Question

A researcher collected data on a sample of 301 patients. The variable of interest was the percentage of subjects with impaired fasting glucose (IFG). IFG refers to a metabolic stage intermediate between normal glucose homeostasis and diabetes. In the study, 24 patients were classified in the IFG stage. The article cites population estimates for IFG among all patients in that area as 6.3 percent. Is there sufficient evidence to indicate that the population of patients has a prevalence of IFG different from 6.3 percent? Use a level of significance of 0.01 to conduct the test.

Answer 1

$$H _ {0}: p = 0. 0 6 3;$$ $$H _ {1}: p \neq 0. 0 6 3;$$ $$p - h a t = \frac {2 4}{3 0 1} = 0. 0 7 9 7;$$ $$p = \frac {6 . 3}{1 0 0} = 0. 0 6 3;$$ $$z = \frac {\left(0 . 0 7 9 7 - . 0 6 3\right)}{\sqrt {0 . 0 6 3 \cdot \frac {\left(1 - 0 . 0 6 3\right)}{3 0 1}}} = 1. 1 9 2 5;$$

critical _ $z = + / - 2.58$ at 0.01 level and two-tailed test.

Hence, computed $z = 1.1925$ is in the interval $(-2.58, 2.58)$ . So, fail to reject $H_0$ .

And we can make next conclusion: that there is no sufficient evidence to indicate that the population of patients has a prevalence of IFG difference from $6.3\%$ .