Answer
$X_0^2=4.673$ with a P-value $\gtα$. Do not reject the null hypothesis.
There is not enough evidence to conclude that the proportion of subjects within each treatment group who experienced dizziness are not the same.
Work Step by Step
$H_0:$ the proportion of subjects within each treatment group who experienced dizziness are the same.
versus
$H_1:$ the proportion of subjects within each treatment group who experienced dizziness are not the same.
In MINITAB, enter the given values:
1-C1 = 83, 1-C2 = 32, 1-C3 = 36, 1-C4 = 5, 1-C5 = 8
2-C1 = 4063, 2-C2 = 1832, 2-C3 = 1330, 2-C4 = 382, 2-C5 = 337
Select Stat -> Tables -> Chi-Square Test for Association
Select "Summarized data in a two-way table"
In columns containing the table enter: C1 C2 C3 C4 C5
Click OK.
$X_0^2=4.673$ with a P-value $=0.323\gtα=0.01$
