Final answer:
To test the hypothesis that more than 35% of students have kids with a significance level of 0.09, a Z-test for proportion is used, and the null hypothesis is rejected only if the p-value is less than 0.09.
Step-by-step explanation:
The subject of this question is to test a hypothesis regarding the proportion of students who have kids, to determine if it is greater than or equal to 35% using a significance level of 0.09. In the given sample, 292 out of 894 students have kids. To perform this test, we'll use a Z-test for proportion.
Here are the detailed steps:
Set up the null hypothesis (H0): p ≥ 0.35 against the alternative hypothesis (Ha): p < 0.35.
Calculate the sample proportion (p = 292 / 894).
Find the standard error (SE = sqrt(p(1-p)/n)), where p is the population proportion under the null hypothesis and n is the sample size.
Calculate the z-score (z = (p - p)/SE).
Determine the p-value from the Z distribution.
Compare the p-value to the significance level (α), and if p-value < α, reject the null hypothesis.
If the calculated p-value is less than 0.09, we have enough evidence to reject the null hypothesis and conclude that the proportion is less than 35%.