Question

A car insurance company performed a study to determine whether an association exists between age and the frequency of car accidents. They obtained the following sample data.

		under 25	25-45	over 45	total
number of	0	74	90	84	248
accidents in	1	19	8	12	39
past 3 years	>1	7	2	4	13
	total	100	100	100	300

Conduct a test, at the 5% significance level, to determine whether the data provide sufficient evidence to conclude that an association exists between age and frequency of car accidents.

(a) State the null and alternative hypotheses for this test.

(b) Calculate:

(i) the expected frequency, E31, for the number of accidents is > 1 and age is under 25 years?

(ii) the chi-square contribution, χ2, for the number of accidents is > 1 and age is under 25 years?

(iii) the p-value for this test?

(c) Given that the chi-square test statistic, χ2, is 9.273, state the conclusion for the test.

(d) The sample data was inputted into a statistical software for analysis. The output generated included the following warning: Warning: 3 cells (33.3%) have expected count less than 5. The minimum expected count is 4.33.

(i) Why was this warning displayed?

(ii) How does this affect the conclusion stated in part (c)?

(iii) Suggest how this problem could be resolved?

Answer 1

Given table data is as below

MATRIX	col1	col2	col3	TOTALS
row 1	74	90	84	248
row 2	19	8	12	39
row 3	7	2	4	13
TOTALS	100	100	100	N = 300

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calculation formula for E table matrix

E-TABLE	col1	col2	col3
row 1	row1*col1/N	row1*col2/N	row1*col3/N
row 2	row2*col1/N	row2*col2/N	row2*col3/N
row 3	row3*col1/N	row3*col2/N	row3*col3/N

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expected frequecies calculated by applying E - table matrix formulae

E-TABLE	col1	col2	col3
row 1	82.667	82.667	82.667
row 2	13	13	13
row 3	4.333	4.333	4.333

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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

Oi	Ei	Oi-Ei	(Oi-Ei)^2	(Oi-Ei)^2/Ei
74	82.667	-8.667	75.117	0.909
90	82.667	7.333	53.773	0.65
84	82.667	1.333	1.777	0.021
19	13	6	36	2.769
8	13	-5	25	1.923
12	13	-1	1	0.077
7	4.333	2.667	7.113	1.642
2	4.333	-2.333	5.443	1.256
4	4.333	-0.333	0.111	0.026
				chisqr^2 o = 9.273

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set up null vs alternative as

null, Ho: no association exists between age and frequency of car accidents

alternative, H1: exists a relation b/w age and frequency of car accidents OR an association exists between age and frequency of car accidents

level of significance, alpha = 0.05

from standard normal table, chi square value at right tailed, chisqr^2 alpha/2 =9.488

since our test is right tailed,reject Ho when chisqr^2 o > 9.488

we use test statistic chisqr^2 o = Σ(Oi-Ei)^2/Ei

from the table , chisqr^2 o = 9.273

critical value

the value of |chisqr^2 alpha| at los 0.05 with d.f (r-1)(c-1)= ( 3 -1 ) * ( 3 - 1 ) = 2 * 2 = 4 is 9.488

we got | chisqr^2| =9.273 & | chisqr^2 alpha | =9.488

make decision

hence value of | chisqr^2 o | < | chisqr^2 alpha | and here we do not reject Ho

chisqr^2 p_value =0.055

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A.

null, Ho: no association exists between age and frequency of car accidents

alternative, H1: exists a relation b/w age and frequency of car accidents OR an association exists between age and frequency of car accidents

B.

(i) the expected frequency, E31 = 4.333 > 1 and age is under 25 years

(ii)test statistic: 9.273

(iii)p-value:0.055

C.

test statistic: 9.273, decision: do not reject Ho

D.

i)
Warning: 3 cells (33.3%) have expected count less than 5, because
three of the expected values are 4.33 <5, and chi square rules explains that
every provided cell observation values should be greater than 5
ii)
In chi-square test we compare observed frequency (that we measure directly
from the data) with the expected frequency. We calculate expected frequency
(in Chi-square for independence) forr each cell. The assumption suggests that no
cell should have expected frequency of less than 5. Or else. the test is not valid.
iii)
It is possible to ‘poor or ‘collapse’ categories into fewer. but this must only be
done if it is meaningful to group the data in this way. Else use Fisher’s exact
test.