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Welcome to MM150 – Unit 9 Seminar
• Instructor: Larry Musolino
– Email: [email protected]
• Some Administrative Items
– Reminder that the Final Project is due by Tuesday
Mar 6th , 2012, 11:59pm ET. Post to Dropbox
• Click on DROPBOX Link at top of page
• Under Basket, select Unit 9 Final Project
– Unit 9 Seminar is the final seminar for the course
(there is no Unit 10 seminar).
– Last Day of the Course is Tues March 13, 2012
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 1
Reminder on Final Project
• For more details on Final Project, see Unit 6 or
Unit 7 topic “Final Project”
– You will select a topic in the course and discuss a potential
application for this concept in your chosen profession. Do
not be afraid to "think outside the box" when discussing an
application in your profession. This also could be an
example of how the concept learned is fundamental to
understanding a more complex concept.
• Check out the example Final Projects, posted
as powerpoint and MS-Word example
documents
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 2
Reminder on Final Project (cont’d)
Final Project Instructions:
• Choose either Microsoft Word or Microsoft PowerPoint in which to create your
project. This course does not teach PowerPoint, so if you choose this format, do
so only because you are already comfortable with it or know someone who can
help you learn to use it.
• Create 5 slides or pages.
• On slides or pages 1 and 2, provide your name, the project title, and the course
and section number, and introduce your chosen profession and give a brief
overview of the concept you will apply to the profession.
• On slides or pages 3 and 4, describe how the concept can apply to you chosen
profession. You will not need to “do the math”; simply describe how you would
use it and provide examples of situations in which you would use the concept you
have chosen.
• On slide or page 5, provide any resources you have used to give credit to others’
ideas and information. YOU MUST HAVE A REFERENCE PAGE !!!
• Check spelling and grammar and visit the Writing Center if needed.
• Submit your final project to the Unit 9 dropbox for grading. You will have an
opportunity to share with your classmates at the Math Fair in Unit 10
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Slide 13 - 3
Seminar Schedule
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Slide 13 - 4
MM150 Unit 9 Seminar
Statistics – Part II
•
•
•
•
9.1 – Measures of Central Tendency
9.2 – Measures of Dispersion
9.3 – The Normal Distribution
9.4 – Linear Regression and Correlation
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Slide 13 - 5
9.1
Measures of Central Tendency:
•Mean
•Median
•Mode
•Midrange
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 6
Definitions
• An average is a number that is representative
of a group of data.
• The arithmetic mean, or simply the mean is
symbolized by x , when it is a sample of a
population or by the Greek letter mu, , when
it is the entire population.
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Slide 13 - 7
Mean
• The mean, is the sum of the data divided by
the number of pieces of data. The formula for
calculating the mean is
Sx
x
n
• where Sx represents the sum of all the data
and n represents the number of pieces of
data.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 8
Example-find the mean
• Find the mean amount of money parents
spent on new school supplies and clothes if 5
parents randomly surveyed replied as follows:
$327 $465 $672 $150 $230
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Slide 13 - 9
Solution
327  465  672  150  230 1844
x

