Religion. A worldwide poll on religion was conducted by WIN-Gallup International and published as the document Global Index of Religiosity and Atheism. One question involved religious belief and educational attainment. The following data is based on the answers to that question.

a. At the 5% significance level, do the data provide sufficient evidence to conclude that an association exists between religiosity and education?

b. Repeat part (a) at the 1% significance level.

Check whether or not the data provide sufficient evidence to conclude that an association exists between religiosity and education.

Step 1:

The test hypotheses are given below:

Null hypothesis:

H0 : There is no association exists between religiosity and education.

Alternative hypothesis:

H1 : There is an association exists between religiosity and education.

Step 2: Decide the level of significance.

Here, the level of significance is, 5%.

Step 3:

Find the expected frequency and test statistic.

MINITAB procedure:

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

Step 2: In Columns containing the table, enter the column of Basic, Secondary and Advanced.

Step 3: In Rows, select RELIGIOUSITY.

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

Step 5: Click OK.

Pearson Chi-square $=13.322,\mathrm{DF}=6$, p-Value =0.038

Likelihood Ratio Chi-Square=12.814, DE=6,P-Value=0.046

From the MINITAB output, the value of the chi-square statistic is 13.322

Step 4:

Find the P-value.

From the MINITAB output, the P-value is 0.038

Step 5:

Rejection rule:

If P-value $\le \alpha$, then reject the null hypothesis.

Here, the P-value is lesser than the level of significance.

That is, P-value (=0.038)<(=0.05) .

Therefore, the null hypothesis is rejected at 5% level.

Thus, the results are statistically significant at 5% level of significance.

Thus, the data provide sufficient evidence to conclude that an association exists between religiosity and education at the 5% significance level.

Decide the significance level.

Here, the level of significance is $\alpha =0.01$.

p-value approach:

From part (a)., the value of the chi-square statistic is $13.322$and the P-value is $0.038$.

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

That is, P-value $(=0.038)>\alpha (=0.01)$.

Therefore, the null hypothesis is not rejected at 1% level.

Thus, the results are not statistically significant at 1% level of significance.