Transcript Chi-Square Practice File

8/27/2015
Quick Question to Start Our Day
Briefly (2-3 sentences) describe the
difference between a reductionist
approach and a systems approach to
biology
Quantitative Skills
…with a little dab of critical thinking!
What are the three most important
things I need to study?
1.
2.
3.
Experimental Design
Math
• Mean, median, mode, range
• SD
• Rate
• Chi Square
• Intrinsic rate of increase
• Percent error
• Hardy-Weinberg
• Water potential
• ΔG = ΔH – TΔS
Making Connections
Design An Experiment
Scientific Method
Experimental Design
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Observation
Hypothesis
Experiment
Analysis
Conclusion
Dependent Variable
Independent Variable
Constants
Control
Multiple Trials
Correct Equipment/subjects
Statistical Analysis
The following statistics suggest that 16-year-olds are safer
than people in their twenties, and that octogenarians are
very safe.
Is this true?
Solution: NO!
As the following graph shows, the reason 16-year-old and
octogenarians appear to be safe drivers is that they don't drive nearly
as much as people in other age groups.
Can we conclude that United, American, and Delta are
the worst airlines and Alaska, Southwest, and
Continental are the best?
Most complaints 2000
United Airlines
252
American Airlines
162
Delta Airlines
119
Fewest complaints 2000
Alaskan Airlines
13
South West Airlines 22
Continental Airlines 20
Solution: NO!
The airlines that had the most complaints also had the
most passengers. As the following graph shows, rates
and percentages are often more informative than raw
numbers.
These two histograms have the same mean of five but they
differ in their variation. Variance (s2) is the degree that that
measurements deviate from the average
The standard deviation, or ‘s’, for the two diagrams below are
different.
s=2.053 for #1
s=1.046 for #2
Histogram #1 has more deviation or variation from the mean of 5,
than histogram #2
The standard deviation,
s for the two diagrams
below are different.
s=2.053 for #1
s=1.046 for #2 which
makes sense because
histogram #1 has more
deviation or variation
from the mean of 5.
Is there variation?
• Where is it?
– Among individuals
– Among populations
• What causes it?
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Snp’s
Copy number variation
Epigenetics
Mitosis/Meiosis
Sexual reproduction
Etc.
• How is it maintained?
• How do we measure it?
How do we measure variation in our
data?
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Mean
Mode
Median
Variance
Standard deviation
Mean is the average, so…
97, 93, 90, 88, 88, 85, 83, 82, 77, 75, 74, 71, 69, 66, 59
• The mean is 1197 / 15 = 79.8
• When would this be beneficial to use?
Median is the middle point, so…
97, 93, 90, 88, 88, 85, 83, 82, 77, 75, 74, 71, 69, 66, 17
• 82 is directly in the middle of the set
• When would this be beneficial to use?
Mode is the most used value, so…
97, 93, 90, 88, 88, 85, 83, 82, 77, 75, 74, 71, 69, 66, 59
• 88 is the most used value in the set
• When would this be beneficial to use?
Tutorial: Chi-Square
Distribution
Purpose
• To measure discontinuous categorical/binned data in which a
number of subjects fall into categories
• We want to compare our observed data to what we expect to
see. Due to chance? Due to association?
• When can we use the Chi-Square Test?
– Testing outcome of Mendelian Crosses, Testing Independence – Is one
factor associated with another?, Testing a population for expected
proportions
Assumptions:
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1 or more categories
Independent observations
A sample size of at least 10
Random sampling
All observations must be used
For the test to be accurate, the expected frequency
should be at least 5
Conducting Chi-Square Analysis
1) Make a hypothesis based on your basic biological
question
2) Determine the expected frequencies
3) Create a table with observed frequencies, expected
frequencies, and chi-square values using the formula:
(O-E)2
E
4) Find the degrees of freedom: (n-1)
5) Find the chi-square statistic in the Chi-Square
Distribution table
6) If chi-square statistic > your calculated chi-square value,
you do not reject your null hypothesis and vice versa.
Example 1: Testing for Proportions
HO: Horned lizards eat equal amounts of leaf cutter, carpenter and black ants.
HA: Horned lizards eat more amounts of one species of ants than the others.
Leaf Cutter
Ants
Carpenter
Ants
Black Ants
Total
Observed
25
18
17
60
Expected
20
20
20
60
O-E
5
-2
-3
0
(O-E)2
E
1.25
0.2
0.45
χ2 = 1.90
χ2 = Sum of all: (O-E)2
E
Calculate degrees of freedom: (n-1) = 3-1 = 2
Under a critical value of your choice (e.g. α = 0.05 or 95% confidence),
look up Chi-square statistic on a Chi-square distribution table.
Example 1: Testing for Proportions
χ2α=0.05 = 5.991
Example 1: Testing for Proportions
Leaf Cutter
Ants
Carpenter
Ants
Black Ants
Total
Observed
25
18
17
60
Expected
20
20
20
60
O-E
5
-2
-3
0
(O-E)2
E
1.25
0.2
0.45
χ2 = 1.90
Chi-square statistic: χ2 = 5.991
Our calculated value: χ2 = 1.90
*If chi-square statistic > your calculated value, then you do not reject your
null hypothesis. There is a significant difference that is not due to chance.
5.991 > 1.90 ∴ We do not reject our null hypothesis.
STANDARD DEVIATION
Let’s calculate a quick one!
Six students took the same quiz and scored the
following:
5, 6, 7, 8, 9, 10
• Find the mean
• 5+6+7+8+9+10 = 45/6 = 7.5
Let’s calculate a quick one!
