Download MM150 Unit 9 Seminar

MM150 Unit 9 Seminar
4 Measures of Central Tendency
Mean – To find the arithmetic mean, or mean, sum the data scores
and then divide by the number of data scores.
Example: Find the mean of the data scores 5, 7, 2, 9, 8
5 + 6 + 2 + 9 + 8 = 30 = 6
5
5
Median – To find the median, put the data scores in ascending or
descending order and then find the middle data score. If there are an
even number of data scores, after ranking the scores, find the mean
of the middle two.
Example: Find the median of the data scores 5, 7, 2, 9, 8
Put the scores in ascending order: 2, 5, 7, 8, 9
Example: Find the median of the data scores 4, 7, 2, 9
Put the scores in ascending order: 2, 4, 7, 9
Find the mean of 4 and 7: (4 + 7)/2 = 11/2 = 5.5
4 Measures of Central Tendency
Mode – The mode is the data score that occurs most frequently.
Example: Find the mode of the data scores 6, 4, 9, 8, 6, 5
It may help to put the scores in ascending order: 4, 5, 6, 6, 8, 9
You can see that the data score 6 occurs most often.
•You can have data sets that don’t have a mode (each score occurs once)
and you can have data sets that are bimodal – which means they have 2
modes.
Midrange – The midrange is the value halfway between the greatest and
least data score. To find it, take the mean of the greatest and least data
score.
Example: Find the midrange of the data scores 6, 4, 9, 8, 6, 5
It may help to put the scores is ascending order: 4, 5, 6, 6, 8, 9
The midrange is (4 + 9)/2 = 13/2 = 6.5
*Please read on page 362 of your text when each is the ‘better’ average.
Mean Example
Todd is taking a math class where his end of term grade is
based on 4 exams, each having the same number of points
and weighted the same. He scored 98, 82, and 87 on the first
three exams in his class. What does he need to score on the
4th exam to get at least a 90% for the final grade? The
instructor uses mean as the average.
98 + 82 + 87 + x = 90
4
267 + x = 90
4
267 + x = 360
X = 93
Todd must score a 93 or higher on the test.
EVERYONE Solution
Mean: 10 + 11 + 11 + 12 + 13 + 15 = 72 = 12
6
6
Median: 10, 11, 11, 12, 13, 15
There are an even number of data scores: 11 + 12 = 23 = 11.5
2
2
Mode: 10, 11, 11, 12, 13, 15
The data score 11 occurs most often.
Midrange: 10 + 15 = 25 = 12.5
2
2
2 Measures of Position
Percentile – There are 99 percentiles that divide the data up into
100 equal parts.
Quartile – Quartiles divide data into 4 equal parts, called quartiles.
The first quartile is at 25%, the second at 50%, and the third at
75%.
Determine Q1, Q2, and Q3 of the data below:
15, 10, 19, 18, 11, 15, 13, 18, 19, 17, 19, 15, 16, 13, 15, 16, 13, 12,
14
First put the data in ascending order
10, 11, 12, 13, 13, 13, 14, 15, 15, 15, 15, 16, 16, 17, 18, 18, 19, 19,
19
•Find the median, 15 is Q2
•Find the median of the lower half for Q1, 13
•Find the median of the upper half for Q3, 18
2 Measures of Dispersion
Range – The range is the difference between the greatest and
least data score.
Example: Find the range of the data scores 55, 59, 51,
64, 60
Put the data scores in ascending order 51, 55, 59, 60, 64
The range is 64 – 51 = 13
Standard deviation – The standard deviation tells us how much
the data differ from the mean.
See the next powerpoint slide for an example of standard
deviation.
Standard Deviation Example
Find the standard deviation of 11, 15, 18, 9, 12
1.Find the mean of the data scores. 11+15+18+9+12 = 65 = 13
5
5
2. Make a chart with 3 columns
3.
4.
5.
Data
Data – Mean
(Data – Mean)2
11
11- 13 = -2
4
15
15 – 13 = 2
4
18
18 – 13 = 5
25
9
9 – 13 = -4
16
12
12 – 13 = -1
1
6.
