/*! 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 3 The accompanying observations on... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The accompanying observations on stabilized viscosity (cP) for specimens of a certain grade of asphalt with \(18 \%\) rubber added are from the article "Viscosity Characteristics of Rubber-Modified Asphalts" (J. of Materials in Civil Engr., 1996: 153-156): \(27812900 \quad 3013 \quad 2856 \quad 2888\) a. What are the values of the sample mean and sample median? b. Calculate the sample variance using the computational formula. [Hint; First subtract a convenient number from each observation.]

Short Answer

Expert verified
The sample mean is 2887.6, the median is 2888, and the variance is 13216.5.

Step by step solution

01

Calculate the Sample Mean

To calculate the sample mean, sum up all the observations and divide by the number of observations. The observations are: 2781, 2900, 3013, 2856, 2888. First, sum them: \[ 2781 + 2900 + 3013 + 2856 + 2888 = 14438 \]There are 5 observations. So the sample mean \( \bar{x} \) is:\[ \bar{x} = \frac{14438}{5} = 2887.6 \]
02

Determine the Sample Median

The sample median is the middle value when the data is sorted. Sort the observations: 2781, 2856, 2888, 2900, 3013. Since there are 5 observations, the middle value is the third one: 2888. Thus, the sample median is 2888.
03

Calculate the Sample Variance

The sample variance can be calculated using the formula:\[ s^2 = \frac{\sum{(x_i - \bar{x})^2}}{n-1} \]where \( x_i \) is each observation, \( \bar{x} \) is the mean, and \( n \) is the number of observations. First, let's subtract 2800 from each observation to simplify calculations:- For 2781: \( x_i - 2800 = -19 \)- For 2900: \( x_i - 2800 = 100 \)- For 3013: \( x_i - 2800 = 213 \)- For 2856: \( x_i - 2800 = 56 \)- For 2888: \( x_i - 2800 = 88 \)Now, compute the sum of squares using these adjustments:\[\sum{(x_i - 2800)^2} = (-19)^2 + 100^2 + 213^2 + 56^2 + 88^2 = 361 + 10000 + 45369 + 3136 + 7744 = 66610\]Correcting for subtracting 2800, we add 2800 back, and adjust for the sample size (5):Variance:\[ s^2 = \frac{66610 + (5 \times 2800^2 - 14438^2)}{4} = \frac{66610 + 19600000 - 20852644}{4} = \frac{52866}{4} = 13216.5 \]The sample variance is approximately 13216.5.

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.

Sample Mean
The sample mean is a fundamental concept in statistics that represents the average value of a dataset. Calculating it involves two main steps. First, you sum up all the data points. For the observations 2781, 2900, 3013, 2856, and 2888 given in the exercise, you add them together to get a sum of 14438.

Next, you divide this total by the number of observations, which in this case is 5. This division gives you the sample mean. So, the calculation is \( \bar{x} = \frac{14438}{5} = 2887.6 \).

#### Why It Matters
The sample mean provides a central value for the dataset, offering a quick preview of the data's general magnitude. It is a simple yet powerful indicator of the overall behavior of the data and plays a crucial role in other statistical calculations and analyses.
Sample Median
The sample median is essential for understanding the position of the data. To find the median, you first need to organize the data points in ascending order, as done in the exercise: 2781, 2856, 2888, 2900, and 3013. Once sorted, you identify the middle value of the dataset.

Since there are 5 values, the median is the third one, which is 2888.

#### Importance of the Median
The median is particularly useful because it is not affected by extreme values. While the mean can shift significantly with very high or low values, the median stays stable, providing a more robust center for the data in situations with outliers.

It's an excellent metric for understanding where the "middle" of the data truly lies, especially in skewed distributions.
Sample Variance
Sample variance measures the spread or dispersion of the data points in a dataset. To find the sample variance, begin by using the formula: \[ s^2 = \frac{\sum{(x_i - \bar{x})^2}}{n-1} \] where \(x_i\) are data points, \(\bar{x}\) is the sample mean, and \(n\) is the number of observations.

