Chapter 6.1, Problem 4E

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

Prior to obtaining data, denote the beam strengths by X₁,..., X_m and the cylinder strengths by Y₁, ..., Y_n. Suppose that the X_i’s constitute a random sample from a distribution with mean μ₁ and standard deviation σ₁ and that the Y_i’s form a random sample (independent of the X_i’s) from another distribution with mean μ₂ and standard deviation σ₂.

1. a. Use rules of expected value to show that $\bar{X}$ - $\bar{Y}$ is an unbiased estimator of μ₁ – μ₂. Calculate the estimate for the given data.
2. 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.
3. c. Calculate a point estimate of the ratio σ₁/σ₂ of the two standard deviations.
4. d. Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference X – Y between beam strength and cylinder strength.

To determine

Show that $\overline{X}-\overline{Y}$ is an unbiased estimator of ${\mu }_1-{\mu }_2$.

Find the point estimate of the population mean for the data.

Answer to Problem 4E

A point estimate of the population mean for the data is –0.434.

Explanation of Solution

Given info:

Reference- Exercise 1: The data gives 20 sample observations for cylinders. The sample beam strengths are denoted by $X_1,X_2,\mathrm{...},X_m$, taken from a distribution with mean ${\mu }_1$ and standard deviation ${\sigma }_1$. The cylinder strengths are denoted by $Y_1,Y_2,\mathrm{...},Y_n$, taken from a distribution with mean ${\mu }_2$ and standard deviation ${\sigma }_2$. The two distributions are independent.

Calculation:

The sample means for beam strengths and cylinder strengths are respectively $\overline{X}$ and $\overline{Y}$.

An estimator is considered as an unbiased estimator, if and only if the expected value of the estimator is equal to the population characteristic it is estimating.

It is known that the sample mean is an unbiased estimator of the population mean. As a result, $\overline{X}$ and $\overline{Y}$ are unbiased estimators of ${\mu }_1$ and ${\mu }_2$ respectively, that is,

$E\left(\overline{X}\right)={\mu }_1$ and $E\left(\overline{Y}\right)={\mu }_2$.

Consider the expectation of $\overline{X}-\overline{Y}$.

$\begin{array}{c}E\left(\overline{X}-\overline{Y}\right)=E\left(\overline{X}\right)-E\left(\overline{Y}\right)\left(\text{by sum law of expectation}\right)\\ ={\mu }_1-{\mu }_2.\end{array}$

Hence, $\overline{X}-\overline{Y}$ is an unbiased estimator of ${\mu }_1-{\mu }_2$.

From Exercise 1, $\overline{X}=8.141$.

Now, sample mean,

$\overline{y}=\frac{1}{m}{\displaystyle \sum _{i=1}^m{y}_i}$.

For the data on cylinder strengths,

$\overline{y}=\frac{1}{20}{\displaystyle \sum _{i=1}^{20}{y}_i}$.

The following table shows the calculation for ${\displaystyle \sum _{i=1}^{20}{y}_i}$:

$y_i$
6.1
5.8
7.8
7.1
7.2
9.2
6.6
8.3
7.0
8.3
7.8
8.1
7.4
8.5
8.9
9.8
9.7
14.1
12.6
11.2
${\displaystyle \sum _{i=1}^{20}{y}_i}=171.5$

Thus,

$\begin{array}{c}\overline{y}=\frac{1}{20}{\displaystyle \sum _{i=1}^{20}{y}_i}\\ =\frac{1}{20}\times \left(171.5\right)\\ =\mathrm{8.575.}\end{array}$

Hence, a point estimate of the population mean for the data is:

$\begin{array}{c}\overline{X}-\overline{Y}=8.141-8.575\\ =\underset{\_}{-0.434}.\end{array}$

To determine

Find an expression for the variance and standard deviation (standard error) of the estimator in part a.

Calculate the estimated standard error.

