The document describes a chi-square test for homogeneity conducted on data from 100 vehicular accidents to determine if there is a significant difference in fatal injury rates between individuals who wore seatbelts and those who did not. The null hypothesis states there is no significant difference, while the alternative suggests a significant difference, particularly that those without seatbelts are at greater risk. The test results showed a Pearson chi-square value of 26.014 with a p-value < .001, leading to the rejection of the null hypothesis.

1. THE CHI-SQUARE TEST
for Homogeneity of Samples

3. TWO INDEPENDENT SAMPLES
Example :
In a study conducted on the use of seat
belts in preventing fatalities, records of the
last 100 vehicular accidents were
reviewed. These 100 accidents involved
238 persons. Each person was classified
as using or not using seat belts when the
accident happened and as injured fatally or
a survivor.

4. Research Question:
Is there a significant difference in the proportion
of persons who are fatally injured between those
who wear seatbelts and those who do not?
Null Hypothesis:
There is no significant difference in the
proportion of persons who are fatally injured
between those who wear seatbelts and those
who do not.

5. Alternative Hypothesis (Non-directional):
There is a significant difference in the proportion
of persons who are fatally injured between those
who wear seatbelts and those who do not.
Alternative Hypothesis (Directional):
The proportion of persons who are fatally injured
is higher for those who do not wear seatbelts
than those who wear seatbelts.

6. Data:
Injured
Fatally?
Wearing Seat Belt? Total
Yes (1) No (0)
Yes (1) 9 88 97
No (0) 23 118 141
Total 32 206 238

8.  



cells
all
over e
e
o 2
2

For contingency tables, the degrees of freedom is
given by
df= (r-1)(c-1)
where r is the number of categories of the row
variable while c is the number of categories of the
column variable.
Where o is the “observed frequency” and e is the
expected frequency.
Test Statistic:

9. The tabled data is actually a summary of the
responses of the respondents which can be
encoded directly using SPSS
• Use codes for the
responses: 1 – Yes; 2 – No
• Encode the frequency
counts in a column
• Put the corresponding
codes for the two variables
• So that SPSS will read the
data as frequency counts,
we have to set on the
“weight cases” command

10. 1. Click “Data”; then “Weight Cases”
to show the window on top right.
2. Highlight the data “Freq”, Click “Weight cases by” then Click
the arrow button; Click “ok”

11. Check if “Weight On” already appears at the bottom right corner
of the worksheet

12. To Analyze the data: Click “Analyze” ”Descriptive Statistics”
 “Crosstabs” to obtain the window at the right.
Transfer the row variable “Injured Fatally” and the column
variable “Wearing Seatbelts” by highlighting them one at a
time and clicking the arrow button.

13. Click “Statistics” then check “Chi-
square”, then click “Continue”;
Click “Cells” then check “Column”;
then Click “Continue”; then Click
“OK”

14. Output (Descriptive Statistics):
Note that 7.1% of those who wore seatbelts were injured
fatally while 29.9% of those who did not wear seatbelts
were injured fatally. Is 29.9% significantly higher than
7.1%?

15. Output (Inferential Statistics):
If there is no cell with expected frequency of less than 5, we report the
Pearson Chi-square value. Since 2 = 26.014 and the associated
probability is <.001 (which is less than  = .05), the null hypothesis is
rejected.
If there is a cell with expected frequency of less than 5, we have to
report the Fisher’s Exact Test. Note that the Fisher’s Exact Probability is
less than .001. Hence the null hypothesis is rejected.