Elementary Statistics and
Inference
22S:025 or 7P:025
Lecture 5
1
Elementary Statistics and
Inference
22S:025 or 7P:025
Chapter 4 (cont.)
2
5.) Chapter Four (cont.)
E.
The Standard Deviation
ƒ
The Standard Deviation (SD) is an index that describes
the spread of scores in a histogram around the
average (mean).
(mean) The SD is an index in score scale
units.
Researchers often report scores (measures) in terms
of the number of SD units a score is from the average
(mean).
3
1
5.) Chapter Four (cont.)
„
For the HANES data for women ages 18-74 (see page
58) – their average height was 63.5 inches and the SD of
their heights was close to 3 inches.
For a woman who was 66 inches tall
tall, she was .8
8 SD
above average height.
For a woman who was 60 inches tall, she was 1.16 SD
below average height.
4
5.) Chapter Four (cont.)
Example:
Group
Mean
Median
SD
Boys
20
18
6
Girls
20
20
3
„
Boys scores are skewed right, and more variable than
girls. Girls scores are symmetric, and less variable –
even though the averages are the same.
„
See example of types of histograms – page 64, Figure 7.
5
5.) Chapter Four (cont.)
In many histograms about 68% of the scores are within 2
SDs of the mean, and roughly 95% of the scores are
within 3 SDs of the mean.
About 68% of Scores
Mean -1 SD
Mean
Mean +1 SD
About 95% of Scores
Mean -2SD
Mean
Mean +2 SD
6
2
5.) Chapter Four (cont.)
„
In the HANES example for women age 18-74, about
72% of the women had heights within 1 standard
deviation of the mean - See page 68 (Figure 8), and
about 97% had heights within 2 standard deviations of
the mean (Figure 9) – page 69.
Exercise Set D – page 70, #1, 2, 3, 9
7
5.) Chapter Four (cont.)
8
5.) Chapter Four (cont.)
F.
Computing the Standard Deviation
ƒ
For each score, compute the deviation between the
score and the mean, square the difference. Do this for
each score,
score and find the average of the squared
deviations. Then compute the square root of the result.
Example:
Scores:
1, 3, 5, 7, 9
Sum = 25
Mean = 5
9
3
5.) Chapter Four (cont.)
Deviation squared:
SD =
(1 − 5) 2 + (3 − 5) 2 + (5 − 5) 2 + (7 − 5) 2 + (9 − 5) 2
5
SD =
16 + 4 + 0 + 4 + 16
=
5
40
= 8 = 2.8284
5
10
5.) Chapter Four (cont.)
In Symbols:
Σ( X − mean)
Σ( X − X )
=
n
n
2
SD =
2
Or
n ΣX 2 − (ΣX ) 2
n2
ΣX 2 = 1 + 9 + 25 + 49 + 81 = 165
ΣX = 25
SD =
SD =
(5)(165) − (25) 2
825 − 625
200
=
=
= 8 = 2.8284
5⋅5
25
25
11
5.) Chapter Four (cont.)
Exercise Set E – pages 72-74. 1, 2, 3, 4, 8, 11, 12
Note: The SD computed with a calculator is slightly
different from the value mentioned above
SD + =
n
⋅ ( SD)
n −1
Review Exercises – page 74-76. 1, 2, 5, 6, 7, 8, 9
12
4
5.) Chapter Four (cont.)
Computing Mean and Standard Deviation for a
Distribution of Scores:
Note: Assume scores are evenly distributed throughout
the score interval.
f ·X2
X
f
cf
f ·X
10
1
39
10
100
9
2
38
18
162
8
3
36
24
192
7
4
33
28
196
6
9
29
54
324
5
8
20
40
200
4
6
12
24
96
3
4
6
12
36
2
2
2
N=39
4
8
Sum=214
Sum of each score
squared = 1314
13
5.) Chapter Four (cont.)
ΣX 214
=
= 5.49
n
39
2
2
(39)(1314) − (214)2 = 51,246 − 45,796 = 3.58
nΣX − (ΣX )
S2 =
=
2
(39) ⋅ (39)
(39)(39)
n
S = 1.89
mean = X =
S+ =
n
39
⋅S =
⋅ (1.89) = 1.91
n −1
38
14
5.) Chapter Four (cont.)
For Grouped Histogram (frequency distribution):
f · X2
X (class)
f
cf
f·X
19-21
2
32
40
800
16-18
4
30
68
1,156
,
13-15
6
26
84
1,176
10-12
8
20
88
968
7-9
6
12
48
384
4-6
4
6
20
100
2
2
1-3
N=32
4
8
Sum=352
Sum=4,592
15
5
5.) Chapter Four (cont.)
Note: Assume scores in a class interval are evenly
distributed, and use midpoint of interval to calculate
mean and SD.
Σf ⋅ X 352
=
= 11.00
n
32
2
nΣfX 2 − (ΣfX ) 2 32(4592) − (352 )
approx. var iance = S 2 ≈
=
(32)(32)
(n ) ⋅ (n )
146,944 − 123,904 23,040
2
S ≈
=
= 22.5
(32)(32)
1,024
S ≈ 4.74
approx. mean = X ≈
16
5.) Chapter Four (cont.)
Effect on Mean and SD if add constant to each score or
multiply each score by constant:
= C + X = constant plus original mean
„
M
„
S c + x = S x = standard deviation not affected
c+ x
example: If X = 30 and S = 3
Add 10 to each score
Subtract 3 from each score
Mean = 40
Mean = 27
SD=3
SD=3
17
5.) Chapter Four (cont.)
„
„
M cx = C ⋅ X =multiply each score by constant, the new mean
is the constant times the original mean
S cx = C ⋅ S x =multiply each score by constant, the new SD is
the absolute value of the constant times the original SD
Example: M X = 30
SX = 3
M 4 X = 4(30) = 120,
S 4 X = 4(3) = 12
M −2 X = (−2)(30) = −60,
S −2 X = − 2 ⋅ 3 = 6
18
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