AP Statistics Section 14.
The main objective of Chapter 14 is to test
claims about qualitative data consisting of
frequency counts for different categories. In
section 14.1, we consider multinomial
experiments which are defined in much the
same way as binomial experiments (section 8.1)
except that a multinomial experiment has more
than two categories.
A multinomial experiment meets the
following criteria:
1. The number of trials is fixed
2. The trials are independent
3. The outcome of each trial can be
classified into exactly one of several
different categories
4. The probabilities for the different
categories remain the same for each trial
The goal of section 14.1 is to consider a
method for testing a claim that the
frequencies observed in the different
categories “fit” a particular distribution.
The method is consequently called a
goodness-of fit test.
A goodness-of-fit test is used to
test a hypothesis that an observed
frequency distribution fits some
claimed form.
We will use the following notation:
O represents the observed
frequency in a particular category
E represents the expected
frequency in a particular category
k represents the number of
different categories
n represents the number of trials
Let’s discuss E, the expected frequency, further.
If the expected frequencies in the various
categories are equal, then
n
E=
k
If the expected frequencies in the various
categories are not equal, then
E = np
Sample frequencies typically deviate, at least
somewhat, from the values that we would
expect. The question we want to answer is, “are
the differences between the observed values, O,
and the expected values, E, statistically
significant?”
The Chi-Square (  2 ) Test for Goodness of Fit
To test the hypotheses
equal the hypothesized proportions
H0: the actual population proportions _________________________
at least two of the actual population proportions differ from their
Ha: _______________________________________
hypothesized proportions
Test Statistic:
2

O  E
2
X =
E
Use the  2 distribution with _____
k  1 degrees of freedom, written
as  2(k-1).
2
2


X
P-value = P(________)
Conditions: All individual expected counts are at least _____
1 and not
more than 20% of the counts are less than _____.
5
Example 1: Four car-pooling students missed their statistics
test and gave a flat tire as their excuse. At the make-up test
the instructor asked them to identify the tire that went flat. If
they didn’t really have a flat tire, would they be able to
randomly identify the same tire? The instructor asked his 40
other students to identify the tire they would select and the
results are given in the following table.
Tire selected
# selected
left-front
11
right-front
15
left-rear
8
right-rear
6
Use a .05 significance level to test the claim that the results fit
a uniform distribution.
Hypothesis:
The population of interest is ________________
all statistics students
PLF  PRF  PLR  PRR  .25
H0: _____________________
at least two of the population proportions differ from .25
Ha: ________________________________________
each P is the proportion of students picking that tire
where _____________________________________
Conditions:
SRS : This is actually a convenience sample, so results may not
generalize to the population.
All expected counts are greater than 5. (i.e. 40  10)
4
No reason not to expect trails are not independent.
Calculations:
2
2
2
2








11

10
15

10
8

10
6

10
2 



10
10
10
10
 2  4.6
D of F  4 - 1  3
P - value : Table : between .20 and .25
Calc. : .204
4.6
 2 cdf (lower , upper , DofF )
Conclusions:
With a p - value of .204, we fail to reject the H 0 at the .05 significance level.
We conclude that the proportions for the 4 tires picked by stastistic students
do not differ from .25.
Example 2: Among drivers who have had a car crash in
the last year, 88 are randomly selected and categorized
by age.
Age
under 25
# of drivers
36
25 - 44
21
45 - 64
12
over 64
19
If all ages have the same crash rate, we would expect,
because of the age distribution of drivers, the four
categories to have rates of 16%, 44%, 27% and 13%
respectively. At the .05 significance level, test the claim
that the distribution of crashes conforms to the
distribution of ages.
Hypothesis:
The population of interest is _________
all drivers
P 25  .16, P25 44  .44, P4564  .27, P65  .13
H0: ________________________________________
Ha: ________________________________________
at least two of the population proportions differ from these
each P is the proportion of crashes in each age group
where _____________________________________
Conditions:
No reason not to assume the sample is an SRS.
All expected counts are greater than 5, the smallest being
(88)(.13)  11.44.
Reasonable to assume sample results are independent. If sampling
done w/o replacement then N  10n
Calculations:
2


36

14
.
08
2 
     53.05
14.08
p  value  1.79 x10 11
53.05
Conclusions:
Our p - value is less than the .05 significance level so we reject the H 0
and conclude that in at least two of the categories the proportion of
crashes differs from the proportion of drivers in that category.
Properties of the Chi-Square Distributions
1. The total area under a chi-square density curve is
equal to ______.
1
2. Chi-Squared distributions take only positive values.
3. Each chi-square curve is __________________.
right  skewed
The curve becomes more symmetrical and looks more
Normal as the number of degrees of freedom
increases.
Follow-up Analysis
If we find significance in a chi-square test for goodness
of fit, we can conclude that our variable has a
distribution different from the specified one. In this
case, it is always a good idea to determine which
categories of the variable provide the greatest
differences between observed and expected counts.
This category is called the largest ____________of
component the
chi-square statistic. For example 2, the largest
component was __________
under 25