/*! 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 48 The article "A Thin-Fim Oxygen U... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article "A Thin-Fim Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants" (Lubric. Engrg., 1984: 75-83) reported the following data on oxidation-induction time (min) for various commercial oils: \(\begin{array}{lllllllllll}87 & 103 & 130 & 160 & 180 & 195 & 132 & 145 & 211 & 105 & 145 \\ 153 & 152 & 138 & 87 & 99 & 93 & 119 & 129 & & & \end{array}\) a. Calculate the sample variance and standard deviation. b. If the observations were reexpressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the reexpression.

Short Answer

Expert verified
The sample variance and standard deviation are calculated from deviations. Reexpressing in hours divides variance by 3600 and standard deviation by 60.

Step by step solution

01

Calculate the Mean

First, determine the mean (average) of the data set. Add up all the observed values and divide by the number of observations. The mean \( \bar{x} \) is calculated as:\[ \bar{x} = \frac{87 + 103 + 130 + 160 + 180 + 195 + 132 + 145 + 211 + 105 + 145 + 153 + 152 + 138 + 87 + 99 + 93 + 119 + 129}{19} \]
02

Compute Each Deviation from the Mean

Subtract the mean \( \bar{x} \) from each observation to find the deviation from the mean for each value. These deviations will be used to calculate the variance.
03

Square Each Deviation

Square each of the deviations obtained in Step 2. This step is necessary to eliminate negative values and to prepare for calculating the variance.
04

Determine the Sample Variance

Calculate the sum of the squared deviations obtained in Step 3. Then divide by \( n-1 \), where \( n \) is the number of observations. The variance \( s^2 \) is given by:\[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} \]
05

Calculate the Sample Standard Deviation

The standard deviation \( s \) is the square root of the variance. This value represents the average distance of each data point from the mean.\[ s = \sqrt{s^2} \]
06

Convert Observations to Hours and Impact on Variance and Standard Deviation

To convert minutes to hours, divide each observation by 60. When reexpressing in hours, the variance \( s^2_{hours} \) becomes \( \frac{s^2}{3600} \) and the standard deviation \( s_{hours} \) becomes \( \frac{s}{60} \) due to the properties of variance and standard deviation.

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.

Data Transformation
When dealing with data, especially when you need to compare or apply across different units, transforming the data is a handy tool. Let’s say you have a list of times measured in minutes, but you’d rather work with hours. This involves transforming each data point from minutes to hours by dividing by 60.

Data transformation does not just stop with converting units. The influence of this transformation extends to how we predict changes in statistical parameters like variance and standard deviation.
  • When you transform each data value from one unit to another, you multiply the variance by the square of the conversion factor.
  • For standard deviation, you multiply by the conversion factor itself.
In our scenario, transforming minutes into hours involves dividing by 60, which affects both the variance and standard deviation.
Mean Calculation
The mean, often referred to as the average, is a fundamental measure in statistics, representing the center of a data set. It's computed by taking the sum of all values and dividing by the number of values.

For example, if you have a data set representing oxidation times:
- You add up all the oxidation times. - Divide this sum by the number of data points in the set.

In a practical sense, the mean tells you about the typical value in your data set. In the context of our oil oxidation example, it helps you understand what a common oxidation time might be, providing a benchmark on which other statistical calculations, like variance and standard deviation, are based. Knowing how to calculate the mean correctly is crucial because it directly impacts subsequent calculations.
Deviation from the Mean
Deviation from the mean is crucial for understanding variability in your dataset. It tells you how far each data point is from the mean value. To calculate it, simply subtract the mean from each data point.

Calculating deviations involves:
  • Taking each value in your dataset.
  • Subtracting the mean from each value to get the deviation.
This step highlights how each observation varies from the central point, the mean. If once deviation values are summed, they should theoretically equal zero, showing balance around the mean. Understanding deviations is key to computing the sample variance and standard deviation, which further tell us about variation and spread. The deviation is fundamental because it leads to squaring deviations for variance calculations, which removes any negative signs.
Squaring Deviations
After calculating deviations, the next essential step is to square these deviations. Squaring serves two purposes:

  • It removes negative signs, as deviations can be negative when the data value is less than the mean.
  • It emphasizes larger deviations, giving more weight to values that are further from the mean.

With these squared deviations, you sum them up. This total is essential when calculating variance and, subsequently, standard deviation. For variance, sum the squared deviations and divide by the number of observations minus one, often denoted as \( n-1 \), for a sample, not the entire population.

Squaring the deviations is a critical step because variance is a cornerstone of statistical analysis, representing the data's spread. Consequently, its square root, the standard deviation, offers insight into the average distance from the mean for data points, balancing the units back to the original data scale.

