Copyright © 2014 Pearson Education, Inc. All rights reserved Chapter 3 Numerical Summaries of...

Copyright © 2014 Pearson Education, Inc. All rights reserved

Chapter 3

Numerical Summaries of Center and Variation

3 - 2 Copyright © 2014 Pearson Education, Inc. All rights reserved

Learning Objectives

Understand how measures of center and spread are used to describe characteristics of real-life samples of data.

Understand when it is appropriate to use the mean and standard deviation and when it is better to use the median and interquartile range.

Understand the mean as the balancing point of the distribution of a sample of data and the median as the point that has roughly 50% of the distribution below it.

Be able to write comparisons between samples of data in context.


3.1

Summaries for Symmetric

Distributions


Summaries for Symmetric Distribution

The mean describes the center. A numerical summary. The balancing point for the distribution. Can be used as a typical value for symmetric mound

shaped distributions.

The standard deviation describes the spread. A numerical summary Measures a typical distance of the observations from the

mean. Measures the variability when the distribution is symmetric


The Mean as a Balancing Point

If we place a finger at the mean, the histogram will balance perfectly.

Major League Baseball 2010


Skewness and the Mean

For a skewed right histogram, the mean is to the right of the typical value.

Major League Baseball 2010


Symmetric Distributions and the Mean

For a symmetric distribution, the mean is at the center.


The Formula for the Mean

To calculate the mean, use the formula:

Σ, read “sigma”, means “add”. x represents all of the data values. n represents the sample size. represents the sample mean.

xx

n

x


Calculating the Sample Mean

Find the mean of the number of siblings for the 8 students questioned: 3,2,2,1,2,3,5,2

The sample size: n = 8.

3 2 2 1 2 3 5 2

8

xx

n

2.5


Standard Deviation

The Standard Deviation, s, is a measure of the spread.

It represents a typical distance from the mean of the observations.

For mound shaped distributions, the majority of the observations are less than one standard deviation from the mean.

The square of the standard deviation is called the variance.


Put the Following in Order From Smallest Standard Deviation to Largest

Solution: (c), (b), (a)


The Standard Deviation and the Mean

In San Francisco, the mean high temperature is 65 degrees and the standard deviation is 8 degrees. In Provo the mean is 67 and the standard deviation is 21. Is a high temperature of 52 rarer in San Francisco or in Provo?

SF: 65 – 8 = 57, Provo: 67 – 21 = 46 Since 52 degrees is within one standard deviation of

Provo’s mean and not of San Francisco’s mean, a temperature of 52 is rarer in San Francisco.


Using StatCrunch to Find the Mean and Standard Deviation

Enter Data, then go to Stat →Summary Stats → Columns

Click on the variable name andhit Calculate.

This calculates the mean,standard deviation and other statistics that will be used later.


3.2

What’s Unusual? The Empirical Rule

and z-Scores


The Empirical Rule Graphically


Empirical Rule

The Empirical Rule: If a distribution is unimodal and symmetric, then Approximately 68% of the observations (roughly

two-thirds) will be within one standard deviation of the mean.

Approximately 95% of the observations will be within two standard deviations of the mean.

Nearly all the observations will be within three standard deviations of the mean.


Empirical Rule Example

The mean body weight for women between 18 and 25 years old is 134 lbs and the standard deviation is 26 lbs. Assume a mound shaped distribution.

134 – 26 = 108, 134 + 26 = 160 About 68% of women in this age group weigh

between 108 and 160 lbs. 134 – 2(26) = 82, 134 + 2(26) = 186 About 95% weigh between 82 and 186 lbs. Almost all weigh between 56 and 212 lbs.


Using the Empirical Rule

High temperatures in San Francisco follow a unimodal and symmetric distribution with mean 65 degrees and standard deviation 8 degrees. Give a range of temperatures that includes the middle 95% of high temperature days in San Francisco.

65 – 2(8) = 49, 65 + 2(8) = 81 About 95% of all days in San Francisco have

high temperatures between 49 and 81 degrees.


