Question

1. what percentage of college students have placed a bet on a sporting event? to determine this, jane surveys a random sample of 600 college students from a university that enrolls 40,000 students, and ted surveys a random sample of 1,200 college students from a university that enrolls 20,000 students. who will have the larger estimated margin of error?

a. without knowing the exact survey results obtained by jane and ted, this question cannot be answered.
b. ted because he is sampling from a smaller university.
c. jane because she is taking a smaller sample.
d. ted because he is taking a larger sample.
e. jane because she is sampling from a larger university.

Explanation:

Step1: Recall margin - of - error formula

The margin of error for a proportion in a simple random sample is approximately $E = z\sqrt{\frac{p(1 - p)}{n}}$ (where $z$ is the z - score related to the confidence level, $p$ is the sample proportion, and $n$ is the sample size). When comparing two samples, assuming the same confidence level and similar population characteristics (unknown true proportion $p$ is assumed to be similar in both cases), the margin of error is inversely proportional to the square - root of the sample size $n$.

Step2: Compare sample sizes

Jane has a sample size $n_1=600$ and Ted has a sample size $n_2 = 1200$. Since the margin of error $E\propto\frac{1}{\sqrt{n}}$, a smaller sample size will result in a larger margin of error.

Answer:

C. Jane because she is taking a smaller sample.