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 fixed2. The trials are independent3. The outcome of each trial can be classified into exactly one of several different categories4. 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 E = If the expected frequencies in the various categories are not equal, then

E =

kn

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 ( ) Test for Goodness of FitTo test the hypotheses H0: the actual population proportions _________________________ Ha: _______________________________________ from their hypothesized proportions Test Statistic: = Use the distribution with _____ degrees of freedom, written as (k-1). P-value = P(________) Conditions: All individual expected counts are at least _____ and not more than 20% of the counts are less than _____.

sproportion edhypothesiz theequaldiffer sproportion population actual theof least twoat

2

2X EEO 2

2

21k

22 X

15

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 left-front right-front left-rear right-rear # selected 11 15 8

6

Use a .05 significance level to test the claim that the results fit a uniform distribution.

Hypothesis: The population of interest is ________________

H0: _____________________

Ha: ________________________________________

where _____________________________________

students statistics all

25.PLF RRLRRF PPP

.25 fromdiffer sproportion population theof least twoat

that tirepicking students of proportion theis Peach

Conditions:

.population the togeneralizenotmay results so sample, econvenienc aactually is This :SRS

)10440 (i.e. 5.an greater th are counts expected All

t.independennot are ilsexpect tra not toreason No

Calculations:

10106

10108

101015

101011 2222

2

6.42

6.4

3 1-4 F of D

.204 :Calc. .25 and .20between :Table :value-P

),,(2 DofFupperlowercdf

Conclusions:

.25. fromdiffer not do students stastisticby picked tires4 for the sproportion that theconclude We level. cesignifican .05 at the H reject the tofail we.204, of value-p aWith 0

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 25 - 44 45 - 64 over 64 # of drivers 36 21 12 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 _________

H0: ________________________________________

Ha: ________________________________________

where _____________________________________

drivers all

13.,27.,44.,16.P 656445442525 PPP

thesefromdiffer sproportion population theof least twoat

group ageeach in crashes of proportion theis Peach

Conditions:

10nNt then replacemen w/odonesampling If t.independen are results sample assume toReasonable

11.44.(88)(.13)beingsmallest the5,an greater th are counts expected All

SRS.an is sample theassume not toreason No

Calculations:

05.5308.14

08.1436 22

05.53

111079.1 xvaluep

Conclusions:

category.in that drivers of proportion thefrom differs crashes of proportion thecategories theof least twoat in that conclude and

H reject the weso level cesignifican .05 than theless is value-pOur 0

Properties of the Chi-Square Distributions 1. The total area under a chi-square density curve is equal to ______.

2. Chi-Squared distributions take only positive values.

3. Each chi-square curve is __________________. The curve becomes more symmetrical and looks more Normal as the number of degrees of freedom increases.

1

skewedright

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 the chi-square statistic. For example 2, the largest

component was __________

component

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