 368.8
5
5
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 10
Median
• The median is the value in the middle of a set
of ranked (ordered) data.
• Example: Determine the median of
$327 $465 $672 $150 $230.
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Slide 13 - 11
Solution
Rank the data from smallest to largest.
$150 $230 $327 $465 $672
middle value
(median)
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Slide 13 - 12
Example: Median (even data)
• Determine the median of the following set of
data: 8, 15, 9, 3, 4, 7, 11, 12, 6, 4.
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Slide 13 - 13
Solution
Rank the data:
3 4 4 6 7 8 9 11 12 15
There are 10 pieces of data so the median will
lie halfway between the two middle pieces the 7
and 8.
The median is (7 + 8)/2 = 7.5
3 4 4 6 7 8 9 11 12 15
(median) middle value
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 14
Summary to Find Median of Dataset
• If the number of datapoints is odd:
– The median is the middle value in the ordered
dataset.
• If the number of datapoints is even:
– The median is then the mean of the two middle
values in the ordered dataset.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 15
You Try It #1
• (A) Find the median of the following dataset:
10, 3, -4, 75, 420, 39, 6, 8
• (B) Find the median of the following dataset:
28, 19, 13, 19, 17, 1, 279
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 16
You Try It #1 - Solution
• (A) Find the median of the following dataset:
10, 3, -4, 75, 420, 39, 6, 8
Solution: Note there are eight datapoints.
Order the data: -4, 3, 6, 8, 10, 39, 75, 420
Median = (8+10)/2 = 18/2 = 9
• (B) Find the median of the following dataset:
28, 19, 13, 19, 17, 1, 279
Solution: Note there are seven datapoints.
Order the data: 1, 13, 17, 19, 19, 28, 279
Median = 19
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 17
Mode
• The mode is the piece of data that occurs
most frequently.
• Example: Determine the mode of the data set:
3, 4, 4, 6, 7, 8, 9, 11, 12, 15.
• Solution: The mode is 4 since it occurs twice
and the other values only occur once.
3, 4, 4, 6, 7, 8, 9, 11, 12, 15
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Slide 13 - 18
More on Mode
• If each piece of data occurs only once , then there is
no mode.
3, 16, 4, 6, 7, 8, 9, 11, 12, 15
• If two values occur in data set more often then
others then we say there are two modes (called
bimodal) – (Note some books refer to this situation
as “no mode”)
3, 16, 4, 6, 7, 7, 9, 12, 12, 15
Our text indicates there are two modes: 7 and 12
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 19
You Try It #2
• (A) Find the mode of the following dataset:
4, 7, 9, 11, 3, 7, 11, 9, 4, -2, 13, 4
• (B) Find the mode of the following dataset:
8, 11, 2, 13, 17, 9
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 20
You Try It #2 - Solution
• (A) Find the mode of the following dataset:
4, 7, 9, 11, 3, 7, 11, 9, 4, -2, 13, 4
Solution: Since the datavalue “4” occurs most often
(three times in the dataset), thus mode is 4.
• (B) Find the mode of the following dataset:
8, 11, 2, 13, 17, 9
Solution: There is no most frequently occurring
datavalue, thus there is no mode.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 21
Midrange
• The midrange is the value halfway between
the lowest (L) and highest (H) values in a set of
data.
lowest value + highest value
Midrange 
2
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Slide 13 - 22
Example
• Find the midrange of the data set $327, $465,
$672, $150, $230.
150 + 672 822
Midrange 

 411
2
2
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 23
Example
• The weights of eight Labrador retrievers
rounded to the nearest pound are 85, 92, 88,
75, 94, 88, 84, and 101. Determine the
a) mean
c) mode
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b) median
d) midrange
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Example--dog weights
85, 92, 88, 75, 94, 88, 84, 101
a. Mean
85  92  88  75  94  88  84  101 707
x

