/*! 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 5 In August 2009 , Harris Interact... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In August 2009 , Harris Interactive released the results of the "Great Schools" survey. In this survey, 1086 parents of children attending a public or private school were asked approximately how much time they spent volunteering at school per month over the last school year. For this sample, the mean number of hours per month was \(5.6\) hours and the median number of hours was \(1.0 .\) What does the large difference between the mean and median tell you about this data set?

Short Answer

Expert verified
The large difference between the mean and median in this dataset indicates a right-skewed distribution, meaning there's a minority of parents who volunteer significantly more hours than the majority, pulling the mean up.

Step by step solution

01

Identify the given mean and median

The mean number of volunteering hours per month is \(5.6\) hours and the median is \(1.0\) hour.
02

Understanding mean and median

The mean is the average of all numbers in the set, calculated by adding up all the figures and then dividing by the total number of data points. The median is the middle value in a dataset when it is ordered from smallest to largest. If there's an even number of data points, the median is the average of the two middle numbers.
03

Understanding the difference

In this dataset, the mean is much higher than the median. This indicates that there are some very high values in the data set (parents who volunteer many hours) further away from the median and the rest of the numbers. These high numbers are influencing the mean to be greater than the median.
04

Interpret the skewness

This kind of distribution, where the mean is greater than the median, indicates a right-skewed distribution. This tells us that there is a minority of parents who volunteer many more hours than the majority, therefore, the data is not evenly distributed.

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.

Mean vs Median
Understanding the difference between mean and median is fundamental in the analysis of data. The mean, often referred to as the average, represents the sum of all values divided by the count of the data. It can be influenced by extreme values, which can push it higher or lower than the median. Meanwhile, the median is the middle value when the data is ordered, exemplifying the center of a data set. It's less affected by outliers and can often give a better representation of a typical value in a skewed distribution.

In the 'Great Schools' survey scenario, the mean volunteering time is significantly higher than the median. This discrepancy can suggest that while most parents volunteer a modest amount, a few parents may contribute an exceptionally large number of hours, disproportionately increasing the mean. Therefore, when dealing with statistical data that may have outliers, it's crucial to consider both the mean and the median to get an accurate picture of the data.
Right-Skewed Distribution
When analyzing data, the way that values are spread out can reveal vital information about the dataset's overall shape. A right-skewed distribution, also known as positive skewness, is characterized by a long tail extending toward the higher values on the right side of the distribution plot. It implies that a subset of larger-than-average values skews the mean upwards, away from the median. This skewed pattern appears when there are outliers or a group of high values.

In our survey example, the presence of parents who volunteer for many more hours than the average illustrates this kind of distribution. A small number of parents with high volunteering times are increasing the survey's average, causing it not to be representative of the typical parent's contribution.
Interpret Data Skewness
Data skewness is an indicator of the symmetry, or lack thereof, in a distribution of data. Skewness is essential for understanding the underlying structure in data and how it deviates from the normal distribution. It affects the mean more than the median, which is why comparing the two can provide insights into the data's symmetry. A right-skewed data set suggests that there's a pull of data points towards larger values, highlighting the presence of outliers. Similarly, left-skewed data points towards the smaller values.

When the mean is greater than the median, as in the survey data where the mean is 5.6 hours and the median is 1.0 hour, it indicates right-skewness. This suggests that data points are not spread out evenly, and interpretations should take into account the possibility of atypical high values influencing the mean.
Statistical Data Analysis
Statistical data analysis encompasses a range of techniques to explore, describe, and infer patterns from data. Descriptive statistics like mean and median offer a snapshot of the data's central tendency. The comparison between these measures is just the tip of the iceberg in data analysis. For a robust analysis, one might employ additional tools, such as variance and standard deviation to gauge data spread, or skewness and kurtosis measures to understand the shape of the data.

When faced with skewed distributions, it becomes especially important to look beyond simple averages. Illustrated by the parents' volunteering time, such distributions can mislead stakeholders if viewed without considering the measure of spread and skewness. Thus, careful statistical analysis allows stakeholders to make informed decisions based on a comprehensive understanding of the data.

