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BA501 : The Class : Stats : Measures : Examples
Discriptive Measures: Examples
Examples

§ 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

Note (click me)

Question 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

Note (click me)

Question 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 )

Note (click me)

Question 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.

Note (click me)

Question 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

Note (click me)

Question 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

S = 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)

Question 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

Note (click me)

Question 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)

Question 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).

Note (click me)

Question 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.

Note (click me)

Question 99.7% of the data lies between xbar plus and minus 3 standard deviations for any distribution. True False (click one)


You should now:

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.

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