Empirical Rule Example

Daily cash register receipts at a local store follow a mound shaped distribution with mean $9,200 and standard deviation $150. The day a new employee was hired the store took in $4,500. Should the manager be concerned?

9200 – 3(150) = 4700 Yes, the manager should be concerned, since it is

highly unlikely that such a low receipt total for the day would happen by random chance alone.


The Trouble With Evaluating if a Data Value is Unusual

Is 2 less than the mean male height short? 2 feet shorter is much shorter. 2 millimeters shorter is not much shorter. Instead, statisticians normalize the values by

citing the Z-Score.


The Z-Score

The Z-Score measures the number of standard deviations the value is from the mean.

The resulting units are called Standard Units. The Z-Score is used to compare values

measured in different units such as feet and millimeters.


The Z-Score Formula x xz

s

The mean price for a loaf of bread is $3.12 and the standard deviation is $0.89. Find the z-Score for a loaf of bread that costs $2.00.

The z-Score is about -1.26.

2.00 3.121.26

0.89z


Comparing values

What is more unusual: a value of 0.26 from a distribution with mean 0.37 and standard deviation 0.03 or a value of 45 from a distribution with mean 38 and standard deviation 4?

0.26

0.26 0.373.67

0.03z

45

45 381.75

4z

The value of 0.26 is more unusual since it has a z-score that is farther from 0.


3.3

Summaries for Skewed Distributions


Skewness and the Trouble with the Mean

For a skewed distribution, the mean gets “pulled” towards the tail.

The mean is also “pulled” towards outliers. For a skewed distribution or a distribution

with only upper or only lower outliers, the mean does not represent a typical value.


The Median to Represent the Center

The middle value, called the median is often a better representation of the center.

The median is defined by the middle number or the average of the two middle numbers if the sample size is even.

The median cuts the data in half. Typically half the values are below the median and half are above.


Median vs. Mean

The median income of $18,000 better represents the typical income than much higher mean income.

The right tail greatly increases the mean but only slightly increases the median.


Calculating the Median

Sort the data from largest to smallest. If the set contains an odd number of observed

values, the median is the middle observed value. If the set contains an even number of observed

values, the median is the average of the two middle observed values. This places the median precisely halfway between the two middle values.


Example

The following data represent eight home prices in thousands of dollars. Find the median: 123, 457, 278, 184, 216, 336, 192, 184

First sort from smallest to largest: 123, 184, 184, 192, 216, 278, 336, 457

Since there are an even number of numbers take the average of the middle two: 192 216

2042


Quartiles

The First Quartile (Q1) is the value such that 25% of the data lie at or below this value.

Q1 is roughly the median of the lower half of the data.

The Third Quartile (Q3) is the value such that 75% of the data lie at or below this value.

Q3 is roughly the median of the upper half of the data.


The Interquartile Range (IQR)

The Interquartile Range (IQR) represents the range of the middle 50% of the data.

Cut the ordered data into four equal parts. The distance taken up by the middle two parts is the interquartile range.

IQR = Q3 – Q1


Interpreting Q1, Q3, and IQR

The first quartile for birth weights is 3.1 kg and the third quartile is 3.7 kg. Interpret Q1, Q3, and the IQR.

Q1 = 3.1. This means that 25% of all babies are born weighing at or below 3.1 kg.

Q3 = 3.7. This means that 75% of all babies are born weighing at or below 3.7 kg.

Q3 – Q1 = 0.6. The middle half of all birth weights has a range of 0.6 kg.


How the Quartiles and IQR are Used

Quartiles and the IQR are primarily used when there are large data sets, for example: National Exam Scores Physical Measurements: Weight, Height

Cholesterol Levels, BMI, etc. Income of state residents Time to run one mile


The Range

The range is the distance spanned by the entire data set.

Range = Maximum ˗ Minimum The range is easy to calculate, but is subject to

peculiarities of the data set and is very sensitive to outliers.

A smaller sample size is likely to produce a smaller range. The range of a sample is a poor predictor of the range for the population.


3.4

Comparing Measures of Center


Mean and Standard Deviation or Median and IQR?

Use the mean and standard deviation when the distribution is mound shaped.