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 chapter introduction gave the accompanying data on the percentage of those eligible for a lowincome subsidy who had signed up for a Medicare drug plan in each of 49 states (information was not available for Vermont) and the District of Columbia (USA Today. May \(9.2006\) ). \(\begin{array}{llllllll}24 & 27 & 12 & 38 & 21 & 26 & 23 & 33 \\ 19 & 19 & 26 & 28 & 16 & 21 & 28 & 20 \\ 21 & 41 & 22 & 16 & 29 & 26 & 22 & 16 \\ 27 & 22 & 19 & 22 & 22 & 22 & 30 & 20 \\ 21 & 34 & 26 & 20 & 25 & 19 & 17 & 21 \\ 27 & 19 & 27 & 34 & 20 & 30 & 20 & 21\end{array}\) 19 18 ( $$ 14 $$ a. Compute the mean for this data set. b. The article stated that nationwide, \(24 \%\) of those eligible had signed up. Explain why the mean of this data set from Part (a) is not equal to 24 . (No information was available for Vermont, but that is not the reason that the mean differs-the \(24 \%\) was calculated excluding Vermont.)

Research by the Food and Drug Administration (FDA) shows that acrylamide (a possible cancer-causing substance) forms in high-carbohydrate foods cooked at high temperatures and that acrylamide levels can vary widely even within the same brand of food (Associated Press, December 6, 2002). FDA scientists analyzed McDonald's French fries purchased at seven different locations and found the following acrylamide levels: \(\begin{array}{lllllll}497 & 193 & 328 & 155 & 326 & 245 & 270\end{array}\) a. Compute the mean acrylamide level and the seven deviations from the mean. b. Verify that, except for the effect of rounding, the sum of the deviations from mean is equal to 0 for this data set. (If you rounded the sample mean or the deviations, your sum may not be exactly zero, but it should be close to zero if you have computed the deviations correctly.) c. Calculate the variance and standard deviation for this data set.

USA Today (May 9,2006 ) published the accompanying average weekday circulation for the 6 -month period ending March 31,2006, for the top 20 newspapers in the country: \(\begin{array}{rrrrr}2,272,815 & 2,049,786 & 1,142,464 & 851,832 & 724,242 \\\ 708,477 & 673,379 & 579,079 & 513,387 & 438,722 \\ 427,771 & 398,329 & 398,246 & 397,288 & 365,011 \\ 362,964 & 350,457 & 345,861 & 343,163 & 323,031\end{array}\) a. Do you think the mean or the median will be larger for this data set? Explain. b. Compute the values of the mean and the median of this data set. c. Of the mean and median, which does the best job of describing a typical value for this data set? d. Explain why it would not be reasonable to generalize from this sample of 20 newspapers to the population of all daily newspapers in the United States.

The National Climate Data Center gave the accompanying annual rainfall (in inches) for Medford, Oregon, from 1950 to 2008 (www.ncdcnoaa.gov/oa/ dimate/research/cag3/dty.html): \(\begin{array}{lllllllll}28.84 & 20.15 & 18.88 & 25.72 & 16.42 & 20.18 & 28.96 & 20.72 & 23.58 \\ 10.62 & 20.85 & 19.86 & 23.34 & 19.08 & 29.23 & 18.32 & 21.27 & 18.93 \\ 15.47 & 20.68 & 23.43 & 19.55 & 20.82 & 19.04 & 18.77 & 19.63 & 12.39 \\ 22.39 & 15.95 & 20.46 & 16.05 & 22.08 & 19.44 & 30.38 & 18.79 & 10.89 \\ 17.25 & 14.95 & 13.86 & 15.30 & 13.71 & 14.68 & 15.16 & 16.77 & 12.33 \\ 21.93 & 31.57 & 18.13 & 28.87 & 16.69 & 18.81 & 15.15 & 18.16 & 19.99\end{array}\) \(18.13\) \(21.99\) \(\begin{array}{cc}19.00 & 23.97\end{array}\) \(\begin{array}{rr}17.25 & 14.07\end{array}\) a. Compute the quartiles and the interquartile range. b. Are there outliers in this data set? If so, which observations are mild outliers? Which are extreme outliers? c. Draw a boxplot for this data set that shows outliers.

The Los Angeles Times (July 17. 1995) reported that in a sample of 364 lawsuits in which punitive damages were awarded, the sample median damage award was \(\$ 50,000\), and the sample mean was \(\$ 775,000\). What does this suggest about the distribution of values in the sample?

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.