\left( {\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{n = 1}}}^{\rm{m}} {{{\rm{Y}}_{\rm{i}}}} } \right)\end{array}\)

\(\begin{array}{c}\mathop {\rm{ = }}\limits^{{\rm{(2)}}} \frac{{\rm{1}}}{{{{\rm{m}}^{\rm{2}}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{m}} {\rm{V}} \left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ + }}\frac{{\rm{1}}}{{{{\rm{n}}^{\rm{2}}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{V}} \left( {{{\rm{Y}}_{\rm{i}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{\rm{1}}}{{{{\rm{m}}^{\rm{2}}}}}{\rm{ \times m \times V}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ + }}\frac{{\rm{1}}}{{{{\rm{n}}^{\rm{2}}}}}{\rm{ \times n \times V}}\left( {{{\rm{Y}}_{\rm{1}}}} \right)\\{\rm{ = }}\frac{{{\rm{\sigma }}_{\rm{1}}^{\rm{2}}}}{{\rm{m}}}{\rm{ + }}\frac{{{\rm{\sigma }}_{\rm{2}}^{\rm{2}}}}{{\rm{n}}}\end{array}\)

(2): Independence.

The standard deviation is

\(\begin{array}{l}{{\rm{\sigma }}_{{\rm{\bar X}}}}{\rm{ - \bar Y = }}\sqrt {{\rm{V(\bar X - \bar Y)}}} \\{\rm{ = }}\sqrt {\frac{{{\rm{\sigma }}_{\rm{1}}^{\rm{2}}}}{{\rm{m}}}{\rm{ + }}\frac{{{\rm{\sigma }}_{\rm{2}}^{\rm{2}}}}{{\rm{n}}}} \end{array}\)

The sample variances \(\sigma _1^2\)and \(\sigma _2^2\)are required in order to calculate the estimate. For the first time, the sample standard deviation was calculated.

\({{\rm{s}}_{\rm{1}}}{\rm{ = 1}}{\rm{.66}}\)

The Sample Variance is \({s^2}\)is\({s^2} = \frac{1}{{n - 1}} \cdot {S_{xx}}.\)

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 Sample Standard Deviation \({\rm{s}}\)is

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

The squared data points, \(y_i^2\),are

\(37.21,33.64,60.84,50.41,51.84,84.64,43.56,68.89,49,68.89,60.84,65.61,54.76,72.25,79.21,96.04,94.09,198.81,158.76,125.44,\)

Thus, the \({S_{yy}}\)is

\(\begin{array}{c}{{\rm{S}}_{{\rm{yy}}}}{\rm{ = }}\sum {{\rm{y}}_{\rm{i}}^{\rm{2}}} {\rm{ - }}\frac{{\rm{1}}}{{\rm{n}}}{\rm{ \times }}{\left( {\sum {{{\rm{y}}_{\rm{i}}}} } \right)^{\rm{2}}}\\{\rm{ = 37}}{\rm{.21 + 33}}{\rm{.64 + \ldots + 125}}{\rm{.44 - }}\frac{{\rm{1}}}{{{\rm{20}}}}{\rm{ \times (6}}{\rm{.1 + 5}}{\rm{.8 + \ldots + 11}}{\rm{.2}}{{\rm{)}}^{\rm{2}}}\\{\rm{ = 1554}}{\rm{.73 - }}\frac{{\rm{1}}}{{{\rm{16}}}}{\rm{ \times 171}}{\rm{.}}{{\rm{5}}^{\rm{2}}}\\{\rm{ = 84}}{\rm{.1175}}\end{array}\)

And the sample variance is

\(\begin{array}{c}{\rm{s}}_{\rm{2}}^{\rm{2}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{16 - 1}}}}{\rm{ \times 84}}{\rm{.1175}}\\{\rm{ = 4}}{\rm{.427}}\end{array}\)
Also, the sample standard deviation is

\(\begin{array}{c}{\rm{s = }}\sqrt {{{\rm{s}}^{\rm{2}}}} \\{\rm{ = }}\sqrt {{\rm{4}}{\rm{.427}}} \\{\rm{ = 2}}{\rm{.104}}{\rm{.}}\end{array}\)

As a result, the standard error is predicted to be \(\begin{array}{c}{{\rm{s}}_{{\rm{\bar X - \bar Y}}}}{\rm{ = }}\sqrt {\frac{{{\rm{s}}_{\rm{1}}^{\rm{2}}}}{{\rm{m}}}{\rm{ + }}\frac{{{\rm{s}}_{\rm{2}}^{\rm{2}}}}{{\rm{n}}}} \\{\rm{ = }}\sqrt {\frac{{{\rm{1}}{\rm{.6}}{{\rm{6}}^{\rm{2}}}}}{{{\rm{27}}}}{\rm{ + }}\frac{{{\rm{2}}{\rm{.10}}{{\rm{4}}^{\rm{2}}}}}{{{\rm{20}}}}} \\{\rm{ = 0}}{\rm{.5687}}{\rm{.}}\end{array}\)

Hence, the required result is\({\rm{0}}{\rm{.5687}}\).

(c)

The point estimate of the ratio \({\sigma _1}/{\sigma _2}\)

\(\begin{array}{c}\frac{{{s_1}}}{{{s_2}}} = \frac{{1.66}}{{2.104}}\\ = 0.789.\end{array}\)

Thus, the required result is\(0.789\).

(d)

The variance of \(X - Y\)is

\(\begin{array}{c}{\rm{V(X - Y)}}\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{V(X) + ( - 1}}{{\rm{)}}^{\rm{2}}}{\rm{V(Y)}}\\{\rm{ = \sigma }}_{\rm{1}}^{\rm{2}}{\rm{ + \sigma }}_{\rm{2}}^{\rm{2}}\end{array}\)

which yields a point estimate of the variance

\(\begin{array}{c}{\rm{s}}_{\rm{1}}^{\rm{2}}{\rm{ + s}}_{\rm{2}}^{\rm{2}}{\rm{ = 1}}{\rm{.6}}{{\rm{6}}^{\rm{2}}}{\rm{ + 2}}{\rm{.10}}{{\rm{4}}^{\rm{2}}}\\{\rm{ = 7}}{\rm{.1824}}\end{array}\)

Hence, the point estimate of the variance is\({\rm{7}}{\rm{.1824}}\).