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The Chi-square goodness of fit test
• Core issue in statistics: When are you viewing just random noise and when is there a real trend?– Example: To see if squash shape & color are linked genes do a test cross.
Chi-square goodness of fit
1 : 1 : 1 : 1 ????
xGgLl ggll
• Your response variable is count data.
When to use a chi-square test
• You have more than one category of the response variable.
• You have a hypothesis for the responses you expect.
• You want to know if the difference between the responses you observe and the responses you expect is significant or not.
• Example hypothesis: “The MSU football team will win every single game this season.” So, according to my hypothesis, I expect MSU’s chance of winning any game is ___%.
100%Does this mean MSU will win 100% of their games?
• Example hypothesis: “The MSU football team’s number of wins and losses will be random.” So, according to this new hypothesis, I expect the team’s chance of winning any game is ___%.
50%
Turn a hypothesis into a numberYour hypothesis tells you what you expect any given response (observation) to be.
Turn your expectation into a fraction or percentage.
Hyp.: “People over the age of 60 are 50% more likely to attend a baseball game than younger people.” So, according to my hypothesis if I go to a baseball game and find out the ages for all the fans in the audience, I expect the odds of any one fan being > 60 to be…
x+ (x-50) = 100, solve for x.75% or 3 out of 4.What are the odds a fan will be < 60 years old?
Turn a hypothesis into a number
• “Pre-hypothesis”: Given the choice, people prefer red and blue m&m’s over the other 4 colors.
• But don’t know how strong their preference might be. So test the “null hypothesis”—People choose m&m colors at random, i.e. they don’t show preference. (vs. “alternative” or “experimental” hypothesis).
• So, according to my null hypothesis, if I hand around a bowl of m&ms, I expect the chance of each color being chosen is…
1/6 or 16.67%.• Use chi square test to see if what you actually observe is significantly different
from 1/6.
Turn a hypothesis into a number
The chi-square test
Game
% fans > 60 years old
1 69
2 80
3 20
4 55
5 67
6 76
7 47
8 81
9 70
10 68
Game
% fans > 60 years old
1 75
2 75
3 75
4 75
5 75
6 75
7 75
8 75
9 75
10 75
Observed Expected
The chi-square test determines whether or not the difference between the responses you observe and the responses you expect is significant.
Significant = not due to random chance alone.
Calculate the “strength of the difference”, get a value that tells you the probability the difference is due to chance (random noise) alone.
If this probability is small (<5%), we conclude there is a significant difference (the difference is not simply due to chance) between obs and exp values.
Interpreting the chi-square test
Game
% fans > 60 years old
1 69
2 80
3 20
4 55
5 67
6 76
7 47
8 81
9 70
10 68
Game
% fans > 60 years old
1 75
2 75
3 75
4 75
5 75
6 75
7 75
8 75
9 75
10 75
Observed ExpectedHypothesis: “People over the age of 60 are 50% more likely to attend a baseball game than younger people.”
If the test tells you your data are not significantly different from what you expect, (your data have a “good fit” to the expected values), you support the hypothesis.Note: no statistical test ever proves a hypothesis!
If the test tells you your data are significantly different from what you expect, you reject the hypothesis.
≈≠
What is chi-square?
“Chi-square” symbol is χ2 (Greek).
χ2 = (Observed – Expected)2
Expected
Observed Expected Obs-Exp (Obs-Exp)2 (Obs-Exp)2
ExpCategory 1
Category 2
…χ2 total
Degrees of Freedom
Based on your hypothesis!
“Sum of”
Σ
Number of categories minus 1 = N-1
Example problem #1
A university biology department would like to hire a new professor. They advertised the opening and received 220 applications, 25% of which came from women. The department came up with a “short list” of their favorite 25 candidates, 5 women and 20 men, for the job. You want to know if there is evidence for the search committee being biased against women. Note: If the committee is unbiased the proportion of women in the short list should match the proportion of women in all the applications. Define your hypothesis. Set up table.
Observed Expected Obs-Exp (Obs-Exp)2 (Obs-Exp)2
Exp
χ2 totalDegrees of Freedom
Women
Men
520
6.2518.75
-1.251.25
1.56251.5625
0.250.080.33
1
Women: 25 * 0.25 =
Men: 25 * 0.75 =
25 = 25
Chi-square probability table
Probabilities
Observed values are significantly different from
expected (differences not just due to random chance).
