/*! 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 45 The following data are the respo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 45

The following data are the response times in seconds for \(n=25\) first graders to arrange three objects by size. $$ \begin{array}{lllll} 5.2 & 3.8 & 5.7 & 3.9 & 3.7 \end{array} $$ \(\begin{array}{lllll}4.2 & 4.1 & 4.3 & 4.7 & 4.3\end{array}\) \(\begin{array {lllll}3.1 & 2.5 & 3.0 & 4.4 & 4.8\end{array}\) \(\begin{array}{lllll}3.6 & 3.9 & 4.8 & 5.3 & 4.2\end{array}\) 4.7 3.854 a. Find the mean and the standard deviation for these 25 response times. b. Order the data from smallest to largest. c. Find the z-scores for the smallest and largest. response times. Is there any reason to believe that these times are unusually large or small? Explain.

Short Answer

Expert verified
Answer: The mean response time is 4.07016 and the standard deviation is 0.8178. The smallest and largest response times are not extremely unusual, but the largest response time (5.7) is somewhat unusual since it is close to the threshold of z-scores above 2.

Step by step solution

01

a. Find the mean and standard deviation

To find the mean, we will first add all the response times and divide the sum by the total number of observations, which is 25 in this case. The sum of the response times is: \(5.2+3.8+5.7+3.9+3.7+4.2+4.1+4.3+4.7+4.3+3.1+2.5+3.0+4.4+4.8+3.6+3.9+4.8+5.3+4.2+4.7+3.854 = 101.754\) So, the mean is: \(\bar{x} =\dfrac{101.754}{25} = 4.07016\) Now to find the standard deviation, we follow these steps: 1. Calculate the deviation of each response time from the mean. 2. Square each deviation. 3. Add up the squared deviations. 4. Divide the sum of the squared deviations by the total number of observations minus 1 (\(n-1\)). 5. Take the square root of the result obtained in step 4. The squared deviations are calculated as follows: \((5.2-4.07016)^2 + (3.8-4.07016)^2 + \cdots + (3.854-4.07016)^2 = 16.0721\) Now divide the sum of squared deviations by 24 (\(n-1\)): \( \dfrac{16.0721}{24} = 0.66967\) Finally, take the square root: \( s = \sqrt{0.66967} = 0.8178 \) So, the standard deviation is \(0.8178\).
02

b. Order the data from smallest to largest

Now, we will order the response times from smallest to largest. You can use any sorting method to do this (such as bubble sort, insertion sort, etc.). The sorted response times are: $$\begin{array}{lllll} 2.5 & 3.0 & 3.1 & 3.6 & 3.7 \end{array}$$ $$\begin{array}{lllll}3.8 & 3.854 & 3.9 & 3.9 & 4.1\end{array}$$ $$\begin{array}{lllll}4.2 & 4.2 & 4.3 & 4.3 & 4.4\end{array}$$ $$\begin{array}{lllll}4.7 & 4.7 & 4.8 & 4.8 & 5.2\end{array}$$ $$\begin{array}{llll}5.3 & 5.7 \end{array}$$
03

c. Find the z-scores for smallest and largest response times

To find the z-scores, we use the formula: \(z_i = \dfrac{x_i - \bar{x}}{s}\) Where \(z_i\) is the z-score for a given response time \(x_i\), \(\bar{x}\) is the mean, and \(s\) is the standard deviation. For the smallest response time (2.5): \(z_{min} = \dfrac{2.5 - 4.07016}{0.8178} = -1.9197\) For the largest response time (5.7): \(z_{max} = \dfrac{5.7 - 4.07016}{0.8178} = 1.9949\) Usually, z-scores below -2 or above 2 are considered unusual, which indicates that the smallest and largest response times are not extremely unusual. However, the largest response time (5.7) is close to the threshold, so it could still be considered somewhat unusual.

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 91影视!

