Descriptive Statistics and Graphing. The Normal Distribution If the frequency (or number) of data...

Descriptive Statistics and Graphing

The Normal Distribution

If the frequency (or number) of data points is plotted on the Y-axis, a bell-shaped curve may be produced.

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Slow FastRunning Speed of People

Skewed Distribution (Not Normal)

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Slow FastRunning Speed of People

Central Tendency

Mean – average Median – middle value Mode – most common value

Disadvantages - Mean

Influenced by outliers Example- Population estimates of waterfowl on seven

lakes:

400

200

220

210

340

250

44,000

Mean = 6,517

Disadvantages - Median

The central number may not be representative, particularly with small samples.

Example: 0, 0, 1, 2, 480, 500

Range

100, 75, 50, 25, 0

52, 51, 50, 49, 48

Standard Deviation

The standard deviation describes the “spread” of data points. It is useful if the data fit a normal distribution.

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Slow FastRunning Speed of People

Calculating the Standard Deviation

Data25262631353638

Sum = 217Mean = 217/7 = 31

1) Calculate the mean

Data25262631353638

Sum = 217Mean = 217/7 = 31

2) Calculate deviation from the mean

Data Deviation25 31 - 25 = 626 31 - 26 = 526 31 - 26 = 531 31 - 31 = 035 31 - 35 = -436 31 - 36 = -538 31 - 38 = -7

Sum = 217Mean = 217/7 = 31

3) Square the deviations

Data Deviation Deviation squared25 31 - 25 = 6 3626 31 - 26 = 5 2526 31 - 26 = 5 2531 31 - 31 = 0 035 31 - 35 = -4 1636 31 - 36 = -5 2538 31 - 38 = -7 49

Sum = 217Mean = 217/7 = 31

4) Sum the squared deviations

Data Deviation Deviation squared25 31 - 25 = 6 3626 31 - 26 = 5 2526 31 - 26 = 5 2531 31 - 31 = 0 035 31 - 35 = -4 1636 31 - 36 = -5 2538 31 - 38 = -7 49

Sum = 217Mean = 217/7 = 31

Sum 176

5) Divide by n-1

Data Deviation Deviation squared25 31 - 25 = 6 3626 31 - 26 = 5 2526 31 - 26 = 5 2531 31 - 31 = 0 035 31 - 35 = -4 1636 31 - 36 = -5 2538 31 - 38 = -7 49

Sum = 217Mean = 217/7 = 31

Sum 176Sum/(n-1) = 29.33333 = Variance

6) Take square root of variance

Data Deviation Deviation squared25 31 - 25 = 6 3626 31 - 26 = 5 2526 31 - 26 = 5 2531 31 - 31 = 0 035 31 - 35 = -4 1636 31 - 36 = -5 2538 31 - 38 = -7 49

Sum = 217Mean = 217/7 = 31

Sum 176Sum/n-1) = 29.33333 = VarianceSqrt = 5.416026 = Standard Deviation

Normal Distribution

68% of the data points are within 1 standard deviation of the mean:= mean + or – S.D.

In the previous example, this is 31 + or – 5.4231+5.42 = 36.4231-5.42 = 25.58

Therefore 68% of the data will fall between 25.58 and 36.42.

Normal Distribution

Approximately 95% of the data points are within 2 standard deviations of the mean:= mean + or – 2 S.D.

In the previous example, this is 31 + or – (2 X 5.42)31 + (2*5.42) = 41.8431 – (2*5.42) = 20.16.

Therefore 95% of the data points fall between 20.16 and 41.84.

Approximately 99% of the data points fall within three standard deviations of the mean.

Heart Rate (beats per minute)

Speed (kilometers per hour)

11 12 13 14 15 16 17

Variables

130

135

140

145

150

155

160

165

0 5 10 15 20 25 30

165

160

155

150

145

140

135

130

Number of mammals in a 1.2 ha woodlot in Clinton County, NY

Grey squirrel – 8

Red squirrels – 4

Chipmunks – 17

White-footed mice – 26

White-tailed deer – 2

Create a Graph Website:http://nces.ed.gov/nceskids/graphing/

Bar Graph - Mammals

0

6

12

18

24

30

Greysquirrels

RedSquirrels

Chipmunks White-footedmice

White-taileddeer

8

4

17

pH of an a pond in Clinton County, NY on 5/11/05

1:00 AM – 5.2 3:00 AM – 5.1 5:00 AM – 5.1 7:00 AM – 6.0 9:00 AM – 6.6 11:00 AM – 6.9 1:00 PM – 7.0 3:00 PM – 7.0 5:00 PM – 6.6 7:00 PM – 5.9 9:00 PM – 5.3 11:00 PM – 5.2

Line Graph - pH

4.5

5.1

5.7

6.3

6.9

7.5

1:00AM

3:00AM

5:00AM

7:00AM

9:00AM

11:00AM

1:00PM

3:00PM

5:00PM

7:00PM

Number of bird species observed in 9 woodlots in January 2006 in Clinton County, NY

Size of Woodlot (ha) # Bird Species

3.3 11

8 13

3.6 13

13 14

1.1 7

11 14

7.4 12

6.6 14

8 12

Number of bird species observed in 9 woodlots in January 2006 in Clinton County, NY

6

7

8

9

10

11

12

13

14

15

16

1 2.4 3.8 5.2 6.6 8 9.4 10.8 12.2

Number of bird species observed in 9 woodlots in January 2006 in Clinton County, NY

6

7

8

9

10

11

12

13

14

15

16

1 2.4 3.8 5.2 6.6 8 9.4 10.8 12.2

Number of bird species observed in 9 woodlots in January 2006 in Clinton County, NY

6

7

8

9

10

11

12

13

14

15

16

1 2.4 3.8 5.2 6.6 8 9.4 10.8 12.2

Group 1 - HormoneWeight (grams)

Group 2 – No HormoneWeight (grams)

12.5 8

13 8.5

12 8

12 8

13 7.5

14 10.5

13 7

10.5 8.5

9.5 6.5

11 7

Statistical Testing

Mean = 12.05 7.95

Conclusion

P is the probability that the difference is due to chance.

If p > 0.05, conclude that the difference is due to chance.

If p < 0.05, conclude that the difference is real (not due to chance).

t-test

http://faculty.clintoncc.suny.edu/faculty/Michael.Gregory/files/shared%20files/Statistics/t-test.xls