Chi square test of Froot Loops

Using Chi-Square Statistical Analysis: practice with Froot Loops™
A chi-square is a statistical tool that helps us to decide if the observed ratio is close enough to the expected
ratio to be acceptable. Chi-square analysis can be used in any area, not just genetics. Whenever you have to
determine if an expected ratio fits an observed ratio, you can use the chi-square.
Analysis of Results
The chi-square ( ) test is used as an analytical tool to test the validity of a null hypothesis, which states that
there is no statistically significant difference between the observed results of your experiment and the expected
results. When there is little difference between the observed results and the expected results, you obtain a very
low chi-square value; your hypothesis is supported.
The formula for chi-square is:
where:
O = observed number of individuals
E= expected number of individuals
Statisticians have developed chi-square tables, based upon the probabilities that a particular chi-square value
will come about purely by chance. There are two "features" to consider.
A. Significance Level…. Scientists like to use the level of 5% (0.05) as our significant "cut-off". Any chi-square
larger than the value from the 5% on the Table indicates an experiment in which the ratios observed are so far
off the ratios expected that we have to conclude that the ratios expected are wrong!
B. Degrees of Freedom… In order to interpret the results of the chi-square computation, one must use a table of
chi-square values. The degrees of freedom of the results of a cross is equal to the phenotypic categories
minus one (df = n-1). The more "classes" (categories) the more likely that a statistical "blip" will increase the
acceptable limits of the chi-square.
Table 1: CHI-SQUARE DISTRIBUTION TABLE
Accept Hypothesis
Reject Hypothesis
Probability (p)
Degrees
of
Freedom
1
2
3
4
5
6
7
8
9
10
0.95
0.004
0.10
0.35
0.71
1.14
1.63
2.17
2.73
3.32
3.94
0.90 0.80 0.70 0.50
0.02
0.21
0.58
1.06
1.61
2.20
2.83
3.49
4.17
4.86
0.06
0.45
1.01
1.65
2.34
3.07
3.82
4.59
5.38
6.18
0.15
0.71
1.42
2.20
3.00
3.83
4.67
5.53
6.39
7.27
0.30
0.20
0.46 1.07
1.64
1.39 2.41
3.22
2.37 3.66
4.64
3.36 4.88
5.99
4.35 6.06
7.29
5.35 7.23
8.56
6.35 8.38
9.80
7.34 9.52 11.03
8.34 10.66 12.24
9.34 11.78 13.44
0.10
0.05
0.01
0.001
2.71
4.60
6.25
7.78
9.24
10.64
12.02
13.36
14.68
15.99
3.84
5.99
7.82
9.49
11.07
12.59
14.07
15.51
16.92
18.31
6.64
9.21
11.34
13.38
15.09
16.81
18.48
20.09
21.67
23.21
10.83
13.82
16.27
18.47
20.52
22.46
24.32
26.12
27.88
29.59
Problem:
Null hypothesis:
Worksheet:
Classes (colors)
Observed #
Expected #
O-E
(O-E)
2
total
2
(O-E) /E
Sum =
X2 = _________
•
How many degrees of freedom do you use for these data? ___________
•
What is the “critical value” for a p of 0.05? ___________
•
Is your x2 value greater or less than this value? GREATER / LESS
•
Do you accept or reject the null hypothesis based on this? Circle one: ACCEPT / REJECT
•
Is there a statistically significant difference between the number of different colors in your container of
cereal? YES / NO