In each of the given Exercises, 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. For each exercise, 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.

given,

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. For each exercise, perform the following tasks.

MINITAB procedure:
<Step 1> Choose Stat > Tables > Chi-Square test for association.
<Step 2> In Columns containing the table, enter the column of A and B.
<Step 3> In Rows, select y.
<Step 4> Under Statistics, we select the Chi-square test, Display counts in each cell, Display marginal counts and expected cell counts.
<Step 5> Click OK.

Chi-Square Test for Association: y, Worksheet columns

So, from the MINITAB output, the value of the chi-square statistic is 0.794.

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 the test statistic is 0.794 and the p-value is 0.373.

Here, the p-value is greater than the level of significance.

That is, $p-value(=0.373)>\alpha (=0.05)$.

Therefore, the null hypothesis is not rejected at the 5% level.

Thus, the data do not provide sufficient evidence that concludes the two-variable are associated at the 5% significance level.