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A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the following data on time (in seconds) to complete the escape ("Oxygen Consumption and Ventilation During Escape from an Offshore Platform," Ergonomics [1997]: \(281-292\) ): \(\begin{array}{lllllllll}389 & 356 & 359 & 363 & 375 & 424 & 325 & 394 & 402 \\\ 373 & 373 & 370 & 364 & 366 & 364 & 325 & 339 & 393 \\ 392 & 369 & 374 & 359 & 356 & 403 & 334 & 397 & \end{array}\) a. Construct a dotplot of the data. Will the mean or the median be larger for this data set? b. Calculate the values of the mean and median. c. By how much could the largest time be increased without affecting the value of the sample median? By how much could this value be decreased without affecting the value of the median?

Short Answer

Expert verified
a. The mean would be larger than the median given our distribution of values. b. After calculating, we find the mean is approximately 372.08 while the median is 371. c. The maximum value can be increased or decreased infinitesimally without affecting the median.

Step by step solution

01

Construct a Dotplot

The Dotplot is a graphical representation of data using dots. We order the data from least to most and represent each value as a dot over a number line.
02

Compare Mean and Median Estimate

Looking at the dotplot we see that our data is a bit skewed to the right. Since the mean is affected by extreme values in the data set (in this case, high escape times) and the median is not, the mean will likely be higher than the median.
03

Calculate Mean

In order to find the mean of the data, we must add all of the times together and then divide by the total number of times. So we calculate \( (389+356+359+363+375+424+325+394+402+373+373+370+364+366+364+325+339+393+392+369+374+359+356+403+334+397) / 26 \)
04

Calculate Median

To find the median, we first need to order the data from least to greatest. If our number of elements were odd, we would choose the middle element. However, since we have an even number of elements, we will need to take the average of the middle two numbers, which will be the 13th and 14th value when ordered.
05

Determine change in maximum value without affecting median

This largest time value can change as much as it wants without affecting the median since the median is determined from the middle values. As long as the changes don't push other values to become new max or min values, this will not change our median.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dotplot Construction
Visualizing data can provide immediate insights that might not be apparent from looking at raw numbers alone. A dotplot is a simple graphical tool used in statistics education to help students understand the distribution of a dataset. To construct a dotplot, values are listed along a number line. Each observation in the dataset is then marked above its corresponding value with a dot. Multiple dots stacked vertically represent frequency of occurrences for the same value.

For the offshore oil workers' escape times, you would start by drawing a horizontal line and marking it with time intervals based on the data's range. Then, place a dot for each worker's time directly above the appropriate time interval on the line. If another worker has the same time, you would stack the dot above the first one, and so on. This visual representation allows us to quickly grasp how the times are distributed and to spot any clustering or outliers in the data.
Mean vs Median
In any dataset, central tendency can be measured in different ways, the most common being the mean and median. The mean is simply the average of all values, calculated by summing up the data and dividing by the number of observations. It is very sensitive to outliers, which can skew the mean away from the center of the data.

The median, on the other hand, is the middle value when the data is ordered from smallest to largest. If there is an even number of observations, the median is the average of the two middle values. Unlike the mean, the median is resistant to outliers and remains unchanged even if the largest or smallest values are altered, provided they don't cross the median's position in the ordered list. This resistance to skew makes the median a better measure of central tendency for skewed distributions.
Calculate Mean
Calculating the mean is a fundamental skill in statistics. To compute the mean time for the oil workers' escape exercise, sum all the individual times and then divide by the number of workers. This calculation will give you the average time taken by a worker to complete the escape exercise. In mathematical terms, the mean, often denoted as \( \bar{x} \), is found using the formula: \[ \bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_i \], where \( x_i \) represents each value in the dataset and \( n \) is the total number of observations. This aggregation of data provides a single value that best represents the center of the data.
Calculate Median
The median is another measure of central tendency that's particularly useful for skewed data, as it isn't affected by extreme values. To find the median of the escape times, we first sort the times from the shortest to the longest. Since there are an even number of workers, we calculate the median by taking the average of the 13th and 14th values in the ordered list. This position ensures that half of the workers have completion times above the median and half below, making the median a measurement of the data's central point that divides the dataset into two equally sized groups.
Data Analysis
In the context of the offshore oil workers' exercise, data analysis involves examining the escape times to draw conclusions. After computing measures like the mean and median, it is worthwhile to reflect on their meanings. While the mean gives us an average time, the median provides a value that isn't swayed by the slower or faster escape times, which could be outliers. Further analysis might include looking at the range of the data, calculating the mode, or determining the standard deviation, all of which can give more insight into the data's variability and the reliability of the average escape times. The robustness of the median is also worth noting; the largest value can increase or decrease significantly without affecting the median, as long as the reordered list's central values remain unchanged, preserving the integrity of the dataset's middle ground.

