91Ó°ÊÓ

a.

Check whether or not the data provide sufficient evidence to conclude that an association exists between completion of treatment and practice type at 5% significance level.

Step 1:

The test hypotheses are given below:

Null hypothesis:

H0 : There is no association exists between completion of treatment and practice type.

Alternative hypothesis:

H1 : There is an association exists between completion of treatment and practice type.

Step 2: Decide the level of significance.

Here, the level of significance is, 1%.

Step 3:

Find the expected frequency and test statistic.

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 Yes and No.

Step 3: Click OK.

Step 4:

Find the P-value.

From the MINITAB output, the P-value is 0.015

Step 5:

Rejection rule:

If P-value $≤\mathrm{Î\pm }$, then reject the null hypothesis.

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

That is, P-value $(=0.015)<\mathrm{Î\pm }(=0.05)$.

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

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

The data do not provide sufficient evidence to conclude that an association exists between completion of treatment and practice type at the 1% significance level.