A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a 95 % confidence interval to estimate the true proportion of students on financial aid. Express the answer in the form $\hat { p } \pm E$ and round to the nearest thousandth.
A) $0.59 \pm 0.068$
B) $0.59 \pm 0.002$
C) $0.59 \pm 0.474$
D) $0.59 \pm 0.005$

Question

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a  95 %  confidence interval to estimate the true proportion of students on financial aid. Express the answer in the form $\hat { p } \pm E$ and round to the nearest thousandth.
A) $0.59 \pm 0.068$
B) $0.59 \pm 0.002$
C) $0.59 \pm 0.474$
D) $0.59 \pm 0.005$

Quizplus · Accepted Answer

The sample proportion ($\hat { p }$) is calculated as the number of students receiving financial aid divided by the total number of students sampled, which is $118/200 = 0.59$. To calculate the margin of error (E) for a 95% confidence interval, the formula $E = Z \cdot \sqrt{\frac{\hat { p }(1 - \hat { p })}{n}}$ is used, where $Z$ is the Z-score corresponding to a 95% confidence level (approximately 1.96), $\hat { p }$ is the sample proportion, and $n$ is the sample size. Substituting the given values, we find $E \approx 0.068$, leading to a confidence interval of $0.59 \pm 0.068$.

A University Dean Is Interested in Determining the Proportion of Students