1. Statistical Analysis: Chi Square

2. “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

3. Null Hypothesis- Ho
(in a statistical test) the hypothesis that there is no significant statistical difference between specified populations
any observed difference being due to sampling or experimental error.

4. 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

5. 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

6. Degrees of Freedom df P = 0.05 P = 0.01 P = 0.001 1 3.84 6.64 10.83 2
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

7. Why subtract 1?
The degrees of freedom are the number of values that are free to vary in the data set. For example:
Q. Pick three numbers that have a mean (average) of 10. A. Some sets of numbers you might pick: 9, 10, 11 or 8, 10, 12 or 5, 10, 15. Once you have chosen the first two numbers in the set, the third is fixed. In other words, you can’t choose the third item in the set. The only numbers that are free to vary are the first two. So degrees of freedom for a set of three numbers is TWO.

8. Practice Problem
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 and homozygous cyclops. The second generation is heterozygous. The data from the F2 generation was 104 wild type and 260 cyclops. ( since it is autosomal dominant you do not need to worry about the sex of the flies) Does this data support or reject the hypothesis? Use chi square to prove your position.

9. χ2 = Expected: E = cyclops e = normal P = EE x ee F1 = all Ee
F2 = Ee x Ee
3: cyclops:1 normal
Out of 364 offspring we should get an expected ratio of
273 cyclops: 91 normal
What we actually got was 260 cyclops: 104 normal
χ2 =
( )2
(104-91)2 +
273 91
= 169/ /91 =
= 2.48

10. χ2 = 2.48 Do we accept the hypothesis?
Reject
χ2 = Do we accept the hypothesis?
Accepted: pattern of inheritance is autosomal dominant

11.                    
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. Let’s test this next class!