Davey wonders what proportion of the students at his school is satisfied with the quality of meals provided by the cafeteria. He interviews an SRS of 50 of the 750 students living in the dormitory. Thirty-eight of them are not satisfied with the quality of meals.

a) Describe the population in this situation and explain in words what the parameter p represents.

b) Give the numerical value of the statistic that estimates p.

c) Find a 95% confidence interval for the true proportion that are satisfied with the quality of meals.

d) What sample size must Davey take to provide a 95% confidence interval with margin of error ±3 percentage points? Comment on this value.

Kasey -- you've posted several multi-part statistics questions. Please post your tentative answers and tell us which parts you don't understand. We'll be glad to help you if we know more precisely which parts you need help with.

I'll give you some hints to get started, then let you take it from there.

This problem would be considered a binomial distribution because there are only two answers: satisfied and not satisfied with the quality of the meals.

The point estimate of the binomial parameter p can be shown proportionally from the data given.

To find a 95% confidence interval for the true proportion who are satisfied:
CI95 = p + or - 1.96 (sqrt of pq/n)
...where sqrt = square root, p = x/n, and q = 1 - p (n = sample size).
Remember that you are looking for the true proportion who ARE satisfied.

To find sample size:
n = [1.96^2 * p * q]/E^2
...where 1.96 represents the 95% confidence interval; ^2 = squared; E = .03 (3 percentage points).

I hope this will help.

THANKS ! ! ! It does.

a) The population in this situation consists of all students in the school. The parameter p represents the proportion of students in the entire school population who are satisfied with the quality of meals provided by the cafeteria.

b) To estimate the parameter p, we can use the sample statistic. In this case, the sample statistic is the proportion of students in the sample who are not satisfied with the quality of meals. So, the numerical value of the statistic that estimates p is 38/50 or 0.76.

c) To find a 95% confidence interval for the true proportion of students who are satisfied with the quality of meals, we can use the formula:
CI95 = p ± 1.96 * √(p * q / n)

First, we need to calculate the value of q, which is 1 - p.
q = 1 - 0.76 = 0.24

Now, we can calculate the confidence interval:
CI95 = 0.76 ± 1.96 * √(0.76 * 0.24 / 50)
CI95 = 0.76 ± 1.96 * √(0.1824 / 50)
CI95 = 0.76 ± 1.96 * √0.003648
CI95 = 0.76 ± 1.96 * 0.0604
CI95 ≈ 0.76 ± 0.1182
CI95 ≈ (0.6418, 0.8782)

Therefore, the 95% confidence interval for the true proportion of students who are satisfied with the quality of meals is approximately (0.6418, 0.8782).

d) To find the sample size needed to provide a 95% confidence interval with a margin of error of ±3 percentage points (0.03), we can use the formula:
n = (1.96^2 * p * q) / E^2

Substituting the values:
n = (1.96^2 * 0.76 * 0.24) / 0.03^2
n = (3.8416 * 0.1824) / 0.0009
n ≈ 7.0039 / 0.0009
n ≈ 7782.1

Therefore, Davey would need to take a sample size of approximately 7783 students to provide a 95% confidence interval with a margin of error of ±3 percentage points.

Note: The sample size calculation is always rounded up to the next whole number to ensure that it is large enough.