Question

3.6 To determine if there is an association between gender and accident circumstances, 146 people who had been injured in an accident were randomly selected from medical records and categorised by gender and the circumstance. The results are shown below:

	At work	At home	car	Other	total
female	4	28	6	24	62
male	18	26	4	36	84
total	22	54	10	60	146

a) State the null and alternative hypotheses for this test of association/independence.

b) How many degrees of freedom are there?

c) Find χ² and the P-value.

d) State your conclusion (using α = 0.05) in plain English.

Showing all working out and dont use technology ie. spss or excel.

Answer 1

Solution:

a) State the null and alternative hypotheses for this test of association/independence.

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H₀: There is no association between gender and accident circumstances.

Alternative hypothesis: H_a: There is an association between gender and accident circumstances.

We are given level of significance = α = 0.05

b) How many degrees of freedom are there?

We are given

Number of rows = r = 2

Number of columns = c = 4

Degrees of freedom = df = (r – 1)*(c – 1) = 1*3 = 3

α = 0.05

Critical value = 7.814728

(by using Chi square table or excel)

c) Find χ² and the P-value.

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	At work	At home	Car	Other	Total
Female	4	28	6	24	62
Male	18	26	4	36	84
Total	22	54	10	60	146

Expected Frequencies
	Column variable
Row variable	At work	At home	Car	Other	Total
Female	9.342466	22.93151	4.246575	25.47945	62
Male	12.65753	31.06849	5.753425	34.52055	84
Total	22	54	10	60	146

Calculations
(O - E)
-5.34247	5.068493	1.753425	-1.47945
5.342466	-5.06849	-1.75342	1.479452

(O - E)^2/E
3.055076	1.120276	0.723995	0.085904
2.254937	0.826871	0.534377	0.063405

Chi square = ∑[(O – E)^2/E] = 8.66484

d) State your conclusion (using α = 0.05) in plain English.

P-value = 0.034095

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that there is an association between gender and accident circumstances.