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 p +- E and round to the nearest thousandth.

Answers:
a) 0.59 +- 0.474
b) 0.59 +- 0.002
c) 0.59 +- 0.005
d) 0.59 +-0.068

To estimate the true proportion of students on financial aid with a 95% confidence interval, we can use the formula:

p ± E = p ± Z * sqrt((p * (1-p)) / n)

Where:
p = sample proportion (118/200)
E = margin of error
Z = z-score corresponding to the confidence level (for a 95% confidence level, the z-score is approximately 1.96)
n = sample size (200)

First, calculate the sample proportion:
p = 118/200 = 0.59

Next, calculate the margin of error:
E = 1.96 * sqrt((0.59 * (1-0.59)) / 200)
E ≈ 0.068

Now we can construct the confidence interval:
p ± E = 0.59 ± 0.068

Rounding to the nearest thousandth, we have:
0.59 ± 0.068

Therefore, the answer is option d) 0.59 ± 0.068.

To calculate the 95% confidence interval for the proportion of students receiving financial aid, we can use the formula:

CI = p +- Z * sqrt((p * (1 - p)) / n)

where:
- p is the proportion of students receiving financial aid (118/200 = 0.59)
- Z is the Z-score corresponding to a 95% confidence level (approximately 1.96 for a large sample size)
- n is the sample size (200)

Let's calculate the confidence interval:

CI = 0.59 +- 1.96 * sqrt((0.59 * (1 - 0.59)) / 200)

CI = 0.59 +- 1.96 * sqrt(0.236 / 200)

CI = 0.59 +- 1.96 * 0.034

CI = 0.59 +- 0.067

Rounding to the nearest thousandth, the answer is "0.59 +- 0.068".

Therefore, the correct answer is d) 0.59 +- 0.068.