Use the Median and IQR when the distribution is skewed left or skewed right.

If the distribution is not unimodal, it may be better to split the data.


Song lengths

The mean is influenced greatly by the right tail. The median isn’t.

The median of 226 seconds better represents the typical song. The IQR of 117 seconds covers the high bars of the histogram.

Song lengths are skewed right because there are many short songs, no negative length songs, but a few long songs.


San Francisco

Temperatures The distribution is

approximately mound shaped.

With mound shaped distributions the mean and median are nearly the same number.

The mean is preferred over the median if they are close together. One standard deviation from the mean gives a lower bound of 57

and an upper bound of 73. This covers the high bars of the histogram.


The Effect of Outliers

Number of employees at several businesses on main street: 6, 7, 14, 18, 23, 25, 26 Mean = 17 Median = 18

If the 26 employee business is turned into a Wal-Mart: 6, 7, 14, 18, 23, 25, 334 Mean = 61 Median = 18

Conclusion: The mean is strongly affected by outliers, while the median is not affected by outliers.


Affected by Outliers?

Affected by Outliers: Mean Standard Deviation Range

Not Affected by Outliers: Median Interquartile Range (IQR)


Bimodal Distributions

For most bimodal distributions, neither the mean nor the median represent typical values.

Investigate further to see if there are two separate sub-populations.

Consider separating the two populations and present their graphs and statistics individually.


Trouble with Bimodal Distributions

There are two typical values. Neither the mean nor the median describe the typical values. The data should be separated out by lunch customers and

dinner customers.


Separating Lunch and Dinner

Displaying the data with two histograms allows a comparison between lunch and dinner.


Separating Lunch and Dinner

The Lunch distribution is mound shaped and the Dinner distribution is skewed right.

Do not compare the mean of one data set with the median of another. Use the medians for comparisons. Lunch median is $8 and Dinner

median is $22


3.5

Using Boxplots for Displaying Summaries


The Five Point Summary

When the data are partitioned into four equal segments, five important numbers arise. They are called the Five Point Summary: Minimum – Smallest Value First Quartile (Q1) – The Median of the Lower Half Median – The Middle Number or Center Third Quartile – The Median of the Upper Half Maximum – Largest Value


How many Boyfriends/Girlfriends?

The results of a survey asking how many boyfriends/girlfriends people have had is shown below: 0, 1, 1, 2, 3, 4, 4, 5, 6, 8,10

The five point summary is: Minimum = 0 Median = 4 Maximum = 10

Q1 = 1 Q3 = 6


Potential Outliers

A Potential Outlier is a data value that is a distance of more than 1.5 interquartile ranges below the first quartile or above the third quartile.1. Calculate IQR = Q3 – Q1

2. Find m = Q1 – (1.5)(IQR)

3. Find M = Q3 + (1.5)(IQR)

4. Any values less than m or more than M are potential outliers.


Finding Possible Outliers

The first quartile, Q1, for triglycerides is 109 mg/dL. The third quartile, Q2, is 150 mg/dL. Determine which if any of the following triglyceride readings are potential outliers:

38, 200, 225 IQR = 150 – 109 = 41 Q1 – (1.5)(IQR) = 109 – (1.5)(41) = 47.5 Q3 – (1.5)(IQR) = 150 + (1.5)(41) = 211.5

38 and 225 are potential Outliers since 38 < 47.5 and 225 > 211.5.


Boxplots

A Boxplot is a chart that visually displays Q1, the median, Q3, and the potential outliers.

To create a boxplot:1. Plot the potential outliers

2. Draw small vertical line segments at Q1, Q3, and the median.

3. Draw a box with base from Q1 to Q3.

4. Sketch horizontal line segments from the ends of the box to the smallest and largest values that are not potential outliers.


Box Plot


Interpreting a Boxplot

What percent of students scored below 83%? Answer: 25%

What percent of students scored between 83% and 92%? Answer: 50%


Comparing Distributions with Boxplots

Both cities have similar typical temperatures. Both cities have fairly symmetric distributions. Provo has a much greater variation in

temperatures than San Francisco.


