Question

In given exercise use either the critical-value approach or the P-value approach to perform a chi square independence lest. provided the conditions for using the test are met.

Diabetes in Native Americans. Preventable chronic diseases are increasing rapidly in Native American populations, particularly diabetes. F. Gilliland et al. examined the diabetes issue in the paper "Preventative Health Care among Rural American Indians in New Mexico" (Preventative Medicine, Vol. 28, pp. 194-202). Following is a contingency table showing cross-classification of educational attainment and diabetic state for a sample of 1273 Native Americans (HS is high school).

At the 1% significance level, do the data provide sufficient evidence to conclude that an association exists between education level and diabetic state for native Americans?

Short answer

At the 1% significance level, the data provide sufficient evidence to conclude that there is an association between education level and diabetic state.

Step-by-step solution

Step 1. Given

Step 2. The null and alternate hypotheses are 

$H_0$: There is no association between education level and diabetic state.

$H_a$: There is an association between education level and diabetic state.

Step 3. 

The following contingency table contains the expected frequencies corresponding to the

Observed frequencies

${(O-E)}^2$

Step 4.

•All expected frequencies are greater than 1, we can verify by referring to the above table.

• At most 20% of the expected frequencies are less than 5. In fact, none of the expected frequencies are less than 5, as we can verify by referring to the above table.

All assumptions are satisfied.

Step 5.

The test is to be performed at the 1% significance level, so$\alpha =0.01$.

Step 6.

$\raisebox{1ex}{${(O-E)}^2$}\!\left/ \!\raisebox{-1ex}{$E$}\right.$

the value of test statistic is${\mathcal{X}}^2=17.517$

Step 7.

the row variable has 4 values and column variable has 2 values, therefore r = 4, c = 2 and

$$\begin{gathered} {\mathcal{X}}^2=17.517df \\ =(r-1).(c-1) \\ =(4-1).(2-1) \\ =3\times 1 \end{gathered}$$

For, $\alpha =0.01$, Table VII reveals that the critical value is =3.

Step 8.

${{\mathcal{X}}_{0.01}}^2=11.345$

Step 9.

we see the value of test statistic falls in the rejection region. Thus we reject the test results are statistically significant at 1% level.

Step 10.

At the 1% significance level, the data provide sufficient evidence to conclude that there is an association between education level and diabetic state.