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.