A survey of an urban university (population of 25,450) showed that 878 of 1,119 students sampled supported a fee increase to fund improvements to the student recreation center. Using the 95% level of confidence, what is the confidence interval for the proportion of students supporting the fee increase?

[0.761, 0.809]
[0.910, 0.973]
[0.929, 0.976]
[0.991, 0.962]

[0.761, 0.809]

Thank you!!

To determine the confidence interval for the proportion of students supporting the fee increase, we need to use the formula for a confidence interval:

Confidence Interval = Sample Proportion ± Margin of Error

Where the margin of error is calculated using the formula:

Margin of Error = Z * √(p * q / n)

In this case, we have the following information:
- Sample Proportion: 878 out of 1,119
- Population size: 25,450
- Confidence level: 95%

To calculate the confidence interval, we need to find the margin of error using the formula above. First, we need to find the standard deviation, which can be approximated using the formula:

Standard Deviation ≈ √(p * q / n)

Where p is the sample proportion, q is the complement of the sample proportion (1 - p), and n is the sample size.

p = 878 / 1,119 = 0.784
q = 1 - p = 1 - 0.784 = 0.216
n = 1,119

Now, we can calculate the standard deviation:

Standard Deviation ≈ √(0.784 * 0.216 / 1,119) ≈ 0.0142

Since the population is larger than 20, we can use the Z-distribution table for a 95% confidence level to find the Z-value. The Z-value for a 95% level of confidence is approximately 1.96.

Margin of Error = 1.96 * 0.0142 ≈ 0.0279

Finally, we can calculate the confidence interval:

Confidence Interval = Sample Proportion ± Margin of Error
Confidence Interval = 0.784 ± 0.0279

Therefore, the confidence interval for the proportion of students supporting the fee increase is approximately [0.7561, 0.8101].

None of the provided answer choices match the calculated confidence interval.