Reject hypothesis.
Reject hyp.
Observed values not significantly different from expected
(differences due to random chance). Support hypothesis.
Chi-square probability table
Probabilities
Observed values are significantly different from
expected (differences not just due to random chance).
Reject hypothesis.
Reject hyp.
Observed values not significantly different from expected
(differences due to random chance). Support hypothesis.
Probability range: 0.5 < p < 0.6Means that there is a 50-60% probability that the difference between obs & exp values are from random chance alone.
So, is the department biased against women
applicants?
Example problem #2Work in groups
Example problem #2Hypothesis: Body color and wing size are unlinked genes.
Expected ratio?9:3:3:1.
Expected values:
Gray Normal wings (GgWw): 9/16 * 102 = 57.375
Gray Vestigial wings (Ggww):3/16 * 102 = 19.125
Ebony Normal wings (ggWw):3/16 * 102 = 19.125
Ebony Vestigial (ggww): 1/16 * 102 = 6.375
Observed Expected Obs-Exp (Obs-Exp)2 (Obs-Exp)2
Exp
χ2 totalDegrees of Freedom
Gray Norm.Gray Vest.
5316
57.37519.125
-4.375-3.125
19.1419.766
0.3330.5111.8050.414
Ebony Norm.Ebony Vest.
258
19.1256.375
5.8751.625
34.5162.641
3.0633
102 = 102
Chi-square probability table
Probabilities
Support hypothesis. Reject hyp.
Probability range: 0.3 < p < 0.4Means that there is a 30-40% probability that the difference between obs & exp values are from random chance alone.
Biology?
Example problem #3Using Chi-square to test for
linked genes
Example problem #31. Hypothesis:
Squash color and shape are not linked genes. OR Squash color and shape are linked genes.
2. Describe the phenotypes and circle the recombinants.
LlGg llGg
llgg Llgg
3. If the 2 genes are not linked the expected ratio is:
1:1:1:1
4. If the two genes are linked the expected phenotype ratio is:
1:0:0:1
Example problem #3If you tested the hypothesis that squash shapre and color ARE LINKED (1:1:1:1) :5. Calculate the expected number of offspring for each phenotype:Wild Wild (LlGg) :
509/4 = 127.25Wild Orange (Llgg) :
127.25Round Wild (llGg) :
127.25Round Orange (llgg) :
127.25
Observed Expected Obs-Exp (Obs-Exp)2 (Obs-Exp)2
Exp
χ2 totalDegrees of Freedom
Wild WildWild Orange
22817
127.25127.25
100.75-110.25
10150.5612155.06
79.895.588.7
105.3Round WildRound Orange
21243
127.25127.25
-106.25115.75
11289.0613398.06
369.33
Chi-square probability table
Probabilities
Support hypothesis. Reject hyp.
Probability range: 0.3 < p < 0.4
Statistical meaning: 30-40% probability that the difference between obs & exp
values are from random chance alone. The obs and exp values are not significantly different. Support hypothesis.
Biological meaning?
Example problem #3If you tested the hypothesis that squash shapre and color ARE NOT LINKED (1:0:0:1) :5. Calculate the expected number of offspring for each phenotype:Wild Wild (LlGg) :
509/2 = 254.5Wild Orange (Llgg) :
0Round Wild (llGg) :
0Round Orange (llgg) :
509/2=254.5
Observed Expected Obs-Exp (Obs-Exp)2 (Obs-Exp)2
Exp
χ2 totalDegrees of Freedom
Wild WildWild Orange
22817
254.50
-26.517
702.25289
2.76(Undef.) 0(Undef.) 0
0.52Round WildRound Orange
21243
0254.5
21
-11.5
441132.25
3.283
Chi-square probability table
Probabilities
Support hypothesis. Reject hyp.
Probability range: p < 0.01
Statistical meaning: < 1% probability that the difference between obs & exp
values are from random chance alone. The obs and exp values are significantly different. Reject hypothesis.
Biological meaning?
Example problem #3
Hypothesis not linked p<0.01 Reject hypothesis
Hypothesis linked 0.3 < p < 0.4, in other words, p > 0.05 Support hypothesis
Are these test results in agreement?
So do these data show that the genes are linked or not?
If you weren’t very confident in your test results, what could you do next to improve your confidence?