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

In a study of author productivity ("Lotka's Test," Collection Manage., 1982: 111-118), a large number of authors were classified according to the number of articles they had published during a certain period. The results were presented in the accompanying frequency distribution: \(\begin{array}{lrrrrrrrrr}\begin{array}{l}\text { Number of } \\ \text { papers }\end{array} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & \\ \text { Frequency } & 784 & 204 & 127 & 50 & 33 & 28 & 19 & 19 & \\ \text { Number of } & & & & & & & & & \\ \text { papers } & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\\ \text { Frequency } & 6 & 7 & 6 & 7 & 4 & 4 & 5 & 3 & 3\end{array}\) a. Construct a histogram corresponding to this frequency distribution. What is the most interesting feature of the shape of the distribution? b. What proportion of these authors published at least five papers? At least ten papers? More than ten papers? c. Suppose the five 15 's, three 16 's, and three 17 's had been lumped into a single category displayed as \(" \geq 15 . "\) Would you be able to draw a histogram? Explain. d. Suppose that instead of the values 15,16 , and 17 being listed separately, they had been combined into a \(15-17\) category with frequency 11. Would you be able to draw a histogram? Explain.

Unlike most packaged food products, alcohol beverage container labels are not required to show calorie or nutrient content. The article "What Am I Drinking? The Effects of Serving Facts Information on Alcohol Beverage Containers" (J. of Consumer Affairs, 2008: 81-99) reported on a pilot study in which each individual in a sample was asked to estimate the calorie content of a \(12 \mathrm{oZ}\) can of light beer known to contain 103 cal. The following information appeared in the article: \begin{tabular}{lr} \hline Class & Percentage \\ \hline \(0-<50\) & 7 \\ \(50-<75\) & 9 \\ \(75-<100\) & 23 \\ \(100-<125\) & 31 \\ \(125-<150\) & 12 \\ \(150-<200\) & 3 \\ \(200-<300\) & 12 \\ \(300-<500\) & 3 \\ \hline \end{tabular} a. Construct a histogram of the data and comment on any interesting features. b. What proportion of the estimates were at least \(100 ?\) Less than \(200 ?\)

The value of Young's modulus (GPa) was determined for cast plates consisting of certain intermetallic substrates, resulting in the following sample observations ("Strength and Modulus of a Molybdenum-Coated Ti-25Al-10Nb-3U-1Mo Intermetallic," J. Mater. Engrg. Perform., 1997: 46-50): \(116.4\) \(115.9\) \(114.6\) \(115.2\) \(115.8\) a. Calculate \(\bar{x}\) and the deviations from the mean. b. Use the deviations calculated in part (a) to obtain the sample variance and the sample standard deviation. c. Calculate \(s^{2}\) by using the computational formula for the numerator \(S_{x x}\). d. Subtract 100 from each observation to obtain a sample of transformed values. Now calculate the sample variance of these transformed values, and compare it to \(s^{2}\) for the original data. State the general principle.

The accompanying data set consists of observations on shower-flow rate (L/min) for a sample of \(n=129\) houses in Perth, Australia ("An Application of Bayes Methodology to the Analysis of Diary Records in a Water Use Study," J. Amer. Statist. Assoc., 1987: 705-711): \(\begin{array}{rrrrrrrrrr}4.6 & 12.3 & 7.1 & 7.0 & 4.0 & 9.2 & 6.7 & 6.9 & 11.5 & 5.1 \\ 11.2 & 10.5 & 14.3 & 8.0 & 8.8 & 6.4 & 5.1 & 5.6 & 9.6 & 7.5 \\\ 7.5 & 6.2 & 5.8 & 2.3 & 3.4 & 10.4 & 9.8 & 6.6 & 3.7 & 6.4 \\ 8.3 & 6.5 & 7.6 & 9.3 & 9.2 & 7.3 & 5.0 & 6.3 & 13.8 & 6.2 \\ 5.4 & 4.8 & 7.5 & 6.0 & 6.9 & 10.8 & 7.5 & 6.6 & 5.0 & 3.3 \\ 7.6 & 3.9 & 11.9 & 2.2 & 15.0 & 7.2 & 6.1 & 15.3 & 18.9 & 7.2 \\ 5.4 & 5.5 & 4.3 & 9.0 & 12.7 & 11.3 & 7.4 & 5.0 & 3.5 & 8.2 \\ 8.4 & 7.3 & 10.3 & 11.9 & 6.0 & 5.6 & 9.5 & 9.3 & 10.4 & 9.7 \\ 5.1 & 6.7 & 10.2 & 6.2 & 8.4 & 7.0 & 4.8 & 5.6 & 10.5 & 14.6 \\ 10.8 & 15.5 & 7.5 & 6.4 & 3.4 & 5.5 & 6.6 & 5.9 & 15.0 & 9.6 \\ 7.8 & 7.0 & 6.9 & 4.1 & 3.6 & 11.9 & 3.7 & 5.7 & 6.8 & 11.3 \\ 9.3 & 9.6 & 10.4 & 9.3 & 6.9 & 9.8 & 9.1 & 10.6 & 4.5 & 6.2 \\ 8.3 & 3.2 & 4.9 & 5.0 & 6.0 & 8.2 & 6.3 & 3.8 & 6.0 & \end{array}\) a. Construct a stem-and-leaf display of the data. b. What is a typical, or representative, flow rate? c. Does the display appear to be highly concentrated or spread out? d. Does the distribution of values appear to be reasonably symmetric? If not, how would you describe the departure from symmetry? e. Would you describe any observation as being far from the rest of the data (an outlier)?

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

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.