The chi-square test is used to determine if differences in frequencies observed in qualitative variable categories are statistically significant or likely due to chance. An example compares influenza rates in a vaccine vs placebo group in a clinical trial. The expected and observed frequencies are calculated. The chi-square test statistic is greater than the critical value, so the null hypothesis that the vaccine and placebo have the same influenza proportion is rejected. Therefore, the difference is likely due to the vaccine's effectiveness rather than chance.

1. Chi square testChi square test

2. Uses and ApplicationsUses and Applications
 Used when you have frequencyUsed when you have frequency
distribution of qualitative type of variables.distribution of qualitative type of variables.
 Applications:Applications:
 To test goodness of fit.To test goodness of fit.
 In the 2X2 table, it is used to test whetherIn the 2X2 table, it is used to test whether
there is anthere is an associationassociation between the rowbetween the row
and the column variables; ie whether theand the column variables; ie whether the
distribution of individuals among thedistribution of individuals among the
categories of one variable iscategories of one variable is
independentindependent of their distribution amongof their distribution among
the categories of the other.the categories of the other.

3. Example, Influenza vaccination trialExample, Influenza vaccination trial
InfluenzaInfluenza NoNo
influenzainfluenza
TotalTotal
VaccineVaccine 2020
(8.3%)(8.3%)
220220 240240
PlaceboPlacebo 8080
(36.4%)(36.4%)
140140 220220
TotalTotal 100100
(21.7%)(21.7%)
360360 460460

4. The question is:The question is:
 Is the difference (in the percentages ofIs the difference (in the percentages of
influenza) due to vaccination or occurredinfluenza) due to vaccination or occurred
by chance?by chance?
 To answer, a Chi square test is doneTo answer, a Chi square test is done
which compares the observed numbers inwhich compares the observed numbers in
each of the four categories in theeach of the four categories in the
contingency table with the numbers to becontingency table with the numbers to be
expected if there was no difference in theexpected if there was no difference in the
effectiveness between the vaccine andeffectiveness between the vaccine and
placebo.placebo.

5. The expected frequencies:The expected frequencies:
InfluenzaInfluenza NoNo
influenzainfluenza
TotalTotal
VaccineVaccine 52.252.2 187.8187.8 240240
PlaceboPlacebo 47.847.8 172.2172.2 220220
TotalTotal 100100 360360 460460

6. Solution, continuedSolution, continued
 Ho: The proportion of influenza among the vaccineHo: The proportion of influenza among the vaccine
group = The proportion of influenza among thegroup = The proportion of influenza among the
placebo group.placebo group.
 Level of significance (alpha) = 0.05Level of significance (alpha) = 0.05
 D.f. = (No. of rows-1) (No. of columns-1).D.f. = (No. of rows-1) (No. of columns-1).
 Test statistics: Chi square test.Test statistics: Chi square test.
E
EO 2
2 )( −∑
=χ

7. Solution, continuedSolution, continued
09.53
2.172
)2.172140(
8.187
)8.187220(
8.47
)8.4780(
2.52
)2.5220( 2222
2
=
−
+
−
+
−
+
−
=χ
The tabulated value for X2
for 1 d.f. is 3.841
The calculated value (53.09) > tabulated value
Therefore reject Ho and conclude that there is statistically
significant difference between the 2 proportions. This
difference is unlikely to be due to chance.
Therefore the vaccine is effective.
P < 0.05