Six students took the same quiz and scored the
following:
5, 6, 7, 8, 9, 10
• Subtract the mean from each data point
• 5 – 7.5 = -2.5
• 6 – 7.5 = -1.5
• 7 – 7.5 = -0.5
• 8 – 7.5 = 0.5
• 9 – 7.5 = 1.5
• 10 – 7.5 = 2.5
Let’s calculate a quick one!
Six students took the same quiz and scored the
following:
5, 6, 7, 8, 9, 10
• Square each one.
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5 – 7.5 = -2.5
6 – 7.5 = -1.5
7 – 7.5 = -.5
8 – 7.5 = .5
9 – 7.5 = 1.5
10 – 7.5 = 2.5
-2.52 = 6.25
-1.52 = 2.25
-0.52 = 0.25
0.52 = 0.25
1.52 = 2.25
2.52 = 6.25
Let’s calculate a quick one!
Six students took the same quiz and scored the following:
5, 6, 7, 8, 9, 10
• Sum the squares
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5 – 7.5 = -2.5
6 – 7.5 = -1.5
7 – 7.5 = -.5
8 – 7.5 = .5
9 – 7.5 = 1.5
10 – 7.5 = 2.5
-2.52 = 6.25
-1.52 = 2.25
-0.52 = 0.25
0.52 = 0.25
1.52 = 2.25
2.52 = +6.25
17.5
Let’s calculate a quick one!
Six students took the same quiz and scored the following:
5, 6, 7, 8, 9, 10
• Divide the sum of the squares by (n-1) or total number
of data points – 1.
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5 – 7.5 = -2.5
6 – 7.5 = -1.5
7 – 7.5 = -.5
8 – 7.5 = .5
9 – 7.5 = 1.5
10 – 7.5 = 2.5
-2.52 = 6.25
-1.52 = 2.25
-0.52 = 0.25
0.52 = 0.25
1.52 = 2.25
2.52 = +6.25
17.5/5 = 3.5
Let’s calculate a quick one!
Six students took the same quiz and scored the following:
5, 6, 7, 8, 9, 10
• Take the square root of the this number.
–
–
–
–
–
–
5 – 7.5 = -2.5
6 – 7.5 = -1.5
7 – 7.5 = -.5
8 – 7.5 = .5
9 – 7.5 = 1.5
10 – 7.5 = 2.5
-2.52 = 6.25
-1.52 = 2.25
-0.52 = 0.25
0.52 = 0.25
1.52 = 2.25
2.52 = +6.25
17.5/5 = 3.5 = 1.87
Let’s calculate a quick one!
Therefore if six students
took the same quiz and
scored the following:
5, 6, 7, 8, 9, 10
• The average of the
grades was 7.5 ± 1.9
Student Hand Length Variability
Number of Students
7
6
5
4
3
2
1
16
17
18
19
Hand Length (cm)
20
21
Student Hand Length Variability
Number of Students
7
6
15.69
17.09
19.89
18.49
21.29
5
4
3
2
1
16
17
18
19
Hand Length (cm)
20
21
Student Hand Length Variability
Number of Students
7
6
5
4
3
2
1
16
17
18
19
Hand Length (cm)
20
21
Student Hand Length Variability
Number of Students
7
6
5
4
3
2
1
16
17
18
19
Hand Length (cm)
20
21
Student Hand Length Variability
Number of Students
7
6
5
4
3
2
1
16
17
18
19
Hand Length (cm)
20
21
Modes of Natural Selection
Quantitative Analysis
YOU WILL NOT HAVE TO CALCULATE…
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Standard Deviation
Standard Error
Q10
pH
However, you need to be able to understand
them well enough to answer questions
regarding their use in data analysis.
Try this one!
In Caucasian humans, hair straightness or curliness is thought to be
governed by a single pair of alleles showing partial dominance.
Individuals with straight hair are homozygous for the Is allele, while
those with curly hair are homozygous for the Ic allele. Individuals with
wavy hair are heterozygous (IsIc). In a population of 1,000 individuals,
245 were found to have straight hair, 393 had curly hair, and 362 had
wavy hair.
• Calculate the allelic frequencies of the Is and Ic alleles.
• Explain whether or not this population is in Hardy-Weinberg
equilibrium?
– Justify your answer. Your explanation should include a chi-square
goodness of fit test.
Try this one!
Two Siamese and three Persian cats survive a shipwreck and
are carried on driftwood to a previously uninhabited tropical
island. All five cats have normal ears, but one carries the
recessive allele f or folded ears (his genotype is Ff).
• Calculate the frequencies of the dominant and recessive
alleles in the cat population on this island.
• If you assume Hardy-Weinberg equilibrium for these
alleles (admittedly very improbable), calculate the number
of cats you would expect to have folded ears when the
island population reaches 20,000?
Big Idea 1
1. Natural selection
2. Genotypic/Phenotypic variation
3. Genetic drift/Gene flow
4. Phylogenetic tree/Cladogram
5. Adaptation
6. Allele frequency
7. Artificial selection
8. Genetic diversity/Variation
9. Biological evolution
10. Speciation
Big Idea 2
1. Photosynthesis
2. Cellular respiration
3. Homeostasis
4. Changes in free energy
5. Diffusion/Osmosis/Passive/Active transport
6. Eukaryotic organelles
7. Biogeochemical cycles
8. Selectively permeable membranes
9. Feedback mechanisms
10. Biotic/Abiotic interactions
1.
2.
3.
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Mutations
Meiosis/Mitosis
Transcription/Translation
Genotype/Phenotype
DNA/RNA
Processes that increase variation
Genetic disorders
Gene regulation
Cell communication
Viral genetics
Big Idea 4
1. Human organ systems
2. Human organs
3. Organic molecules
4. Structure = Function
5. Community relationships
6. Ecosystem dynamics
7. Population size/growth
8. Competition/Cooperation
9. Autotrophs/Heterotrophs
10. Plant anatomy/behavior