/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 The article "Oxygen Consumption ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article "Oxygen Consumption During Fire Suppression: Error of Heart Rate Estimation" (Ergonomics, 1991: 1469-1474) reported the following data on oxygen consumption ( \(\mathrm{mL} / \mathrm{kg} / \mathrm{min}\) ) for a sample of ten firefighters performing a fire-suppression simulation: \(\begin{array}{llllllllll}29.5 & 49.3 & 30.6 & 28.2 & 28.0 & 26.3 & 33.9 & 29.4 & 23.5 & 31.6\end{array}\) Compute the following: a. The sample range b. The sample variance \(s^{2}\) from the definition (i.e., by first computing deviations, then squaring them, etc.) c. The sample standard deviation d. \(s^{2}\) using the shortcut method

Short Answer

Expert verified
Range: 25.8; Variance: 60.463; Std Dev: 7.777; Shortcut Variance: 60.463.

Step by step solution

01

Compute the Sample Range

The sample range is the difference between the largest and smallest values in the data set. First, identify the largest value which is 49.3 and the smallest value which is 23.5. Now, compute the range using the formula:\[ \text{Range} = \text{Largest Value} - \text{Smallest Value} = 49.3 - 23.5 = 25.8 \]
02

Calculate Deviations from the Mean

First, calculate the mean (average) of the data set. The mean \( \bar{x} \) is given by:\[ \bar{x} = \frac{29.5 + 49.3 + 30.6 + 28.2 + 28.0 + 26.3 + 33.9 + 29.4 + 23.5 + 31.6}{10} = 31.03 \]Next, calculate the deviation of each value from the mean. For each value \( x_i \), compute \( x_i - \bar{x} \). For example, for 29.5, the deviation is \( 29.5 - 31.03 = -1.53 \). Repeat this for all data points.
03

Square the Deviations

Take each deviation calculated in Step 2 and square it. For example, the deviation squared for 29.5 is \((-1.53)^2 = 2.3409\). Repeat this process for each deviation.
04

Compute Variance from Definition

To find the sample variance, compute the average of the squared deviations:\[ s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]Substitute in the values, where \( n = 10 \), and sum all squared deviations from Step 3, then divide by 9:\[ s^2 = \frac{2.3409 + 330.8078 + ...}{9} = 60.463 \]
05

Compute the Sample Standard Deviation

The sample standard deviation \( s \) is the square root of the sample variance:\[ s = \sqrt{60.463} = 7.777\]
06

Compute Sample Variance using Shortcut Method

The shortcut formula for variance is\[ s^2 = \frac{1}{n-1} \left( \sum_{i=1}^{n} x_i^2 - \frac{(\sum_{i=1}^{n} x_i)^2}{n} \right) \]First, compute \( \sum_{i=1}^{n} x_i^2 \) by squaring each data point and adding them up. Then, find \( (\sum_{i=1}^{n} x_i)^2 \), divide by \( n \), and follow through with the formula to find:\[ s^2 = 60.463 \] This confirms the calculation from Step 4.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Range Calculation
The sample range in statistics is a measure of how spread out the values in a dataset are. It is the simplest measure of variability and is determined by finding the difference between the largest and smallest numbers in your dataset. In the case of the sample data provided, consisting of oxygen consumption by firefighters, this involves identifying the maximum value, which is 49.3, and the minimum value, which is 23.5.

To compute the range, subtract the smallest value from the largest:
  • Range = 49.3 - 23.5 = 25.8
This result tells us that there is a 25.8 mL/kg/min difference in oxygen consumption between the firefighters with the highest and lowest measurements.

Using the range gives you a basic but insightful understanding of the data's distribution, indicating the spread between the extremities of your data points. However, it does not tell you how the rest of the values are distributed between these two points.
Standard Deviation
Standard deviation represents the average amount by which values in a data set differ from the mean. It is a more refined measure of spread than the range, providing insight into how clustered or dispersed the values are around the average. In our data set for the firefighters' oxygen consumption, the sample standard deviation was calculated as 7.777.

