3.3 Stats

1. 1
3
6
10
8
7
0
2
4
6
8
10
12
30 33 36 39 42 45
Frequency
Temperature
Temperature
#18

2. 1
3
6
10
8
7
0
2
4
6
8
10
12
30 33 36 39 42 45
Frequency
Temperature
Temperature
#18

3. 49%
39%
27%
20%
16%
9%
0%
10%
20%
30%
40%
50%
60%
Business Engineering Liberal Arts Education Science Social Sciences
Percentages
Majors
Percent of Students with Different Majors#19

4. 25%
30%15%
10%
20%
Color of Cars Preferred by Customers
108°
#20

5. Set 2 Set 1
8, 6, 3, 0 1 0, 2
8, 3, 3 2 2, 2, 4, 6, 7
7, 6, 1, 0 3 1, 4, 5, 9
5 4 9
#21

6. 3-3: Measures of Variation
Objective: To describe data using
measures of variation, such as the
range, variance, and standard
deviation.

7.  A testing lab wishes to test two
experimental brands of outdoor paint to
see how long each will last before fading.
The testing lab makes 6 gallons of each
paint to test. Since different chemical
agents are added to each group and only
six cans are involved, these two groups
constitute two small populations. The
results (in months) are shown. Find the
mean of each group.

8. Brand A Brand B
10 35
60 45
50 30
30 35
40 40
20 25

9.  Mean for brand A:
 Mean for brand B:
6
210


N
X

6
210


N
X


10. Brand A
X X X X X X
10 15 20 25 30 35 40 45 50 55 60
Variation in Paint (in months)
Brand B
X
X X X X X
10 15 20 25 30 35 40 45 50 55 60

11.  Even though the means of the two sets
were the same, the spread or variation,
was very different.

12.  Three common measures of spread or
variability of a set of data:
◦ Range
◦ Variance
◦ Standard Deviation

13.  Range: highest value – lowest value
◦ “R” is the symbol used for the range

14. Brand A Brand B
10 35
60 45
50 30
30 35
40 40
20 25
Range for set A: 60 – 10 = 50 months
Range for set B: 45 – 25 = 20 months

15.  Rounding Rule for the Standard Deviation:
Same as for the mean. Round to one
more decimal place than the original data.

16.  Find the variance and the standard
deviation for the fading time of paint.
 Brand A: 10, 60, 50, 30, 40, 20

17. Step 1- Find the mean for the data
35
6
210
6
204030506010




N
X


18. Step 2: Subtract the Mean from
each data point.
 10 – 35 = -25
 60 – 35 = +25
 50 – 35 = 15
 30 – 35 = -5
 40 – 35 = +5
 20 – 35 = -15

19. Step 3: Square each result.
 Square
 10 – 35 = -25………625
 60 – 35 = +25…….625
 50 – 35 = 15……….225
 30 – 35 = -5…………25
 40 – 35 = +5………..25
 20 – 35 = -15……….225

20. Step 4: Find the sum of the
squares
 625 + 625 + 225 + 25 + 25 + 225 = 1750

21. Step 5: Divide the sum by N to get
the variance.
 1750 ÷ 6 = 291.7
 Variance = 291.7

22. Step 6: Standard Deviation is the
square root of the variance.
1.177.291 

23.  Find the variance and standard deviation
for Brand B: 35, 45, 30, 35, 40, 25
1. Find the mean.
2. Subtract mean from each data value.
3. Square each result.
4. Find the sum of the squares.
5. Divide sum by N to get variance.
6. Take square root to get standard deviation.

24. A B C
X 2. X-μ 3. (X-μ)²
35 35-35=0 0²=0
45 45-35=10 10²=100
30 30-35=-5 (-5) ²=25
35 35-35=0 0²=0
40 40-35=5 5²=25
25 25-35=-10 (-10) ²=100
1. Calculate the mean: 210/6 = 35 months
4. Find the sum of column C: 0+100+25+0+25+100=250
5. Divide sum (step 4) by N to get the variance: 250/6=41.7
6. Take square root of the variance (step 5) to get the
standard deviation:
5.6
6
250


25.  Compare set A to set B
 Any conclusions? (see slide 10)
Set A Set B
Variance 291.7 41.7
Standard Deviation 17.1 6.5

26.  Variance: The average of the squares of
the distance each value point is from the
mean.
 Symbol: σ²
 Population Variance:
 Where X: individual value
μ: population mean
N: population size
N
X 2
2
)( 




