Descriptive Statistics and Graphing
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Transcript of Descriptive Statistics and Graphing
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:400200220210340250
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, NYGrey squirrel – 8Red squirrels – 4 Chipmunks – 17White-footed mice – 26White-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 Species3.3 118 133.6 1313 141.1 711 147.4 126.6 148 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