Answer
$X_0^2=12.646$ with a P-value $=0.013$. Reject the null hypothesis.
There is enough evidence to conclude that the proportion of incidence of salmonella is not equal for each brand.
Work Step by Step
$H_0:$ the proportion of incidence of salmonella is equal for each brand.
versus
$H_1:$ the proportion of incidence of salmonella is not equal for each brand.
In MINITAB, enter the given values:
1-C1 = 8, 1-C2 = 192
2-C1 = 17, 2-C2 = 183
3-C1 = 27, 3-C2 = 173
4-C1 = 14, 4-C2 = 186
5-C1 = 20, 5-C2 = 180
Select Stat -> Tables -> Chi-Square Test for Association
Select "Summarized data in a two-way table"
In columns containing the table enter: C1 C2
Click OK.
$X_0^2=12.646$ with a P-value $=0.013$
