Business 205. Review of last class NOIR Validity Reliability.

Business 205

Review of last class

NOIRValidityReliability

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Frequency, Range, Means, Medians, Modes

GraphsVarianceStandard Deviation

Descriptive Statistics

A way to present quantitative descriptions in a manageable (aka: numerical) way

Example: 52% Females, 48% Males

Qualitative Data

Used to describe sample Class Frequency

How many times it occurs in a given class

Exec Board = Division Head =

Staff =

Name Status Score

Ducky Exec Board 100

Webbie Exec Board 95

Mortimer Exec Board 95

WP Division Head 95

Baby Duck Staff 8

Peepers Staff 75

Qualitative Data

Used to describe sample Class Relative

Frequency Class frequency

divided by total number in sample

Staff = 2/6 = .33 Exec Board = Division Head =

Name Status Score

Ducky Exec Board 100

Webbie Exec Board 95

Mortimer Exec Board 95

WP Division Head 95

Baby Duck Staff 8

Peepers Staff 75

Qualitative Data

Used to describe sample Class Percentage

Multiply class relative frequency by 100.

Staff = .33*100 = 33%

Exec Board = Division Head =

Name Status Score

Ducky Exec Board 100

Webbie Exec Board 95

Mortimer Exec Board 95

WP Division Head 95

Baby Duck Staff 8

Peepers Staff 75

A Bunch of Scores…

Sample Size (n) = 9People/Sample Score

1 1

2 4

3 4

4 6

5 8

6 4

7 3

8 5

9 5

Distributions

An arrangement of scores in order of magnitude

Scores: 1, 4, 4, 6, 8, 3, 4, 5, 2

Distribution: 1, 2, 3, 4, 4, 4, 5, 6, 8

Frequency Distributions

Listing of scores in magnitude with amount of people who received that score

Scores: 1, 4, 4, 6, 8, 3, 4, 5, 2

Distribution: 1, 2, 3, 4, 4, 4, 5, 6, 8

Frequency Distributions

NOTE: The total (f) MUST be equal to the sample size!

In this example we had n = 9 so our f = 9!!!

Score (X) Frequency (f)

1 1

2 1

3 1

4 3

5 1

6 1

8 1

Smart Frequency Distributions

Color f

Pink

Yellow

White

Green

Purple

Orange

Unknown

Total #: ______________

Color f

Pink

Yellow

White

Green

Purple

Orange

Unknown

Total #: ______________

Graphs

Bar GraphsPie ChartsDot PlotsStem-and-Leaf PlotsHistograms

Graphs

Frequency

0

0.5

1

1.5

2

2.5

3

3.5

Exec Board DivisionHead

Staff

Frequency

Bar Chart

Frequency

Exec Board

Division Head

Staff

Pie Chart

Stem-and-Leaf

10 0 9 5 5 5 8 0 7 5 6 5 4 3 2 1

Try it out

Create the following: Histogram Pie chart Stem-and-leaf plot

Person Score

A 98

B 92

C 25

D 10

E 81

F 36

G 36

Central Tendency

Information concerning the average or typical score of the sample interested in.

**Do NOT confuse this with the mean…

Mean

Arithmetic average of all scores

1. Sum ( ) the scores (X).

2. Divide by the sample size (n).

n

X M

Mean Example

Scores: 1, 4, 4, 6, 8, 3, 4, 5, 2Distribution: 1, 2, 3, 4, 4, 4, 5, 6, 8

M = (1, 2, 3, 4, 4, 4, 5, 6, 8)/9= 37/9= 4.1111111

n

X M

Median

The midpoint of all the scores

1. Put all scores in order

2. Find the middle score1. Interpolate score if necessary

Median Example, Non interpolated

Scores: 1, 4, 4, 6, 8, 3, 4, 5, 2Distribution: 1, 2, 3, 4, 4, 4, 5, 6, 8

1, 2, 3, 4, 4, 4, 5, 6, 8

Median = 4

Median Example, interpolated

Scores: 1, 4, 4, 6, 8, 10, 3, 5, 5, 2Distribution: 1, 2, 3, 4, 4, 5, 5, 6, 8, 10

1, 2, 3, 4, 4, 5, 5, 6, 8, 10

Median = (4+5)/2= 9/2=4.5

Median = 4.5

Mode

The score that appears the most

1. Put the scores in order

2. Find the frequencies of the scores

3. Choose the one that appears the most times

Mode Example

4 appears 3 times.

Mode = 4

Score (X) Frequency (f)

1 1

2 1

3 1

4 3

5 1

6 1

8 1

Mode Example

What if they are all the same in frequency?

Mode = ?

Score (X) Frequency (f)

1 1

2 1

3 1

4 1

5 1

6 1

8 1

How did they compare?

Mean = 4.11

Median = 4

Mode = 4

Can you have more than 1 mode?Can you have more than 1 mean?Can you have more than 1 median?

Normal Distribution Curve

Mean = Median = Mode

Positively Skewed

Mode

Median

Mean

Negatively Skewed

Mean

Median

Mode

Skew this

Distribution: 1, 2, 8, 9, 9

Mean =

Median =

Mode =

Graph:

Range (R)

Measurement of the width of scores.

R = high score – low score + 1

Range Example

Scores: 1, 4, 4, 6, 8, 3, 4, 5, 2Distribution: 1, 2, 3, 4, 4, 4, 5, 6, 8

CORRECT: High Score = 8; Low Score = 1R = 8 – 1 + 1

= 8

INCORRECT: High Score = 8; Low Score = 1R = 8-1

= 7

Standard Deviation (SD)

How much scores in a distribution differ from the mean.

1. Find the mean2. Subtract each score

from the mean (x)3. Square each

difference (x2) and sum()

4. Divide the sum by the sample size (n)

5. Take the square root of the number

1 -n

x SD

2

sample

N

x SD

2

population

Standard Deviation Example

Mean = 4.11

x2 = 34.89

= [(34.89)/9]= 1.97

Scores (X) X – Mean x x2

1 1 – 4.11 -3.11 9.6721

2 2 – 4.11 -2.11 4.4521

3 3 – 4.11 -1.11 1.2321

4 4 – 4.11 -.11 .0121

4 4 – 4.11 -.11 .0121

4 4 – 4.11 -.11 .0121

5 5 – 4.11 .89 .7921

6 6 – 4.11 1.89 3.5721

8 8 – 4.11 3.89 15.132

N

x SD

2

population

Standard Deviation Shortcut For Example

Mean = 4.11

X2 = (1+4+9+16+16+16+25+36+64)= 187

SD = [(187/9)-(4.112)]= 1.97

Scores (X) X2

1 1

2 4

3 9

4 16

4 16

4 16

5 25

6 36

8 64

22

Mn

X SD

Variance

How much all the scores in the distribution vary from the mean.

V = SD2

N

SS Vpopulation

1 -n

SS Vsample

In Class Example: Range

You have the following scores:8, 10, 4, 4

R = high score – low score + 1

In Class Example: Distribution

You have the following scores: 8, 10, 4, 4

Score (X) Frequency (f)

In Class Example: Mean, Median, Mode

You have the following scores: 8, 10, 4, 4

Mean

Median

Mode

n

X M

In Class Example: Standard Deviation

You have the following scores: 8, 10, 4, 4

Scores (X)

X – Mean x x2

4

4

8

10

In Class Example: Standard Deviation

You have the following scores:8, 10, 4, 4

1 -n

x SD

2

sample

N

x SD

2

population

In Class Example: Variance

You have the following scores:8, 10, 4, 4

2 V SD

1 -n

SS Vsample

N

SS Vpopulation