27.  Standard Deviation: square root of the
variance.
 Symbol: σ
 Population Standard Deviation:
N
X 

2
2
)( 


28. Sample Variance
1
)( 2
2




n
XX
s
Where
X
X
= individual value
= sample mean
n = sample size

29. Sample Standard Deviation
1
)( 2
2




n
XX
ss
Where
X
X
= individual value
= sample mean
n = sample size

30. Computational Formula for s² and s
 Variance
 Standard Deviation
1
)( 2
2
2











 
n
n
X
X
s
1
)( 2
2











 
n
n
X
X
s

31. Example 3-23, p. 121
 Use the computational formulas for s and
s² to find the standard deviation and the
variance for the amount of European auto
sales (in millions) for a sample of 6 years
shown: 11.2, 11.9, 12.0, 12.8, 13.4,
14.3
 Answers: s²=1.28 million
s = 1.13 million

32. Variance and Standard Deviation
for Grouped Data
 Procedure for finding the variance and
standard deviation for grouped data is
similar to that for finding the mean for
grouped data: use the midpoint.

33. Procedure for Finding the Sample
Variance and Standard Deviation
for Grouped Data
 1) Make a table with the following columns
 2) Multiply: Frequency * Midpoint (column D)
 3) Multiply: Frequency * Midpoint squared
(column E)
 4) Total columns B, D, and E.
◦ Total of B is n.
◦ Total of D is
◦ Total of E is
A B C D E
Class Frequency Midpoint mXf  2
m
Xf 
  )( mXf
  )( 2
mXf

34. Grouped Data-Variance &
Standard Deviation cont’d
 5) Substitute values from step 4 into
 6) Take the square root of the variance
(step 5) to find the standard deviation.
1
)(
)(
2
2
2








 



n
n
Xf
Xf
s
m
m

35. Uses of Variance and Standard
Deviation
 Determine spread of data (large values
mean data is fairly spread out)
 Determine consistency of a variable.
(Nuts & bolts diameters must have small
variance & st. dev.)
 Used to determine how many data values
fall within certain interval. (Chebyshev-
75% within 2 st. dev. of mean).
 Used in inferential statistics (we’ll see how
later).

36. Coefficient of Variation
 Allows comparison of data with different
units (number of sales per salesperson vs.
commissions made by salesperson).
 Coefficient of Variation:
◦ Denoted: Cvar
◦ For Samples:
◦ For Populations:
%100
X
s
CVar
%100


CVar

37. Example-Coefficient of Variation
 Example 3-25
 The mean of the number of sales of cars
over a 3-month period is 87 and the
standard deviation is 5. The mean of the
commissions is $5225 and the standard
deviation is $773. Compare the variations
of the two.
 Sales:
 Commission:
%7.5%100
87
5

X
s
CVar
%8.14%100
5225
773



CVar

38. Range Rule of Thumb
4
range
s 
• Only an approximation
• Use only when distribution is
unimodal and roughly symmetric
• Can be used to find large value and
small value when you know the mean
and the standard deviation
• Large:
• Small:
sX 2
sX 2

39.  For many sets of data, almost all values
fall within 2 standard deviations of the
mean.
 Better approximations can be obtained by
using Chebyshev’s Theorem.

40. Chebyshev’s Theorem
 Specifies the proportions of the spread in
terms of the standard deviation (for any
shaped distribution)
 Theorem states: The proportion of values
from a data set that will fall within k standard
deviations of the mean, will be at least
where k is a number greater than 1 (k is not
necessarily an integer).
2
1
1
k


41. Example of Chebyshev’s Theorem
 What percent of the data in a set should
fall within 3 standard deviations of the
mean?
 So, 89% of the numbers in the set fall
within 3 standard deviations of the mean.
%89
9
8
9
1
1
3
1
1
1
1 22

k

42. Empirical Rule
 Applies only to bell-shaped (normal-shaped)
distributions.
 Rule states:
◦ Approximately 68% of the data values fall within 1
standard deviation of the mean.
◦ Approximately 95% of the data values fall within 2
standard deviations of the mean.
◦ Approximately 99.7% of the data values fall within
3 standard deviations of the mean.
 See Figure 3-4, top of p. 128

43. Homework
p.129-132
#1-5, 7, 11, 13, 19, 31-41 odd