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§ Descriptive Measures §
Lesson 2- Examples
§ Sample Mean § Population Mean § Mode §
§ Median § Range § Sample Variance §
§ Population Variance § Sample Standard Deviation §
§ Z Score § Empirical Rule §
Example 1
Sample Mean
Given the following data on x: Sample Data> 2, 4, 6, 6, 10. Give the hand calculations and solve for the mean using Excel. Explain xbar in terms of the individual deviations of the x values from xbar.
xbar = mean = Sx / n = 28 / 5 = 5.6
where m means sum
x’s are the individual data values
n is the number of data values
The mean is a number when subtracted from each individual x value, the resulting deviations (x – xbar) sum to zero.
|
x |
- xbar |
= d |
|
|
2 |
- 5.6 |
= -3.6 |
|
|
4 |
- 5.6 |
= -1.6 |
|
|
6 |
- 5.6 |
= +0.4 |
|
|
6 |
- 5.6 |
= +0.4 |
|
|
10 |
- 5.6 |
= +4.4 |
S(x - xbar) = Sd = 0
(click me)
Change a single number in a set of data to a larger
number. Only the deviation associated with that number would change and all
other deviations remain the same. True
False
(click one)
Example 2
Population Mean
Given the following data on x from which the sample in Example 1 was taken: Population data>2, 2, 4, 4, 4, 6, 6, 6, 6, 8, 8, 8, 10, 10, 14. Calculate the mean value of x. Give the hand calculations and solve for the mean using Excel.
m = Sx / N = 98 / 15 = 6.53
where x’s are the individual data values
N is the number of data values in the population
(click me)
The population mean is seldom known. True
False
(click one)
Example 3
Mode
Given the following data on x: Sample Data> 2, 4, 6, 6, 10. Calculate the mode value of x? Give the hand calculations and solve for the mode using Excel.
(1) Mo = 6(2) where (2) is the frequency (count)
(2) Another Example: DATA> 2, 4, 6, 6, 10, 10
Mo = 6(2) 10(2) (Bimodal)
(3) Another Example: DATA> 2, 4, 4, 6, 6, 10, 10, 10
Mo = 10 ( 3 )
(click me)
The mode of the following set of data is 5. Data: 1, 3,
5, 7, 9 True
False
(click one)
Example 4
Median
Given the following data on x: Sample Data> 2, 4, 6, 6, 10. Calculate the median value of x. Give the hand calculations and solve for the median using Excel.
n = 5 (odd)
When n is odd, the Md position in order array
= (n + 1) / 2
= ( 5 + 1) / 2 = 6 / 2 = 3rd position.
Md = 6
Another Example: When n is even,
Md = (two middle values) / 2
Data> 2, 4, 6, 10
n = 4 (even)
Md = ( 4 + 6 ) / 2 = 10 / 2 = 5.
(click me)
The median of the
following set of data is 3. Data: 9, 5, 3, 7, 1 True
False
(click one)
Example 5
Range
Given the following data on x: Sample Data> 2, 4, 6, 6, 10. Calculate the range of x values. Give the hand calculations and solve for the range using Excel.
Range = high value - low value
= H - L
Range = H - L = 10 - 2 = 8
(click me)
What does the range measure? (a)
the units of measure that H is above L (b)
the average difference between H and L. (click one)
Example 6
Sample Variance
Given the following data on x: Sample Data> 2, 4, 6, 6, 10. Calculate the variance of x values using the conceptual and then the computing equations. Give the hand calculations and solve for the variance using Excel.
Conceptual Sample Variance
Sample Variance = s² = Sd² / (n - 1)
First calculate the squared deviations of x from xbar.
|
x |
- xbar |
= d |
d2 |
|
|
2 |
- 5.6 |
= -3.6 |
12.96 |
|
|
4 |
- 5.6 |
= -1.6 |
2.56 |
|
|
6 |
- 5.6 |
= +0.4 |
0.16 |
|
|
6 |
- 5.6 |
= +0.4 |
0.16 |
|
|
10 |
- 5.6 |
= +4.4 |
19.36 |
S(x - xbar) = Sd = 0
Sd² = 35.20
Now, average these squared deviations by dividing by n -1.