50
7. Divide 50 by n – 1, where n is the number of data scores. So divide 50 by 4,
which is 12.5
8. Find the square root of the number found in step 7. √12.5 ≈ 3.5355
EVERYONE solution
1. Find the mean. 15 + 16 + 20 + 13 = 64 = 16
4
4
2. Make a 3-column table.
3.
Data
4. Data – Mean
5.(Data – Mean)2
15
-1
1
16
0
0
20
4
16
13
-3
9
6. 26
7. 26/3 = 8.667
8. √8.667 ≈ 2.944
Rectangular Distribution
Frequency
Observed Values
J-Shaped Distributions
Constantly Increasing
Constantly Decreasing
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Skewed Distributions
Mode
Median
Mean
12
Normal Distributions
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Normal Distribution
PROPERTIES OF A NORMAL DISTRIBUTION
1. The graph of a normal distribution is called a normal curve.
2. The normal curve is bell-shaped and symmetric about the mean.
3. The mean, median, and mode of a normal distribution all have the
same value and all occur a the center of the distribution.
EMPIRICAL RULE
In any normal distribution
1. Approximately 68% of all the data lies within one standard deviation
of the mean (in both directions).
2. Approximately 95% of all the data lies within two standard deviations
of the mean (in both directions).
3. Approximately 99.7% of all the data lies within three standard
deviations of the mean (in both directions).
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Z-Score
z = value of a given data score - mean
standard deviation
Example:
A normal distribution has a mean of 50 and a standard deviation of
10. Determine the z-score for the data value of 60.
z60 = value - mean
standard dev.
z60 = 60 - 50
10
z60 = 10 = 1 The score of 60 is one standard deviation
10
above the mean.
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EVERYONE Solution
A normal distribution has a mean of 50 and a standard deviation of
10. Determine the z-score for the data value of 45.
z45 = value - mean
standard dev.
z45 = 45 - 50
10
z45 = -5 = -1 The score of 45 is 0.5 standard deviation
10
2 below the mean.
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Linear Correlation Coefficient
r=
n(∑xy) - (∑x)(∑y)
√[n(∑x2)- (∑x)2]√[n(∑y2) - (∑y)2]
Let’s do Page 407, #24.
The first thing to do is plot the points.
Here we have (6,13), (8,11), (11,9),
(14,10) and (17,7).
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x
y
x2
y2
xy
6
13
36
169
78
8
11
64
121
88
11
9
121
81
99
14
10
196
100
140
17
7
289
49
119
56
50
706
520
524
r=
n(∑xy) - (∑x)(∑y)
√[n(∑x2)- (∑x)2]√[n(∑y2) - (∑y)2]
r=
5(524) - (56)(50)
√[(5)(706) - 562]√[(5)(520) - 502]
r=
2620 - 2800
√[3530 - 3136]√[2600 - 2500]
r=
-180
√394 √100
=
-180
19.8494 * 10
=
|-0.907| = 0.907
Remember n = 5
c) 0.907 > 0.878 so a
correlation exists.
d) 0.907 < 0.959 so a
correlation does not exist.
-180
=
198.494
-0.907
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Linear Regression
Slope-Intercept form of a line:
y = mx + b
where m is the slope and (0,b) is the y-intercept.
Linear Regression Formula
y = mx + b
m = n(∑xy) - (∑x)(∑y)
n(∑x2) - (∑x)2
and
b = ∑y - m(∑x)
n
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Linear Regression Example
y = mx + b
m = n(∑xy) - (∑x)(∑y)
n(∑x2) - (∑x)2
and
b = ∑y - m(∑x)
n
Let’s use the data from Page 407 #24 to save us some work!
n∑xy - (∑x)(∑y) = -180
n(∑x2) - (∑x)2 = 394
∑x = 56
∑y = 50
m = -180/394 = -0.457
b = 50 - (-0.457)(56)
5
=
50 + 25.584
5
=
15.117
y = -0.457x + 15.117
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