 88.375
8
8
b. Median-rank the data
75, 84, 85, 88, 88, 92, 94, 101
The median is 88.
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Slide 13 - 25
Example--dog weights
85, 92, 88, 75, 94, 88, 84, 101
c. Mode-the number that occurs most
frequently. The mode is 88.
d. Midrange = (L + H)/2
= (75 + 101)/2 = 88
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Slide 13 - 26
You Try It #3
•
The salaries of employees of a company are
provided below (in thousands of dollars):
32, 44, 35, 39, 125
(A) Find the mean of this dataset
(B) Find the median of this dataset
(C) Which measurement do you think is the better
measure of central tendency for this dataset?
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 27
You Try It #3 - Solution
•
The salaries of employees of a company are
provided below (in thousands of dollars):
32, 44, 35, 39, 125
(A) Find the mean of this dataset
Solution: Mean = (32 + 44 + 35 + 39 + 125) / 5 = 275 / 5 = 55
(B) Find the median of this dataset
Solution: Rank order the dataset: 32, 35, 39, 44, 125
Median = 39
(C) Which measurement do you think is the better measure
of central tendency for this dataset?
In this case, the median would be better measure of central
tendency since it is not affected by an outlier datapoint (in this
case, 125 is an outlier datavalue since it is significantly different
than the other datavalues in the dataset).
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Slide 13 - 28
Measures of Position
• Measures of position are often used to make
comparisons.
• Two measures of position are percentiles and
quartiles.
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Slide 13 - 29
To Find the Quartiles of a Set of Data
1. Order the data from smallest to largest.
2. Find the median, or 2nd quartile, of the set
of data.
1. If there are an odd number of datapoints, the
median is the middle value.
2. If there are an even number of datapoints, the
median will be the mean of the two middle
pieces of data.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 30
To Find the Quartiles of a Set of Data
continued
3.
The first quartile, Q1, is the median of the lower half
of the data; that is, Q1, is the median of the data less
than Q2.
4.
The third quartile, Q3, is the median of the upper
half of the data; that is, Q3 is the median of the data
greater than Q2.
5.
Note: The Quartiles divide the dataset into four
parts.
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Slide 13 - 31
Example: Quartiles
• The weekly grocery bills for 23 families are as
follows. Determine Q1, Q2, and Q3.
170
330
225
75
95
210
80
225
160
172
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270
170
215
130
190
270
240
310
74
280
270
50
81
Slide 13 - 32
Example: Quartiles continued
• Order the data:
50 75 74 80 81 95 : Q1
130 160 170 170 172
190 : Median (Q2 )
210 215 225 225 240 270 : Q3
270 270 280 310 330
Since there are 23 datapoints, the median will
be the middle datapoint (12th datapoint)
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Slide 13 - 33
Example: Quartiles continued
Q2 is the median of the entire data set which
is 190.
Q1 is the median of the numbers from 50 to
172 which is 95.
Q3 is the median of the numbers from 210 to
330 which is 270.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 34
You Try It #4
• Find the quartiles Q1, Q2, and Q3 of the
following dataset:
11, 7, 5, 9, 42, 27, 35, 19, 39, 36, 23
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 35
You Try It #4 - Solution
• Find the quartiles Q1, Q2, and Q3 of the
following dataset:
11, 7, 5, 9, 42, 27, 35, 19, 39, 36, 23
Solution:
First, rank order the data and find the median (this
is Q2) : 5, 7, 9, 11, 19, 23, 27, 35, 36, 39, 42
Thus Median, Q2 = 23
Now find median of lower set of data: 5, 7, 9, 11, 19
Now find median of upper set of data: 27, 35, 36, 39, 42
Thus, Q1 = 9 Q2 = 23 Q3 = 36
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Slide 13 - 36
9.2
Measures of Dispersion
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Slide 13 - 37
Measures of Dispersion
• Measures of dispersion are used to indicate
the spread of the data.
• The range is the difference between the
highest and lowest values; it indicates the
total spread of the data.
Range = highest value – lowest value
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Slide 13 - 38
Example: Range
• Nine different employees were selected and
the amount of their salary was recorded. Find
the range of the salaries.
$24,000 $32,000 $26,500
$56,000 $48,000 $27,000
$28,500 $34,500 $56,750
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 39
Solution
• Highest Value in Dataset = $56,750
• Lowest Value in Dataset = $24,000
• Range = $56,750  $24,000 = $32,750
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Slide 13 - 40
Standard Deviation
• The standard deviation measures how much
the data differ from the mean. It is symbolized
by “s” when it is calculated for a sample, and
with  (Greek letter sigma) when it is
calculated for a population.
s
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
S xx

2
n 1
Slide 13 - 41
To Find the Standard Deviation of a Set
of Data
1. Find the mean of the set of data.
2. Make a chart having three columns:
DataValue
DataValue  Mean
(DataValue  Mean)2
3. List the data vertically under the column
marked DataValue
4. Subtract the mean from each piece of data
and place the difference in the
DataValue  Mean column.
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Slide 13 - 42
To Find the Standard Deviation of a Set
of Data continued
5. Square the values obtained in the Data 
Mean column and record these values in the
(DataValue  Mean)2 column.
6. Determine the sum of the values in the
(DataValue  Mean)2 column.
7. Divide the sum obtained in step 6 by n  1,
where n is the number of datapoints.
8. Determine the square root of the number
obtained in step 7. This number is then the
standard deviation of the set of data.
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Slide 13 - 43
Example
• Find the standard deviation of the following
prices of selected washing machines:
$280, $217, $665, $684, $939, $299
Find the mean.
280  217  665  684  939  299 3084
x

 514
6
6
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Slide 13 - 44
Example continued, mean = 514
Data
217
280
299
665
684
939
Data  Mean
297
234
215
151
170
425
0
Copyright © 2009 Pearson Education, Inc.
(Data  Mean)2
(297)2 = 88,209
54,756
46,225
22,801
28,900
180,625
421,516
Slide 13 - 45
Example continued, mean = 514
s

