Chapter 7.3, Problem 36E

A normal probability plot of the n = 26 observations on escape time given in Exercise 36 of Chapter 1 shows a substantial linear pattern; the sample mean and sample standard deviation are 370.69 and 24.36, respectively.

1. a. Calculate an upper confidence bound for population mean escape time using a confidence level of 95%.
2. b. Calculate an upper prediction bound for the escape time of a single additional worker using a prediction level of 95%. How does this bound compare with the confidence bound of part (a)?
3. c. Suppose that two additional workers will be chosen to participate in the simulated escape exercise. Denote their escape times by X₂₇ and X₂₈, and let ${\bar{X}}_{\text{new}}$ denote the average of these two values. Modify the formula for a PI for a single x value to obtain a PI for ${\bar{X}}_{\text{new}}$, and calculate a 95% two-sided interval based on the given escape data.

To determine

Find the 95% upper confidence bound of the population mean escape time of additional workers.

Answer to Problem 36E

The 95% upper confidence bound of the population mean escape time of additional workers is 378.8499.

Explanation of Solution

Given info:

The escape times for 26 randomly selected additional workers is specified in an article. The statistics of the sample are $\overline{x}=370.69$ and $s=24.36$. The population of escape times follow normal distribution.

Calculation:

The following assumptions are required for the confidence interval about mean using t-distribution to be valid.

Requirements for t-distribution to construct confidence interval about mean:

• The sample must be drawn using simple random sampling.
• The population standard deviation is unknown.
• Either the population must be approximately normal or the sample size must be greater than 30.

Thus, the requirements must be satisfied for the confidence interval about mean using t-distribution to be valid.

Here, the sample is drawn randomly and the population standard deviation is unknown.

Even though the sample size is not greater than 30, the distribution of the population from which the sample is drawn is approximately normal.

Therefore, application of t-distribution is valid in constructing confidence interval about mean.

Critical value:

For 95% confidence level

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\end{array}$

Degrees of freedom:

The number of samples that are sampled is $n=26$

The degrees of freedom is,

$\begin{array}{c}d.f=n-1\\ =26-1\\ =25\end{array}$

From Table A.5 of the t-distribution in Appendix A, the critical value corresponding to the right tail area 0.05 and 25 degrees of freedom is 1.708.

Thus, the critical value is $\left(t_{\alpha }\right)=1.708$.

Point estimate:

Point estimate is the single value obtained from point estimation on a set of sample data as a parameter for the population.

The point estimate of the population mean $\mu$ is the sample mean $\left(\overline{x}\right)$ and it is an unbiased estimator of population mean.

The point estimate of the population mean escape time of additional workers is $\overline{x}=370.69$.

Upper confidence bound:

The upper bound of the confidence interval is,

$\begin{array}{c}\text{Upper}\text{bound}=\overline{x}+t_{\alpha }\times \frac{s}{\sqrt{n}}\\ =370.69+1.708\times \frac{24.36}{\sqrt{26}}\\ =370.69+\left(8.1599\right)\\ =378.8499\end{array}$

Thus, the upper confidence bound of the population mean escape time of additional workers at 95% confidence level is 378.8499.

Interpretation:

There is 95% confident that the upper confidence bound of the population mean escape time of additional workers will be less than 378.8499.

To determine

Find the 95% upper prediction bound for the escape time of a single additional worker.

Compare the upper prediction bound with the upper confidence bound obtained in part (a).

Answer to Problem 36E

The 95% upper prediction bound for the escape time of a single additional worker is 413.0895.

The 95% upper prediction bound is greater than the 95% upper confidence bound obtained in part (a)

Explanation of Solution

Calculation:

Prediction interval for a single future value:

Prediction interval is used to predict a single value of the focus variable that is to be observed at some future time. In other words it can be said that the prediction interval gives a single future value rather than estimating the mean value of the variable.

