Question

The article from which the data in Exercise 1 was extracted also gave the accompanying strength observations for cylinders:

\(\begin{array}{l}\begin{array}{*{20}{r}}{{\rm{6}}{\rm{.1}}}&{{\rm{5}}{\rm{.8}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{7}}{\rm{.1}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{9}}{\rm{.2}}}&{{\rm{6}}{\rm{.6}}}&{{\rm{8}}{\rm{.3}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{8}}{\rm{.3}}}\\{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\\\begin{array}{*{20}{l}}{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\end{array}\)

Prior to obtaining data, denote the beam strengths by X1, … ,Xm and the cylinder strengths by Y1, . . . , Yn. Suppose that the Xi ’s constitute a random sample from a distribution with mean m1 and standard deviation s1 and that the Yi ’s form a random sample (independent of the Xi ’s) from another distribution with mean m2 and standard deviation\({{\rm{\sigma }}_{\rm{2}}}\).

a. Use rules of expected value to show that \({\rm{\bar X - \bar Y}}\)is an unbiased estimator of \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}\). Calculate the estimate for the given data.

b. Use rules of variance from Chapter 5 to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a), and then compute the estimated standard error.

c. Calculate a point estimate of the ratio \({{\rm{\sigma }}_{\rm{1}}}{\rm{/}}{{\rm{\sigma }}_{\rm{2}}}\)of the two standard deviations.

d. Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference \({\rm{X - Y}}\) between beam strength and cylinder strength.

Short answer

The estimate value is \({\rm{0}}{\rm{.434}}\)

The standard error is predicted to be \({\rm{0}}{\rm{.5687}}{\rm{.}}\)

The point estimate of the ratio \({\sigma _1}/{\sigma _2}\) is \(0.789\).

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

Step-by-step solution

Concept introduction

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.

Estimation of the sample mean for the beam strengths

(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 \(\begin{array}{c}{\rm{\bar x - \bar y = 8}}{\rm{.141 - 8}}{\rm{.575}}\\{\rm{ = 0}}{\rm{.434}}\end{array}\)

Calculation of standard deviation

(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\({\rm{0}}{\rm{.5687}}\).

Finding the point estimate of the ratio

(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\(0.789\).

Finding the point estimate of the variance

(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\({\rm{7}}{\rm{.1824}}\).