In this exercise, simplifying the calculation by temporarily subtracting 2800 from each observation helps manage large numbers. Adjustments included values such as -19 for 2781 and 100 for 2900. After finding these adjusted values, compute their squares and sum them up for 66610.

Understanding variance helps to see how much the data points deviate from that central value, which is the sample mean.

#### Interpreting Variance
A higher sample variance indicates greater variability within the data, while a lower variance signals that the data points are more clustered around the mean. Variance forms the basis for standard deviation, a key descriptive statistic, making it crucial for data interpretation overall.

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

The accompanying specific gravity values for various wood types used in construction appeared in the article "Bolted Connection Design Values Based on European Yield Model" (J. of Structural Engr., 1993: 2169-2186): \(\begin{array}{lllllllll}.31 & .35 & .36 & .36 & .37 & .38 & .40 & .40 & .40 \\\ .41 & .41 & .42 & .42 & .42 & .42 & .42 & .43 & .44 \\ .45 & .46 & .46 & .47 & .48 & .48 & .48 & .51 & .54 \\ .54 & .55 & .58 & .62 & .66 & .66 & .67 & .68 & .75\end{array}\) Construct a stem-and-leaf display using repeated stems (see the previous exercise), and comment on any interesting features of the display.

A sample of 20 glass bottles of a particular type was selected, and the internal pressure strength of each bottle was determined. Consider the following partial sample information: \(\begin{array}{lrrr}\text { median }=202.2 & \text { lower fourth }=196.0 & \\\ \begin{array}{l}\text { upper fourth }=216.8\end{array} & & & \\ & & & \\\ \text { Three smallest observations } & 125.8 & 188.1 & 193.7 \\ \text { Three largest observations } & 221.3 & 230.5 & 250.2\end{array}\) a. Are there any outliers in the sample? Any extreme outliers? b. Construct a boxplot that shows outliers, and comment on any interesting features.

The California State University (CSU) system consists of 23 campuses, from San Diego State in the south to Humboldt State near the Oregon border. A CSU administrator wishes to make an inference about the average distance between the hometowns of students and their campuses. Describe and discuss several different sampling methods that might be employed. Would this be an enumerative or an analytic study? Explain your reasoning.

How does the speed of a runner vary over the course of a marathon (a distance of \(42.195 \mathrm{~km}\) )? Consider determining both the time to run the first \(5 \mathrm{~km}\) and the time to run between the \(35-\mathrm{km}\) and \(40-\mathrm{km}\) points, and then subtracting the former time from the latter time. A positive value of this difference corresponds to a runner slowing down toward the end of the race. The accompanying histogram is based on times of runners who participated in several different Japanese marathons ("Factors Affecting Runners' Marathon Performance," Chance, Fall, 1993: 24-30). What are some interesting features of this histogram? What is a typical difference value? Roughly what proportion of the runners ran the late distance more quickly than the early distance?

Automated electron backscattered diffraction is now being used in the study of fracture phenomena. The following information on misorientation angle (degrees) was extracted from the article "Observations on the Faceted Initiation Site in the Dwell-Fatigue Tested Ti-6242 Alloy: Crystallographic Orientation and Size Effects (Metallurgical and Materials Trans., 2006: 1507-1518). \(\begin{array}{lcccc}\text { Class: } & 0-<5 & 5-<10 & 10-<15 & 15-<20 \\\ \text { Rel freq: } & .177 & .166 & .175 & .136 \\ \text { Class: } & 20-<30 & 30-<40 & 40-<60 & 60-<90 \\ \text { Rel freq: } & .194 & .078 & .044 & .030\end{array}\) a. Is it true that more than \(50 \%\) of the sampled angles are smaller than \(15^{\circ}\), as asserted in the paper? b. What proportion of the sampled angles are at least \(30^{\circ}\) ? c. Roughly what proportion of angles are between \(10^{\circ}\) and \(25^{\circ}\) ? d. Construct a histogram and comment on any interesting features.

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.