Answer to Problem 4E

An expression for the variance of the estimator in part a is $\underset{\_}{V\left(\overline{X}-\overline{Y}\right)=\frac{{\sigma }^2{}_1}{m}+\frac{{\sigma }^2{}_2}{n}}$.

An expression for the standard deviation (standard error) of the estimator in part a is $\underset{\_}{{\sigma }_{\overline{X}-\overline{Y}}=\sqrt{\frac{{\sigma }^2{}_1}{m}+\frac{{\sigma }^2{}_2}{n}}}$.

The estimated standard error is 0.5691.

Explanation of Solution

Calculation:

Consider the variance of the estimator in part a, that is, $\overline{X}-\overline{Y}$.

$\begin{array}{c}V\left(\overline{X}-\overline{Y}\right)=V\left(\overline{X}\right)+V\left(\overline{Y}\right)\left(\text{due to independence of the two distributions}\right)\\ ={\sigma }^2{}_{\overline{X}}+{\sigma }^2{}_{\overline{Y}}.\end{array}$

For a sample of size m, the variance of the mean $\overline{x}$ is:

${\sigma }^2{}_{\overline{X}}=\frac{{\sigma }^2}{\sqrt{m}}$

where ${\sigma }^2$ is the variance of X.

Thus,

$\begin{array}{c}V\left(\overline{X}-\overline{Y}\right)={\sigma }^2{}_{\overline{X}}+{\sigma }^2{}_{\overline{Y}}\\ =\frac{{\sigma }^2{}_1}{\sqrt{m}}+\frac{{\sigma }^2{}_2}{\sqrt{n}}.\end{array}$

Hence, the variance of the estimator in part a is:

$\underset{\_}{V\left(\overline{X}-\overline{Y}\right)=\frac{{\sigma }^2{}_1}{m}+\frac{{\sigma }^2{}_2}{n}}$.

The corresponding standard deviation is:

$\begin{array}{c}{\sigma }_{\overline{X}-\overline{Y}}=\sqrt{V\left(\overline{X}-\overline{Y}\right)}\\ =\sqrt{\frac{{\sigma }^2{}_1}{m}+\frac{{\sigma }^2{}_2}{n}}.\end{array}$

Thus, an expression for the standard deviation (standard error) of the estimator in part a is:

$\underset{\_}{{\sigma }_{\overline{X}-\overline{Y}}=\sqrt{\frac{{\sigma }^2{}_1}{m}+\frac{{\sigma }^2{}_2}{n}}}$.

For a sample of size m from a normally distributed random variable, the standard error of the mean $\overline{x}$ is estimated as:

$\begin{array}{c}{\widehat{\sigma }}_{\overline{X}}=s_{\overline{X}}\\ =\frac{s}{\sqrt{m}},\end{array}$

where s is the sample standard deviation of the observations of X. A point estimate of the population standard deviation is the sample standard deviation.

From Exercise 1, $s_1=1.663$.

Now, sample variance is:

$\begin{array}{c}{s}^2=\frac{1}{m-1}{\left({\displaystyle \sum _{i=1}^m{x}_i}-\overline{x}\right)}^2\\ =\frac{1}{m-1}\left({\displaystyle \sum _{i=1}^m{x}_i{}^2}-m{\overline{x}}^2\right).\end{array}$

For the data,

$\begin{array}{c}{s}^2{}_2=\frac{1}{20-1}{\left({\displaystyle \sum _{i=1}^{20}{y}_i{}^2}-20{\overline{x}}^2\right)}^2\\ =\frac{1}{19}\left({\displaystyle \sum _{i=1}^{20}{y}_i{}^2}-\left(20\times {\left(8.575\right)}^2\right)\right)\\ =\frac{1}{19}\left({\displaystyle \sum _{i=1}^{20}{y}_i{}^2}-1,470.6125\right).\end{array}$

The following table shows the calculation for ${\displaystyle \sum _{i=1}^{20}{y}_i{}^2}$:

