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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.

Short Answer

Expert verified

The data do not provide sufficient evidence to conclude that the two variables are associated at the 5% significance level.

Step by step solution

01

Step 1. Given

02

Step 2. Solution a). Find the Expected frequencies using MINITAB

MINITAB procedure:

Step 1: Choose Stat > Tables > Chi-Square test (Two-Way Table in Worksheet).

Step 2: In Columns containing the table, enter the columns of AandB.

Step 3: Click OK.

03

Step 3. MINITAB output

04

Step 4. Row total, column total and grand total

yABTotal
a53540
b2080100
c3585110
Total50200250
05

Step 5. Expected Frequencies

yABTotal
a83240
b2080100
c2288110
Total50200250
06

Step 6. Solution b)

Determine the value of chi-squared statistic

From the MINITAB output, the value of chi-squared statistic is1.918

07

Step 7. Solution 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:

: The two variables are not associated

Alternative hypothesis:

: The two variables are associated

08

Step 8. Conclusion for 5% significance level

From the output, the value of test statistic is 1.918and the p-value is 0.383.

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

That is, p-value(=0.383)>α(=0.05).

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

Thus, the data do not provide sufficient evidence to conclude that the two variables are associated at the 5% significance level.

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Frequency

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