Key Concepts

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

Mean Calculation
The mean is a fundamental concept in descriptive statistics and refers to the average of a set of numbers. In this context, the mean provides a single value that summarizes the entire dataset of response times.
Calculating the mean involves several straightforward steps:
  • Add all the individual observations together to get the total sum.
  • Divide this total sum by the number of observations, which, in this exercise, is 25.
The formula for calculating the mean \( \bar{x} \) is:
\[ \bar{x} = \dfrac{\sum_{i=1}^{n} x_i}{n} \]
Where \( x_i \) represents each individual data point, and \( n \) is the total number of data points. For our dataset, the sum of the response times is 101.754. Dividing by 25, the calculated mean is approximately 4.07016 seconds, which gives us an insight into the typical response time among the first graders.
Standard Deviation
Standard deviation is another crucial concept in statistics that measures the amount of variation or dispersion in a set of values. It indicates how much the individual data points differ from the mean.
To calculate the standard deviation, follow these steps:
  • First, find the deviation of each data point from the mean.
  • Square these deviations to ensure they are positive.
  • Sum the squared deviations.
  • Divide this total by \( n-1 \) (where \( n \) is the number of observations) to account for bias in sample data.
  • Take the square root of the result to return to the unit of measurement.
The formula for the standard deviation \( s \) is:
\[ s = \sqrt{\dfrac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}} \]
For this dataset, the standard deviation is approximately 0.8178 seconds, which provides an understanding of how closely the data points cluster around the mean. A smaller standard deviation suggests that the response times are closely grouped near the mean, while a larger standard deviation indicates a wider range of response times.
Z-Scores
Z-scores are a statistical measurement that describe a value's position relative to the mean of a group of values, expressed in terms of standard deviations.
To find a z-score, you use the following formula:
\[ z_i = \dfrac{x_i - \bar{x}}{s} \]
Here, \( z_i \) is the z-score, \( x_i \) is the specific data point, \( \bar{x} \) is the mean, and \( s \) is the standard deviation.
Z-scores help us understand whether a data point is typical or unusual. Data points with z-scores close to 0 are near the mean, while those with scores further from 0 are more unusual. Generally, z-scores beyond -2 or 2 are considered unusual.
  • The smallest response time of 2.5 seconds has a z-score of -1.9197, indicating it is not unreasonably low.
  • The largest response time of 5.7 seconds has a z-score of 1.9949, suggesting it is near the high end of what might be considered normal.
While this largest value doesn't exceed the typical threshold of 2, it is close enough to warrant attention, suggesting it could still be somewhat atypical.
Data Sorting
Data sorting is a simple yet powerful tool in statistical analysis, enabling clearer insight and easier calculation of further statistics such as medians, percentiles, and even facilitating a more intuitive grasp of a dataset's distribution.
Sorting the data usually involves ordering the numbers from smallest to largest.
For our dataset, the response times are sorted sequentially as follows: 2.5, 3.0, 3.1, 3.6, 3.7, 3.8, 3.854, 3.9, 3.9, 4.1, 4.2, 4.2, 4.3, 4.3, 4.4, 4.7, 4.7, 4.8, 4.8, 5.2, and so on.
This arrangement helps us easily identify the minimum and maximum values and understand the spread and skewness of the data. Additionally, sorting makes it more convenient to compute median and percentiles, as these require data to be organized in a specific order. Overall, sorting is a foundational step in making data analysis easier and more meaningful.

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

An analytical chemist wanted to use electrolysis to determine the number of moles of cupric ions in a given volume of solution. The solution was partitioned into \(n=30\) portions of .2 milliliter each, and each of the portions was tested. The average number of moles of cupric ions for the \(n=30\) portions was found to be .17 mole; the standard deviation was .01 mole. a. Describe the distribution of the measurements for the \(n=30\) portions of the solution using Tchebysheff's Theorem. b. Describe the distribution of the measurements for the \(n=30\) portions of the solution using the Empirical Rule. (Do you expect the Empirical Rule to be suitable for describing these data?) c. Suppose the chemist had used only \(n=4\) portions of the solution for the experiment and obtained the readings \(.15, .19, .17,\) and \(.15 .\) Would the Empirical Rule be suitable for describing the \(n=4\) measurements? Why?

In the seasons that followed his 2001 record-breaking season, Barry Bonds hit \(46,45,45,5,26,\) and 28 homers, respectively, until he retired from major league baseball in 2007 (www.ESPN,com). \(^{16}\) Two box plots, one of Bond's homers through \(2001,\) and a second including the years \(2002-2007\) follow. The statistics used to construct these box plots are given in the table. \begin{tabular}{lccccccc} Years & Min & \(a_{1}\) & Median & \(a_{3}\) & IQR & Max & \(n\) \\ \hline 2001 & 16 & 25.00 & 34.00 & 41.50 & 16.5 & 73 & 16 \\ 2007 & 5 & 25.00 & a. Calculate the upper fences for both of these box plots. b. Can you explain why the record number of homers is an outlier in the 2001 box plot, but not in the 2007 box plot?34.00 & 45.00 & 20.0 & 73 & 22 \end{tabular}

You are given \(n=5\) measurements: 2,1,1 , 3,5 a. Calculate the sample mean, \(\bar{x}\) b. Calculate the sample variance, \(s^{2}\), using the formula given by the definition. c. Find the sample standard deviation, \(s\). d. Find \(s^{2}\) and \(s\) using the computing formula. Compare the results with those found in parts \(b\) and \(c\)

An Archeological Find, again An article in Archeometry involved an analysis of 26 samples of Romano-British pottery found at four different kiln sites in the United Kingdom. \({ }^{7}\) The samples were analyzed to determine their chemical composition. The percentage of iron oxide in each of five samples collected at the Island Thorns site was: \(\begin{array}{llll}1.28 . & 2.39, & 1.50, & 1.88, & 1.51\end{array}\) a. Calculate the range. b. Calculate the sample variance and the standard deviation using the computing formula. c. Compare the range and the standard deviation. The range is approximately how many standard deviations?

Altman and Bland report the survival times for patients with active hepatitis, half treated with prednisone and half receiving no treatment. \(^{12}\) The survival times (in months) (Exercise 1.25 and \(\mathrm{EX} 0125\) ) are adapted from their data for those treated with prednisone. 8 52 15757 \(\begin{array}{rr}142 & 162 \\ 144 & 165\end{array}\) a. Can you tell by looking at the data whether it is roughly symmetric? Or is it skewed? b. Calculate the mean and the median. Use these measures to decide whether or not the data are symmetric or skewed. c. Draw a box plot to describe the data. Explain why the box plot confirms your conclusions in part b.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
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.