Goodness of fit 0 This test is used to decide whether there is any difference between the observed (experimental) value and the expected.

Goodness of fit
0 This test is used to decide whether there is any
difference between the observed (experimental) value
and the expected (theoretical) value.
Goodness of Fit
Free from Assumptions
0 Chi square goodness of fit test depends only on the set
of observed and expected frequencies and degrees of
freedom. This test does not need any assumption
regarding distribution of the parent population from
which the samples are taken.
0 Since this test does not involve any population
parameters or characteristics, it is termed as nonparametric or distribution free tests. This test is also
sample size independent and can be used for any
sample size.
It is all about expectations
Oi = an observed frequency (i.e. count) for
measurement i
Ei = an expected (theoretical) frequency
for measurement i, asserted by the null
hypothesis.
Expected Value
0 F = the cumulative Distribution function for the distribution
being tested.
0 Yu = the upper limit for class I
0 (maximum possible observations for any category)
0 Yl = the lower limit for class I
0 (minumum possible observations for any category)
0 N = the sample size
Hypothesis testing
Choose a level of alpha – usually 0.05
This implies a 95% level of comfort that the observation is
correct.
Example
The number of cubs delivered to a population of bears in the
wild is tested to see if there is no difference in probability of
twins. (N = 50 females)
Number of
cubs
0
1
2
3
Observed
1
5
35
9
Expected
12.5
12.5
12.5
12.5
Degrees of Freedom = Number of groups – 1
df = 4 – 1 = 3
CHI-SQUARE DISTRIBUTION
TABLE
Decision Rule
0 Based on the alpha and the degrees of freedom, look
up the value in the table.
0 For our example of alpha=.05 and df=3
0 If chi square is greater than 7.82 then reject the
null hypothesis that bears normally birth twins.
Number of
cubs
Calculate the value
0
1
2
3
Observed
1
5
35
9
Expected
12.5
12.5
12.5
12.5
Chi-square =
(1-12.5)2/12.5 + (5-12.5)2/12.5 + (35-12.5)2/12.5 + (912.5)2/12.5 =
10.58 + 4.5 + 40.5 + 0.98 = 56.56
Since 56.56 > 7.82 we reject the null hypothesis that the
number of bear cubs is equally possible for 0-3 cubs
Interpret the result
0 Since we rejected the null hypothesis, what
conclusions (inferences) can we come to?