We have presented a contingency table that gives a cross-classification of a random sample of values for two variables x and y, of a population.

Perform the following tasks

a. Find the expected frequencies Note: You will first need to compute the row totals, column totals, and grand total.

b. Determine the value of the chi-square statistic

c. Decide at the 5% significance level whether the data provide sufficient evidence to conclude that the two variables are associated.

MINITAB procedure:

Step 1: Choose Stat > Tables > Chi-Square test for association.

Step 2. In Columns containing the table, enter the column of A, B and C

Step 3. In Rows, select y

Step 4: Under Statistics, select Chi-square test, Display counts in each cell, Display marginal counts and expected cell counts.

Step 5: Click OK.

Determine the value of chi-squared statistic

From the MINITAB output, the value of chi-squared statistic is $\boxed{12.5}$.

Check whether or not the data provide sufficient evidence to conclude that the two variables are associated at the 5% significance level

The hypotheses are given below

Null hypothesis:

$H_0$: The two variables are not associated

Alternative hypothesis:

$H_1$: The two variables are associated

From the output, the value of test statistic is $\boxed{12.5}$and the p-value is $\boxed{0.002}$.

Here, the p-value is lesser than the level of significance

That is,$p-value(=0.002)<\alpha (=0.05)$.

Therefore, the null hypothesis is rejected at 5% level

Thus, the data provide sufficient evidence to conclude that the two variables are associated at

the 5% significance level