Sample Variance = s² = Sd² / (n - 1)
= ( 35.2 ) / ( 5 - 1 )
= 8.8
Computing Sample Variance
Uses only columns of numbers to do the calculations.
|
x |
x² |
|
|
2 |
4 |
|
|
4 |
16 |
|
|
6 |
36 |
|
|
6 |
36 |
|
|
10 |
100 |
Sx = 28
Sx² = 192
s² = [Sx² - (Sx)² / n] / (n - 1)
= [192 - 28² / 5] / (5 - 1)
= [192 - 156.8 ] / ( 4 )
= [ 35.2 ] / 4
= 8.8
The numerators of s²:
Computing Conceptual
[Sx² - (Sx)² / n] = S[x - xbar]² = 35.2
(click me)
The sample variance is an estimate of the sample mean.
True
False (click one)
Example 7
Population Variance
Given the following data on x from which the sample in Example 1 was taken: Population data>2, 2, 4, 4, 4, 6, 6, 6, 6, 8, 8, 8, 10, 10, 14. Calculate the variance and standard deviation of x. Give the hand calculations and solve for both using Excel.
Population Variance
s² = S(x - m)² / N
= 147.73/15
= 9.85
Population Standard Deviation
s = [s²] = Ö[Sd² / N]
= Ö[9.85]
= 3.14
(click me)
The population standard deviation is in squared units
of measurement. True
False
(click one)
Example 8
Sample Standard Deviation
Given the following data on x: Sample Data> 2, 4, 6, 6, 10. Calculate the standard deviation of x values using the computing equation. Give the hand calculations and solve for the standard deviation using Excel.
s = Ö{[Sx² - (Sx)² / n] / (n - 1)}
= Ö{[192 - 28² / 5] / (5 - 1)}
= Ö{[192 - 156.8] / (4)}
= Ö{[35.2]/4}
= Ö8.8 = Ös²
= 2.97
(click me)
The sample standard deviation is an estimate of: (a) population
standard
deviation (b) population variance
(click one)
Example 9
Z Score
Given the following data on x: Sample Data> 2, 4, 6, 6, 10. Calculate the Z score of x each of the data. Interpret the Z score for the data values 2 and 10. Give the hand calculations and solve for the standard deviation using Excel.
Z = [x - xbar] / s
|
x |
xbar |
s |
[x - xbar] / s |
indiv’l/standard |
Z score |
|
2 |
5.6 |
2.97 |
[ 2 - 5.6 ]/2.97 |
[ -3.6 ]/2.97 |
- 1.21 |
|
4 |
5.6 |
2.97 |
[ 4 - 5.6]/2.97 |
[ -1.6 ]/2.97 |
- 0.54 |
|
6 |
5.6 |
2.97 |
[ 6 - 5.6]/2.97 |
[ +0.4 ]/2.97 |
+ 0.13 |
|
6 |
5.6 |
2.97 |
[ 6 - 5.6]/2.97 |
[ -+0.4 ]/2.97 |
+ 0.13 |
|
10 |
5.6 |
2.97 |
[ 10 - 5.6]/2.97 |
[ +4.4 ]/2.97 |
+1.48 |
The data value 2 with Z = - 1.21 is 1.21 standard deviations (s = 2.97) less than the mean (5.6).
The data value 10 with Z = +1.48 is 1.48 standard deviations (s = 2.97) greater than the mean (5.6).
(click me)
A Z score of zero means that x = mean. True
False
(click one)
Example 10
Empirical Rule
The mean score of students in an statistics class was 75.4, and the standard deviation was 21. Using the empirical rule, estimate the range of scores within which about 95% of the data values are expected to lie.
Rule #2- 95% of data values between
xbar ± 2s
= 75.4 ± 2( 21 )
= 75.4 ± 42
= 33.4 to 117.4
If 117.4 is not a possible score on the exam; then, 33.4 to 100 is the practical range.
(click me)
99.7% of the data lies between xbar plus and minus 3
standard deviations for any distribution. True
False
(click one)
Go on to Excel
and Equations
or
Go back to Lesson 1: Introduction
or
Go back to Descriptive Measures: Assignments and Activities
Please reference "BA501 (your last name) Assignment name and number" in the subject line of either below.
E-mail Dr. James V. Pinto at
BA501@mail.cba.nau.edu
or call (928) 523-7356. Use WebMail for attachments.
Copyright 2002 Northern Arizona University
ALL RIGHTS RESERVED