S xx

2
n 1
421,516

 84303.2  290.35
5
• The standard deviation is $290.35.
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Slide 13 - 46
9.3
The Normal Curve
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Slide 13 - 47
Types of Distributions
• Rectangular Distribution
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• J-shaped distribution
Slide 13 - 48
Types of Distributions continued
• Bimodal
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• Skewed to right
Slide 13 - 49
Types of Distributions continued
• Skewed to left
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• Normal
Slide 13 - 50
Properties of a Normal Distribution
• The graph of a normal distribution is called
the normal curve.
• The normal curve is bell shaped and
symmetric about the mean.
• In a normal distribution, the mean, median,
and mode all have the same value and all
occur at the center of the distribution.
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Slide 13 - 51
Empirical Rule
• Approximately 68% of all the data lie within
one standard deviation of the mean (in both
directions).
• Approximately 95% of all the data lie within
two standard deviations of the mean (in both
directions).
• Approximately 99.7% of all the data lie within
three standard deviations of the mean (in
both directions).
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Slide 13 - 52
Normal Distribution
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Slide 13 - 53
z-Scores
• z-scores determine how far, in terms of
standard deviations, a given score is from the
mean of the distribution.
value of the piece of data - mean x  
z

standard deviation
s
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 54
Example: z-scores
• A normal distribution has a mean of 50 and a
standard deviation of 5. Find z-scores for the
following values.
• a) 55
b) 60
c) 43
55  50 5
• a) z 
 1
5
5
A score of 55 is one standard deviation
above the mean.
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Slide 13 - 55
Example: z-scores continued
60  50 10
• b) z 

2
5
5
A score of 60 is 2 standard deviations above
the mean.
43  50 7

 1.4
• c) z 
5
5
A score of 43 is 1.4 standard deviations below
the mean.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 56
To Find the Percent of Data Between
any Two Values
1.
2.
3.
Draw a diagram of the normal curve,
indicating the area or percent to be
determined.
Use the formula to convert the given
values to z-scores. Indicate these zscores on the diagram.
Look up the percent that corresponds to
each z-score on page 387-388.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 57
To Find the Percent of Data Between
any Two Values continued
4.
a) When finding the percent of data between two zscores on opposite sides of the mean (when one
z-score is positive and the other is negative), you
find the sum of the individual percents.
b) When finding the percent of data between two zscores on the same side of the mean (when both
z-scores are positive or both are negative),
subtract the smaller percent from the larger
percent.
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Slide 13 - 58
To Find the Percent of Data Between
any Two Values continued
c) When finding the percent of data to the right of a
positive z-score or to the left of a negative z-score,
subtract the percent of data between 0 and z from
50%.
d) When finding the percent of data to the left of a
positive z-score or to the right of a negative zscore, add the percent of data between 0 and z to
50%.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 59
Example
Assume that the waiting times for customers at
a popular restaurant before being seated for
lunch are normally distributed with a mean of
12 minutes and a standard deviation of 3 min.
a) Find the percent of customers who wait for at
least 12 minutes before being seated.
b) Find the percent of customers who wait between
9 and 18 minutes before being seated.
c) Find the percent of customers who wait at least
17 minutes before being seated.
d) Find the percent of customers who wait less than
8 minutes before being seated.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 60
Solution
a. wait for at least 12
minutes
Since 12 minutes is the
mean, half, or 50% of
customers wait at least 12
min before being seated.
b. between 9 and 18
minutes
9  12 3
z

