Answer
$X_0^2=330.803$ with a P-value $\ltα$. Reject the null hypothesis.
There is enough evidence to conclude that the proportion of individuals for each level of time spent in bars is not equal for smokers and nonsmokers.
By the graph, we can conclude that most of those with a higher frequency of visits to bars are smokers.
Work Step by Step
$H_0:$ the proportion of individuals for each level of time spent in bars is equal for smokers and nonsmokers.
versus
$H_1:$ the proportion of individuals for each level of time spent in bars is not equal for smokers and nonsmokers.
In MINITAB, enter the given values:
1-C1 = 80, 1-C2 = 409, 1-C3 = 294, 1-C4 = 362, 1-C5 = 433, 1-C6 = 336, 1-C7 = 1265
2-C1 = 57, 2-C2 = 350, 2-C3 = 379, 2-C4 = 471, 2-C5 = 573, 2-C6 = 568, 3-C7 = 3297
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 C6 C7
Click OK.
$X_0^2=330.803$ with a P-value $\lt0.001\ltα=0.05$