The prediction interval is obtained as,

$P.I=\overline{x}\pm t_{\frac{\alpha }{2},n-1}\times s\sqrt{1+\frac{1}{n}}$

Critical value:

For 95% confidence level

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\end{array}$

Degrees of freedom:

The number of samples that are sampled is $n=26$

The degrees of freedom is,

$\begin{array}{c}d.f=n-1\\ =26-1\\ =25\end{array}$

From Table A.5 of the t-distribution in Appendix A, the critical value corresponding to the right tail area 0.05 and 25 degrees of freedom is 1.708.

Thus, the critical value is $\left(t_{\alpha }\right)=1.708$.

Upper prediction bound:

The upper bound of the prediction interval is,

$\begin{array}{c}\text{Upper}\text{bound}=\overline{x}+t_{\alpha ,n-1}\times s\sqrt{1+\frac{1}{n}}\\ =370.69+1.708\times 24.36\sqrt{1+\frac{1}{26}}\\ =370.69+\left(42.3995\right)\\ =413.0895\end{array}$

Thus, the upper prediction bound for the escape time of a single additional worker at 95% confidence level is 413.0895.

Interpretation:

There is 95% confident that the upper prediction bound for the escape time of a single additional worker will be less than 413.0895.

Comparison:

The 95% upper confidence bound of the population mean escape time of additional workers is 378.8499.

The 95% upper prediction bound for the escape time of a single additional worker is 413.0895.

Here, $413.0895>378.8499$.

Thus, the 95% upper prediction bound is greater than the 95% upper confidence bound obtained in part (a)

To determine

Find the modified formula for prediction interval to obtain prediction interval for ${\overline{X}}_{\text{new}}$.

Find the 95% upper prediction interval for ${\overline{X}}_{\text{new}}$ of escape data.

Answer to Problem 36E

The modified formula for prediction interval to obtain prediction interval for ${\overline{X}}_{\text{new}}$ is $P.I=\overline{x}\pm t_{\frac{\alpha }{2},n-1}\times s\sqrt{\frac{1}{2}+\frac{2}{n}}$.

The 95% upper prediction interval for ${\overline{X}}_{\text{new}}$ of escape data is $\underset{\_}{\left(332.5743,408.8057\right)}$.

Explanation of Solution

Given info:

Two additional workers will be chosen to participate in the simulated escape exercise and the escapes times of the two workers are $X_{27}$ and $X_{28}$. The average of the two values is ${\overline{X}}_{\text{new}}$.

Calculation:

Prediction interval for a single future value:

Prediction interval is used to predict a single value of the focus variable that is to be observed at some future time. In other words it can be said that the prediction interval gives a single future value rather than estimating the mean value of the variable.

General setup for prediction interval:

Let $X_1,X_2,\mathrm{...},X_n$ be a random sample from a normal population with mean $\mu$ and variance ${\sigma }^2$.

That is, $X_i\sim N\left(\mu ,{\sigma }^2\right)$.

To predict a value of single future observation $X_{n+1}$ from the above sample the point prediction is $\overline{X}$ and the prediction error $\overline{X}-X_{n+1}$.

Therefore, from the properties of normal distribution. The prediction error $\overline{X}-X_{n+1}$ will also be normally distributed with mean “0” and variance ${\sigma }^2\sqrt{\frac{1}{2}+\frac{2}{n}}$.

That is, $\overline{X}-X_{n+1}\sim N\left(0,{\sigma }^2\sqrt{\frac{1}{2}+\frac{2}{n}}\right)$

From the properties of standard normal distribution the prediction interval for $X_{n+1}$ is,

$P.I=\overline{x}\pm t_{\frac{\alpha }{2},n-1}\times s\sqrt{1+\frac{1}{n}}$

Here, the objective is to find the prediction interval for the single future observation ${\overline{X}}_{\text{new}}$ of the escape times.

It is given that the population of escape times follow normal distribution.

Here, ${\overline{X}}_{\text{new}}$ is the average of $X_{27}$ and $X_{28}$.

