Question

A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data on time (sec) to complete the escape (“Oxygen Consumption and Ventilation During Escape from an Offshore Platform,” Ergonomics, 1997: 281–292):

389 356 359 363 375 424 325 394 402

373 373 370 364 366 364 325 339 393

392 369 374 359 356 403 334 397

a. Construct a stem-and-leaf display of the data. How does it suggest that the sample mean and median will compare?

b. Calculate the values of the sample mean and median.(Hint:\(\sum {{x_i} = } \)9638.)

c. By how much could the largest time, currently 424, be increased without affecting the value of the sample median? By how much could this value be decreased without affecting the value of the sample median?

d. What are the values of \(\bar x\)and \(\tilde x\), when the observations are re expressed in minutes?

Short answer

a.

The stem and leaf plot is given as,

32	55
33	49
34
35	6699
36	34469
37	03345
38	9
39	2347
40	23
41
42	4

b.

The sample mean is 370.692 sec.

The median value is 369.5 sec.

c. The largest value can be increased by any amount but can be decreased by 55 without affecting the value of the sample median.

d. The sample mean is 6.18 min.

The median is 6.16 min.

Step-by-step solution

Given information

The data on time (sec) to complete the escape is provided.

The size of the sample is 26.

Construction of a stem-and-leaf display

A stem-and-leaf display provides a visual representation of the dataset.

The steps to construct a stem-and-leaf display are as follows,

1. Select the leading digit for the stem and trailing digits for the leaves.

2 Represent the stem digits vertically and similarly the trailing digits corresponding to the stem digits.

3) Mention the units for the display.

The stem and leaf display for the provided scenario is,

32	55
33	49
34
35	6699
36	34469
37	03345
38	9
39	2347
40	23
41
42	4

Unit: 35|6=356

It can be observed that the distribution is slightly positively skewed. This implies that the mean value is greater than the median.

Computing the sample mean and median

Let X represents the time (sec).

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{389 + 356 + 359 + ... + 397}}{{26}}\\ &=& \frac{{9638}}{{26}}\\ &=& 370.692\end{array}\)

Thus, the sample mean is 370.692 sec.

The sample median is computed by first ordering the data in ascending order.

The data arranged in ascending order is as follows,

325	325	334	339	356	356	359	359	363	364	364	366
369	370	373	373	374	375	389	392	393	394	397	402
403	424

For the even number of observations, the median value is computed as,

\(\begin{array}{c}\tilde x &=& average\;of\;{\left( {\frac{n}{2}} \right)^{th}}and\;{\left( {\frac{n}{2} + 1} \right)^{th}}\;ordered\;values\\ &=& average\;of\;{\left( {\frac{{26}}{2}} \right)^{th}}and\;{\left( {\frac{{26}}{2} + 1} \right)^{th}}\;ordered\;values\\ &=& average\;of\;{\left( {13} \right)^{th}}and\;{\left( {14} \right)^{th}}\;ordered\;values\\ &=& \frac{{369 + 370}}{2}\\ &=& 369.5\end{array}\)

Thus, the median value is 369.5 sec.

Computing the increase in largest value without affecting the median

Referring to part b, the value of the sample median is 369.5.

The largest value in the dataset is 424.

The largest value can be increased by any amount as the median value is the average of 13^th and 14^th value in the provided scenario. So, the median will not be affected by any amount of increase in the largest value of the dataset.

Computing the decrease in largest value without affecting the median

The largest value cannot be less than the middle values so that the median value is remains unaffected.

Therefore, the largest value can be decreased by 55 (424-370) without affecting the median.

Thus, the largest value can be decreased by 55.

Computing the sample mean and median

d.

Referring to part b,

The sample mean is 370.692 sec.

The median value is 369.5.

When the observations are re-expressed in minutes, the sample mean is computed as,

\(\begin{array}{c}\frac{{\bar x}}{{60}} &=& \frac{{370.692}}{{60}}\\ &=& 6.18\;\min \end{array}\)

The median is computed as,

\(\begin{array}{c}\frac{{\tilde x}}{{60}} &=& \frac{{369.5}}{{60}}\\ &=& 6.16\;\min \end{array}\)

Therefore, the mean and median are 6.18 min and 6.16 min respectively.