 1
3
3
18  12 6
z
 2
3
3
Use Table 13.7 Page 801
34.1% + 47.7% = 81.8%
0.341 + 0.477 = 0.818
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 61
Solution continued
c. at least 17 min
d. less than 8 min
Use table 13.7 page
801.
45.3% is between the
mean and 1.67.
50%  45.3% = 4.7%
Thus, 4.7% of
customers wait at least
17 minutes.
Use table 13.7 page
801.
40.8% is between the
mean and 1.33.
50%  40.8% = 9.2%
Thus, 9.2% of
customers wait less
than 8 minutes.
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Slide 13 - 62
9.4
Linear Correlation and Regression
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Slide 13 - 63
Linear Correlation
•
Linear correlation is used to determine
whether there is a relationship between two
quantities and, if so, how strong the
relationship is.
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Slide 13 - 64
Linear Correlation
– The linear correlation coefficient, r, is a
unitless measure that describes the strength
of the linear relationship between two
variables.
• If the value is positive, as one variable
increases, the other increases.
• If the value is negative, as one variable
increases, the other decreases.
• The variable, r, will always be a value
between –1 and 1 inclusive.
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Slide 13 - 65
Scatter Diagrams
• A visual aid used with correlation is the scatter
diagram, a plot of points (bivariate data).
– The independent variable, x, generally is a quantity
that can be controlled.
– The dependent variable, y, is the other variable.
• The value of r is a measure of how far a set of points
varies from a straight line.
– The greater the spread, the weaker the correlation
and the closer the r value is to 0.
– The smaller the spread, the stronger the correlation
and the closer the r value is to 1.
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Slide 13 - 66
Correlation
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Slide 13 - 67
Correlation
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Slide 13 - 68
Linear Correlation Coefficient
• The formula to calculate the correlation
coefficient (r) is as follows:
r
n   xy     x   y 

n x
Copyright © 2009 Pearson Education, Inc.
2
   x 
2

n y
2
   y 
2
Slide 13 - 69
Example: Words Per Minute versus
Mistakes
There are five applicants applying for a job as a
medical transcriptionist. The following shows
the results of the applicants when asked to type
a chart. Determine the correlation coefficient
between the words per minute typed and the
number of mistakes.
Applicant
Ellen
George
Phillip
Kendra
Nancy
Words per Minute
24
67
53
41
34
Copyright © 2009 Pearson Education, Inc.
Mistakes
8
11
12
10
9
Slide 13 - 70
Solution
• We will call the words typed per minute, x,
and the mistakes, y.
• List the values of x and y and calculate the
necessary sums.
WPM Mistakes
x
y
x2
y2
24
8
576
64
67
11
4489
121
53
12
2809
144
41
10
1681
100
34
9
1156
81
x2 =10,711y2 = 510
x = 219 y = 50
Copyright © 2009 Pearson Education, Inc.

xy
192
737
636
410
306
xy =
Slide 13 - 71
2,281
Solution continued
• The n in the formula represents the number of
pieces of data. Here n = 5.
r

   
n  x   x  n  y   y 
5 2281 219 50 
5 10,711 219  5 510  50 
n  xy   x  y
2
r
Copyright © 2009 Pearson Education, Inc.
2
2
2
2
2
Slide 13 - 72
Solution continued




11,405  10,950

 
5 10,711  47,961 5 510  2500
455
53,555  47,961 2550  2500
455
 0.86
5594 50
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Slide 13 - 73
Solution continued
• Since 0.86 is fairly close to 1, there is a fairly
strong positive correlation.
• This result implies that the more words typed
per minute, the more mistakes made.
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Slide 13 - 74
Linear Regression
• Linear regression is the process of determining
the linear relationship between two variables.
• The line of best fit (regression line or the least
squares line) is the line such that the sum of
the squares of the vertical distances from the
line to the data points (on a scatter diagram) is
a minimum.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 75
The Line of Best Fit
• Equation:
y  mx  b,
m
where

   ,
n  x   x 
n  xy   x  y
2
Copyright © 2009 Pearson Education, Inc.
2
and b 
 
y  m x
n
Slide 13 - 76
Example
•
•
Use the data in the previous example to find
the equation of the line that relates the
number of words per minute and the
number of mistakes made while typing a
chart.
Graph the equation of the line of best fit on
a scatter diagram that illustrates the set of
bivariate points.
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 77
Solution
• From the previous results, we know that
m

   
n  x   x 
n  xy   x  y
2
2
5(2,281)  (219)(50)
m
5(10,711)  219 2
455
m
5594
m  0.081
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 78
Solution
• Now we find the y-intercept, b.
b
b
 
y  m x
n
50  0.081 219
 
5
32.261
b
 6.452
5
Therefore the line of best fit is y = 0.081x + 6.452
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 79
Solution continued
• To graph y = 0.081x + 6.452, plot at least two
points and draw the graph.
x
10
20
30
Copyright © 2009 Pearson Education, Inc.
y
7.262
8.072
8.882
Slide 13 - 80
Solution continued
Copyright © 2009 Pearson Education, Inc.
Slide 13 - 81