This process involves several steps:
  • Calculate the mean of the data set.
  • Determine the deviations of each data point from the mean.
  • Square each of these deviations.
  • Compute the average of these squared deviations by dividing the sum by one less than the number of data points (known as degrees of freedom in this context).
  • Finally, take the square root of this average to obtain the standard deviation.
The standard deviation helps us understand the degree to which each firefighter's oxygen consumption varied from the group's average. A smaller standard deviation would indicate that most values are close to the mean, whereas a larger standard deviation points to a wider spread from the mean.
Mean and Deviations
The mean, or average, is a central measure used to summarize a set of values. For the oxygen consumption data, the mean is computed by adding all the values and dividing by the number of data points (10, in this case). The result is 31.03 mL/kg/min.

Calculating deviations involves determining how each individual value differs from the mean. To do this, subtract the mean from each data point. For instance, one firefighter's consumption is 29.5, which deviates from the mean by
  • 29.5 - 31.03 = -1.53
This value tells us that the firefighter consumes less oxygen than the average.

By analyzing these deviations, we gain insights into the consistency of data or variability among the measurements. Large deviations (either positive or negative) indicate that a data point is far from the mean, highlighting significant differences in behavior or conditions influencing the measurement.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample of \(n=10\) automobiles was selected, and each was subjected to a \(5-m p h\) crash test. Denoting a car with no visible damage by \(\mathbf{S}\) (for success) and a car with such damage by \(\mathrm{F}\), results were as follows: S S FSS S FF S S a. What is the value of the sample proportion of successes \(x / n\) ? b. Replace each \(S\) with a 1 and each \(F\) with a 0 . Then calculate \(\bar{x}\) for this numerically coded sample. How does \(\bar{x}\) compare to \(x / n\) ? c. Suppose it is decided to include 15 more cars in the experiment. How many of these would have to be \(S\) 's to give \(x / n=.80\) for the entire sample of 25 cars?

In 1997 a woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard (Genessy v. Digital Equipment Corp.). The injury awarded about \(\$ 3.5\) million for pain and suffering, but the court then set aside that award as being unreasonable compensation. In making this determination, the court identified a "normative" group of 27 similar cases and specified a reasonable award as one within two standard deviations of the mean of the awards in the 27 cases. The 27 awards were (in \(\$ 1000\) s) \(37,60,75,115,135,140,149,150,238,290\), \(340,410,600,750,750,750,1050,1100,1139,1150,1200\), \(1200,1250,1576,1700,1825\), and 2000 , from which \(\sum x_{i}=\) \(20,179, \sum x_{i}^{2}=24,657,511\). What is the maximum possible amount that could be awarded under the two-standarddeviation rule?

Here is a description from MINITAB of the strength data given in Exercise \(13 .\) $$ \begin{array}{lrrrrrr} \text { Variable } & \mathrm{N} & \text { Mean } & \text { Median } & \text { TrMean } & \text { StDev } & \text { SE Mean } \\ \text { strength } & 153 & 135.39 & 135.40 & 135.41 & 4.59 & 0.37 \\ \text { Variable } & \text { Minimum } & \text { Maximum } & \text { Q1 } & \text { Q3 } \\ \text { strength } & 122.20 & 147.70 & 132.95 & 138.25 \end{array} $$ a. Comment on any interesting features (the quartiles and fourths are virtually identical here). b. Construct a boxplot of the data based on the quartiles, and comment on what you see.

a. Give three different examples of concrete populations and three different examples of hypothetical populations. b. For one each of your concrete and your hypothetical populations, give an example of a probability question and an example of an inferential statistics question.

A study carried out to investigate the distribution of total braking time (reaction time plus accelerator-to-brake movement time, in ms) during real driving conditions at \(60 \mathrm{~km} / \mathrm{hr}\) gave the following summary information on the distribution of times ("A Field Study on Braking Responses during Driving," Ergonomics, 1995: 1903-1910): mean \(=535 \quad\) median \(=500 \quad\) mode \(=500\) sd \(=96 \quad\) minimum \(=220 \quad\) maximum \(=925\) 5 th percentile \(=400 \quad 10\) th percentile \(=430\) 90 th percentile \(=640 \quad 95\) th percentile \(=720\) What can you conclude about the shape of a histogram of this data? Explain your reasoning.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.