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.