That is, ${\overline{X}}_{\text{new}}=\frac{X_{27}+X_{28}}{2}$

Here, the prediction error is $\overline{X}-{\overline{X}}_{\text{new}}$.

From the properties of normal distribution, if $X_i\sim N\left(\mu ,{\sigma }^2\right)$ then $\overline{X}\sim N\left(\mu ,\frac{{\sigma }^2}{n}\right)$.

Mean of prediction interval:

The mean of the prediction error $\overline{X}-{\overline{X}}_{\text{new}}$ is,

$\begin{array}{c}E\left(\overline{X}-{\overline{X}}_{\text{new}}\right)=E\left(\overline{X}\right)-E\left({\overline{X}}_{\text{new}}\right)\\ =\mu -\mu \\ =0\end{array}$

Thus, the mean of the prediction error is “0”.

Variance of prediction interval:

The variance of the prediction error $\overline{X}-{\overline{X}}_{\text{new}}$ is,

$\begin{array}{c}V\left(\overline{X}-{\overline{X}}_{\text{new}}\right)=V\left(\overline{X}\right)+V\left({\overline{X}}_{\text{new}}\right)\\ =V\left(\overline{X}\right)+V\left(\frac{X_{27}+X_{28}}{2}\right)\\ =V\left(\overline{X}\right)+\frac{1}{4}V\left(X_{27}\right)+\frac{1}{4}V\left(X_{28}\right)\\ =\frac{{\sigma }^2}{n}+\frac{{\sigma }^2}{4}+\frac{{\sigma }^2}{4}\end{array}$

                      $={\sigma }^2\left(\frac{1}{2}+\frac{1}{n}\right)$

Thus, the mean of the prediction error is ${\sigma }^2\left(\frac{1}{2}+\frac{1}{n}\right)$.

Test statistic:

The test statistic is,

$T=\frac{\overline{X}-{\overline{X}}_{new}}{s\sqrt{\left(\frac{1}{2}+\frac{1}{n}\right)}}\sim t_{n-1}$

From the properties of prediction interval is,

$P.I=\overline{x}\pm t_{\frac{\alpha }{2},n-1}\times s\sqrt{\frac{1}{2}+\frac{2}{n}}$

Thus, the modified prediction interval is,

$P.I=\overline{x}\pm t_{\frac{\alpha }{2},n-1}\times s\sqrt{\frac{1}{2}+\frac{2}{n}}$

Critical value:

For 95% confidence level

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Degrees of freedom:

The number of samples that are sampled is $n=26$

The degrees of freedom is,

$\begin{array}{c}d.f=n-1\\ =26-1\\ =25\end{array}$

From Table A.5 of the t-distribution in Appendix A, the critical value corresponding to the right tail area 0.025 and 25 degrees of freedom is 2.060.

Thus, the critical value is $\left(t_{\alpha }\right)=2.060$.

Prediction bound:

The 95% prediction interval is,

$\begin{array}{c}P.I=\left(\overline{x}-t_{\alpha ,n-1}\times s\sqrt{\frac{1}{2}+\frac{2}{n}},\overline{x}+t_{\alpha ,n-1}\times s\sqrt{\frac{1}{2}+\frac{2}{n}}\right)\\ =\left(370.69-2.060\times 24.36\sqrt{\frac{1}{2}+\frac{2}{26}},370.69+2.060\times 24.36\sqrt{\frac{1}{2}+\frac{2}{26}}\right)\\ =\left(370.69-\left(38.1157\right),370.69+\left(38.1157\right)\right)\\ =\left(332.5743,408.8057\right)\end{array}$

Thus, the upper prediction interval for ${\overline{X}}_{\text{new}}$ of escape data at 95% confidence level is $\underset{\_}{\left(332.5743,408.8057\right)}$

Interpretation:

There is 95% confident that the future observation of ${\overline{X}}_{\text{new}}$ will lie between 332.5743 and 408.8057.