Download 251dscr_D

251dscr_D 9/25/02 (Open this document in 'Outline' view!)
Appendix D: Review
5. Review
a. Grouped Data.
Consider the following sample:
Class
f
0.5 - 1.5
1.5 - 2.5
2.5 - 3.5
3.5 - 4.5
1
0
1
2
a.
b.
c.
d.
e.
f.
Calculate the mean
Calculate the variance and standard deviation
Calculate and interpret the skewness and relative skewness
Calculate and interpret Pearson's measure of skewness
Calculate the median
Calculate the interquartile range
Solution:
You need the following:
x
Class
(midpoint)
0.5 - 1.5
1
1.5 - 2.5
2
2.5 - 3.5
3
3.5 - 4.5
4
1
0
1
2
4
fx  12 ,
a.
1
1
2
4
 f  n  4, 
 fx
 fx  12  3
Calculate the mean x 
Note:
n
fx
F
f
2
 42 ,
4
b. Calculate the variance and standard deviation s 2 

1
0
3
8
12
fx 3  156 .
 fx
2
 nx 2
n 1

fx2
fx3
1
0
9
32
42
1
0
27
128
156
42  432 6
 2
3
3
s
2

 0.471
x
3
c. Calculate and interpret* the skewness and relative skewness
n
k 3
fx 3  3x
fx 2  2nx 3
(n  1)( n  2)
s  variance  2  1.414 C 





k
4
2
4
 1.414
156  3342   2433   6  4 . g 1  33 
3
3
(3)( 2)
s
2
d. Calculate and interpret* Pearson's measure of skewness
The mode is 4, since that occurs most.

 
SK 
3mean  mode 33  4

 2.121
std .deviation
2
2
e. Calculate the median
 pN  F 
First use position  pn  1 , then use x1 p  L p  
 w to find the value. Here
 f p 
p  .5 . So pn  1  .55  2.5 . This location is above 2 and below 4, so use the class 3.5 to
 .54  2 
4.5. Then x.5  3.5  
1  3.5 .
 2 
f. Calculate the interquartile range
For the first quartile position  pn  1  .255  1.25 . This location is above 1 and below 2 , so
 .25 4   1 
x.75  Q1  2.5  
1  2.5
1


For the third quartile pn  1  .755  3.75 . This location is above 2 and below 4, so use the
use the class 2.5 to 3.5. Then, using the second formula, we find
 .75 4   2 
x.25  Q3  3.5  
1  4.0 .
2


IQR  Q3  Q1  4.0  2.5  1.5
class 3.5 to 4.5. Then, we find
b. Ungrouped Data.
Consider the following sample: 1, 3, 4, 4 -which can also be written as
x
1
x1
x2
x3
x4
3
4
4
a.
b.
c.
d.
e.
f.
Calculate the mean
Calculate the variance, standard deviation and coefficient of variation.
Calculate and interpret the skewness and relative skewness
Calculate and interpret Pearson's measure of skewness
Calculate the median
Calculate the interquartile range
Solution:
You need the following:
x
1
x1
x2
x3
x4
x2
1
9
16
16
42
x 2  42,
3
4
4
12
Note: n  4,
a.
 x  12 , 
 x  12  3
Calculate the mean x 
n
x3
1
27
64
64
156
x 3  156 .

4
3
b. Calculate the variance and standard deviation s 2 
x
2
 nx 2
n 1

42  432 6
 2
3
3
s
2

 0.471
x
3
c. Calculate and interpret* the skewness and relative skewness
n
k 3
x 3  3x
x 2  2nx 3
(n  1)( n  2)
s  variance  2  1.414 C 





k
4
2
4
 1.414
156  3342   2433   6  4 . g 1  33 
3
3
(3)( 2)
s
2
d. Calculate and interpret* Pearson's measure of skewness
The mode is 4, since that occurs most.
3mean  mode 33  4
SK 

 2.121
std .deviation
2
e. Calculate the median
First use position  pn  1  a.b , then use x1 p  xa  .bxa1  xa  to find the value. Here p  .5 . So

 
pn  1  .55  2.5 . Thus a  2 and .b  0.5 . So, using the second formula,
x.5  x 2  0.5x 21  x 2   x 2  0.5x3  x 2   3  0.54  3  3.5.
f. Calculate the interquartile range
For the first quartile position  pn  1  .255  1.25 . Thus a  1 and .b  0.25 . So
x.75  Q1  x1  0.25x11  x1   x1  0.25x 2  x1   1  0.253  1  1.5.
For the third quartile pn  1  .755  3.75 . Thus a  3 and .b  0.75 . So
x.25  Q3  x3  0.75x31  x3   x3  0.75x 4  x3   4  0.754  4  4.
IQR  Q3  Q1  4.0 1.5  2.5
* These negative numbers all indicate skewness to the left.
4