$y_i$	$y_i{}^2$
6.1	37.21
5.8	33.64
7.8	60.84
7.1	50.41
7.2	51.84
9.2	84.64
6.6	43.56
8.3	68.89
7.0	49.0
8.3	68.89
7.8	60.84
8.1	65.61
7.4	54.76
8.5	72.25
8.9	79.21
9.8	96.04
9.7	94.09
14.1	198.81
12.6	158.76
11.2	125.44
Total	${\displaystyle \sum _{i=1}^{20}{y}_i{}^2}=1,554.73$

Thus,

$\begin{array}{c}{s}^2{}_2=\frac{1}{19}\left({\displaystyle \sum _{i=1}^{20}{y}_i{}^2}-29.08\right)\\ =\frac{1}{19}\left(1,554.73-1,470.6125\right)\\ =\frac{84.1175}{19}\\ =\mathrm{4.4272.}\end{array}$

Thus, the estimated standard error is:

$\begin{array}{c}{\widehat{\sigma }}_{\overline{X}-\overline{Y}}=\sqrt{\frac{s^2{}_1}{m}+\frac{s^2{}_2}{n}}\\ =\sqrt{\frac{{\left(1.663\right)}^2}{27}+\frac{4.4272}{20}}\\ =\sqrt{0.1025+0.2214}\\ =\sqrt{0.3239}\end{array}$

$=\underset{\_}{0.5691}$.

To determine

Find a point estimate of the ratio $\frac{{\sigma }_1}{{\sigma }_2}$ of the two standard deviations.

State the estimator used.

Answer to Problem 4E

A point estimate of the ratio $\frac{{\sigma }_1}{{\sigma }_2}$ of the two standard deviations is 0.7904.

Explanation of Solution

Calculation:

A point estimate of the population standard deviation is the sample standard deviation. As a result, a point estimate of the ratio $\frac{{\sigma }_1}{{\sigma }_2}$ is:

$\frac{{\widehat{\sigma }}_1}{{\widehat{\sigma }}_2}=\frac{s_1}{s_2}$.

From part b, $s_1=1.663$. $s_1=1.663$ and $s^2{}_2=4.4272$. Thus,

$\begin{array}{c}{s}_2=\sqrt{s^2{}_2}\\ =\sqrt{4.4272}\\ =\mathrm{2.104.}\end{array}$

Hence, a point estimate of the ratio $\frac{{\sigma }_1}{{\sigma }_2}$ of the two standard deviations is:

$\begin{array}{c}\frac{s_1}{s_2}=\frac{1.663}{2.104}\\ =\underset{\_}{0.7904}.\end{array}$

To determine

Find a point estimate of the variance of the difference $X-Y$ between beam strength and cylinder strength, if a single beam and a single cylinder are randomly selected.

Answer to Problem 4E

A point estimate of the variance of the difference $X-Y$ between beam strength and cylinder strength, when a single beam and a single cylinder are randomly selected is 7.1928.

Explanation of Solution

Calculation:

For independent random variables X and Y, it is known that:

$V\left(X-Y\right)=V\left(X\right)+V\left(Y\right)$.

Here, $V\left(X\right)={\sigma }^2{}_1$ and $V\left(Y\right)={\sigma }^2{}_2$.

A point estimate of the variance of X is ${\widehat{\sigma }}^2{}_1=s^2{}_1$. A point estimate of the variance of Y is ${\widehat{\sigma }}^2{}_2=s^2{}_2$.

Thus, a point estimate of the variance of the difference $X-Y$ between beam strength and cylinder strength, when a single beam and a single cylinder are randomly selected is:

$\begin{array}{c}\widehat{V\left(X+Y\right)}=\widehat{V\left(X\right)}+\widehat{V\left(Y\right)}\\ ={\widehat{\sigma }}^2{}_1+{\widehat{\sigma }}^2{}_2\\ ={\left(1.663\right)}^2+4.4272\\ =2.7656+4.4272\end{array}$

     $=\underset{\_}{7.1928}$.