Question

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.

Short answer

Ans:

(a)

(b) From the MINITAB output, the value of the chi-square statistic is 0.794.

(c) The data do not provide sufficient evidence to conclude that the two-variable are associated at the 5% significance level.

Step-by-step solution

Step 1. Given information.

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.

Step 2. (a) Find the expected frequencies by using MINITAB.

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.

Step 3. Output:

Chi-Square Test for Association: y, Worksheet columns

Step 4. From the MINITAB output, the row totals, column total, and grand total are tabulated given below:

Step 5. From the MINITAB output, the expected frequencies are tabulated given below:

Step 6. (b) Determine the value of the chi-square statistic.

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

Step 7. (c) 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.

Step 8. Conclusion for the 5% significance level:

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.