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Statistical Analysis: Chi Square
AP Biology
Ms. Haut
“Goodness of Fit Test” Chi Square (χ2)
• Statistical analysis used to determine whether data obtained experimentally provides a “good fit” to the expected data
• Used to determine if any deviations from the expected results are due to random chance alone or to other circumstances
Chi Square (χ2)
• Use the equation to test the “null” hypothesis– The prediction that data from the experiment will
match the expected results
• χ2 = Σ (observed results – expected results)2
expected results
• When all else is equal, the value of χ2 increases as the difference between the observed and expected values increase
Chi Square (χ2)
• Once you have calculated the value of χ2, you must determine the probability that the difference between the observed and expected values (χ2) occurred simply by chance (sample error)
• You compare the calculated value to the appropriate value in a “degrees of freedom” table
Degrees of Freedom• Degrees of freedom = # of categories – 1
• Takes into account the natural increase in χ2 as the number of categories increases
df P = 0.05 P = 0.01 P = 0.001
1 3.84 6.64 10.83
2 5.99 9.21 13.82 3 7.82 11.35 16.27 4 9.49 13.28 18.47 5 11.07 15.09 20.52 6 12.59 16.81 22.46 7 14.07 18.48 24.32
• Scientists usually say that if their probability of getting the observed deviation from the expected results by chance is greater that 0.05 (5%), then we can accept the null hypothesis. – The differences we see between what we
expected and what we actually see happened by chance sampling error
• If the probability is less than 0.05, then we reject the null hypothesis– The differences between expected and observed
results did not just happen by chance sampling error
Practice
• A newly identified fruit fly mutant, cyclops eye (large and single in the middle of the head), is hypothesized to be autosomal dominant. The experimenter started with homozygous wild type females and homozygous cyclops males. The data from the F2 generation was 44 wild type males, 60 wild type females, 110 cyclops males and 150 cyclops females. Does this data support or reject the hypothesis? Use chi square to prove your position.
E = cyclopse = normal
P = EE x ee
F1 = all Ee
F2 = Ee x Ee
3: cyclops:1 normal
Expected:
Out of 364 offspring we should get an expected ratio of 273 cyclops: 91 normal
What we actually got was 260 cyclops: 104 normal
2 = (260-273)2
273+ (104-91)2
91
= 169/273 + 169/91= 0.61 + 1.86 = 2.48
29.623.218.316.011.89.347.274.872.5610
27.921.716.914.710.68.346.394.172.099
26.120.115.513.49.57.345.533.491.658
24.318.814.112.08.46.354.672.831.247
22.516.812.610.67.25.353.832.200.876
20.515.111.19.26.14.353.001.610.555
18.513.39.57.84.93.362.021.060.304
16.311.37.86.33.72.371.420.580.123
13.89.26.04.62.41.390.710.210.022
10.86.63.82.71.10.460.150.0160.00021
Degreesof
freedom
0.0010.010.050.100.300.500.700.900.99
Test of Significance: Chi-Square Distribution of X2 Probability, Abridged* Probability
29.623.218.316.011.89.347.274.872.5610
27.921.716.914.710.68.346.394.172.099
26.120.115.513.49.57.345.533.491.658
24.318.814.112.08.46.354.672.831.247
22.516.812.610.67.25.353.832.200.876
20.515.111.19.26.14.353.001.610.555
18.513.39.57.84.93.362.021.060.304
16.311.37.86.33.72.371.420.580.123
13.89.26.04.62.41.390.710.210.022
10.86.63.82.71.10.460.150.0160.00021
Degreesof
freedom
0.0010.010.050.100.300.500.700.900.99
Test of Significance: Chi-Square Distribution of X2 Probability, Abridged* Probability
2 = 2.48 Do we accept the hypothesis?
Accept Reject
Accepted: pattern of inheritance is autosomal dominant
Null Hypothesis:If the Mars Co. M & M sorters are doing their job correctly, then there should be no difference in M & M color ratios between actual store-bought bags of M &Ms and what the Mars Co. claims are the actual ratios.