descriptive statistics and graphing. the normal distribution if the frequency (or number) of data...
TRANSCRIPT
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Descriptive Statistics and Graphing
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The Normal Distribution
If the frequency (or number) of data points is plotted on the Y-axis, a bell-shaped curve may be produced.
#
Slow FastRunning Speed of People
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Skewed Distribution (Not Normal)
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Slow FastRunning Speed of People
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Central Tendency
Mean – average Median – middle value Mode – most common value
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Disadvantages - Mean
Influenced by outliers Example- Population estimates of waterfowl on seven
lakes:
400
200
220
210
340
250
44,000
Mean = 6,517
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Disadvantages - Median
The central number may not be representative, particularly with small samples.
Example: 0, 0, 1, 2, 480, 500
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Range
100, 75, 50, 25, 0
52, 51, 50, 49, 48
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Standard Deviation
The standard deviation describes the “spread” of data points. It is useful if the data fit a normal distribution.
#
Slow FastRunning Speed of People
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Calculating the Standard Deviation
Data25262631353638
Sum = 217Mean = 217/7 = 31
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1) Calculate the mean
Data25262631353638
Sum = 217Mean = 217/7 = 31
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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
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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
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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
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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
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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
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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.
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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.
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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
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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/
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Bar Graph - Mammals
0
6
12
18
24
30
Greysquirrels
RedSquirrels
Chipmunks White-footedmice
White-taileddeer
8
4
17
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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
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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
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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
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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
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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
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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
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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
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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).
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t-test
http://faculty.clintoncc.suny.edu/faculty/Michael.Gregory/files/shared%20files/Statistics/t-test.xls