What Boxplots Show and Don’t Show

Boxplots Show: Typical Range of Values Possible Outliers Variation

Boxplots Don’t Show: Modality Mean Anything for small data sets, especially < 5.


Chapter 3

Case Study


Perceived Risk


Perceived Risk of Appliances

Skewed right for both men and women. Unimodal for both men and women. Women’s typical value slightly higher than men’s. Five Point Summary appropriate for both.


Risk of Appliance: Statistics

Men’s median is 10, women’s median is higher at 15.

The middle 50% of men varied by 20, while the variation was higher, 25 for women.


Perceived Risk X-rays

Relatively symmetric for both men and women. Unimodal for both men and women. Women’s typical value close to men’s. Mean and standard deviation appropriate for both.


Risk of X-rays: Statistics

Men and Women have similar mean and standard deviation risk perception for X-rays.

About 68% of men perceive a risk between 26.8 and 66.8.

About 68% of women perceive a risk between 27 and 68.6.

Mean Standard Deviation

Men 46.8 20

Women

47.8 20.8


Chapter 3

Guided Exercise 1


The mean rate of violent crime in the west was 406 per 100,000 people, and the standard deviation was 177. Assume the distribution is approximately unimodal and symmetric.

Between which two values would you expect to find about 95% of the violent crime rates?


If a western state had a violent crime rate of 584 crimes per 100,000 people, would you consider this unusual?

Would 30 crimes per 100,000 people be unusual?



By the Empirical Rule, about 95% of the data is within two standard deviations of the mean.

This represents the green and blue areas together. The number 583 represents one standard deviation

more than the mean: 406 + 177 = 583.



406 – 177 = 229 406 – 2(177) = 52 406 + 2(177) = 760




95% of the violent crime rates are between 52 and 760 crimes per 100,000 people.


68% of the violent crime rates are between 229 and 583 crimes per 100,000 people.



If a western state had a violent crime rate of 584 crimes per 100,000 people, would you consider this unusual?

No, since 584 is within 2standard deviations ofthe mean.

Would 30 crimes per 100,000 people be unusual? Yes, because less than 5% occur so far from the

mean.


Chapter 3

Guided Exercise 2


The head circumferences in centimetersfor some men and women in a statistics class are given. Men: 58, 60, 62.5, 63, 59.5, 59, 60, 57, 55 Women: 63, 55, 54.5, 53.5, 53, 58.5, 56,

54.5, 55, 56, 56, 54, 56,53, 51 Compare the circumferences of the men’s

and women’s heads.


Histograms of the two sets of Data.


Shapes

The distribution for men is unimodal and not too far from symmetric.

The distribution for women is unimodal and nearly symmetric except one possible outlier.Copyright © 2013 Pearson Education, Inc.. All rights reserved.


Mean and Standard Deviation or Quartiles and IQR?

Since the women’s distribution has a possible outlier, the quartiles and IQR should be used for comparisons.


Compare Centers

The median head circumference for the men was 59.5 cm, and the median head circumference for the women was 55 cm. This shows that the men tended to have larger heads.


Compare Variances

The interquartile range for the head circumferences for the men was 2 cm, and the interquartile range for the women was 2.5 cm. This shows that the women tended to have more variation, as measured by the interquartile range.


Outliers

Men: 58, 60, 62.5, 63, 59.5, 59, 60, 57, 55 Q1 – (1.5)(IQR) = 55, Q3 + (1.5)(IQR) = 63 No Possible outliers for the men.

Women: 63, 55, 54.5, 53.5, 53, 58.5, 56, 54.5, 55, 56, 56, 54, 56, 53, 51 Q1 – (1.5)(IQR) = 49.75, Q3 + (1.5)(IQR) = 59.75 63 is a possible outlier for the women.


Final Comparison

The typical head circumference for men is about 4.5 cm larger than the head circumference for women. The women’s head circumference had slightly more variation than the men’s.

Copyright © 2014 Pearson Education, Inc. All rights reserved Chapter 3 Numerical Summaries of...

Documents

Transcript of Copyright © 2014 Pearson Education, Inc. All rights reserved Chapter 3 Numerical Summaries of...