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Most popular questions from this chapter

The accompanying data on number of cell phone minutes used in one month are consistent with summary statistics from a marketing study of San Diego residents (Tele-Truth, March, 2009).$$ \begin{array}{rrrrrrrrrr} 189 & 0 & 189 & 177 & 106 & 201 & 0 & 212 & 0 & 306 \\ 0 & 0 & 59 & 224 & 0 & 189 & 142 & 83 & 71 & 165 \\ 236 & 0 & 142 & 236 & 130 & & & & & \end{array} $$ Calculate and interpret the values of the median and the interquartile range.

The accompanying data are a subset of data read from a graph in the paper "Ladies First? A Field Study of Discrimination in Coffee Shops" (Applied Economics [April, 2008]). The data are the waiting times (in seconds) between ordering and receiving coffee for 19 female customers at a Boston coffee shop. \(\begin{array}{llllllll}0 & 80 & 80 & 100 & 100 & 100 & 120 & 120\end{array}\) \(\begin{array}{llllllll}120 & 140 & 140 & 150 & 160 & 180 & 200 & 200\end{array}\) $$ \begin{array}{lll} 220 & 240 & 380 \end{array} $$ a. Calculate the mean and standard deviation for this data set. b. Delete the observation of 380 and recalculate the mean and standard deviation. How do these values compare to the values calculated in Part (a)? What does this suggest about using the mean and standard deviation as measures of center and spread for a data set with outliers?

Data on tipping percent for 20 restaurant tables, consistent with summary statistics given in the paper "Beauty and the Labor Market: Evidence from Restaurant Servers"(unpublished manuscript by Matt Parrett, 2007), are: \(\begin{array}{rrrrrrr}0.0 & 5.0 & 45.0 & 32.8 & 13.9 & 10.4 & 55.2 \\ 50.0 & 10.0 & 14.6 & 38.4 & 23.0 & 27.9 & 27.9 \\ 105.0 & 19.0 & 10.0 & 32.1 & 11.1 & 15.0 & \end{array}\) a. Calculate the mean and standard deviation for this data set. b. Delete the observation of 105.0 and recalculate the mean and standard deviation. How do these values compare to the values from Part (a)? What does this suggest about using the mean and standard deviation as measures of center and spread for a data set with outliers?

USA Today (May 9,2006 ) published the accompanying weekday circulation numbers for the top 20 newspapers in the country for the 6 -month period ending March 31,2006: \(\begin{array}{rrrrr}2,272,815 & 2,049,786 & 1,142,464 & 851,832 & 724,242 \\\ 708,477 & 673,379 & 579,079 & 513,387 & 438,722 \\ 427,771 & 398,329 & 398,246 & 397,288 & 365,011 \\ 362,964 & 350,457 & 345,861 & 343,163 & 323,031\end{array}\) Explain why the average may not be the best measure of a typical value for this data set.

The paper "Total Diet Study Statistics on Element Results" (Food and Drug Administration, April 25,2000 ) gave information on sodium content for various types of foods. Twenty-six brands of tomato catsup were analyzed. Data consistent with summary quantities in the paper are Sodium content ( \(\mathrm{mg} / \mathrm{kg}\) ) \(\begin{array}{lrrrrr}12,148 & 10,426 & 10,912 & 9,116 & 13,226 & 11,663 \\ 11,781 & 10,680 & 8,457 & 10,788 & 12,605 & 10,591 \\\ 11,040 & 10,815 & 12,962 & 11,644 & 10,047 & \\ 10,478 & 10,108 & 12,353 & 11,778 & 11,092 & \\ 11,673 & 8,758 & 11,145 & 11,495 & & \end{array}\) Calculate and interpret the values of the quartiles and the interquartile range. (Hint: See Example 3.9 )

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