1. A state’s department of education reports that 12% of the high school students in that state attend private schools. The state university wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a random sample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attend private schools.
a. The admission officers want to estimate the true percentage of private school applicants to within 4%, with 90% confidence. How many applications should they sample?
b. They actually select a random sample of 450 applications, and find that 46 of those students attend private schools. Create the 90% confidence interval and interpret it in this context.
c. Should the admissions officers conclude that the percentage of private school students in their applicant pool is lower than the statewide enrollment rate of 12%? Explain your answer.
a. The admission officers should sample 864 applications in order to estimate the true percentage of private school applicants to within 4%, with 90% confidence.
b. The 90% confidence interval is (0.039, 0.072). This means that we are 90% confident that the true percentage of private school applicants in the applicant pool is between 3.9% and 7.2%.
c. No, the admissions officers should not conclude that the percentage of private school students in their applicant pool is lower than the statewide enrollment rate of 12%. The confidence interval does not include 12%, but it is also not significantly lower than 12%. Therefore, the admissions officers cannot conclude that the percentage of private school students in their applicant pool is lower than the statewide enrollment rate.
a. To estimate the true percentage of private school applicants with 90% confidence and a margin of error of 4%, we can use the formula for sample size calculation for estimation of a proportion:
n = (Z^2 * p * (1-p)) / (E^2)
Where:
n = sample size
Z = z-score corresponding to the desired confidence level (90% confidence level corresponds to a z-score of approximately 1.645)
p = estimated proportion (we can use the statewide enrollment rate of 12% as an estimate)
E = margin of error (4% in this case)
Substituting the values into the formula, we get:
n = (1.645^2 * 0.12 * (1-0.12)) / (0.04^2)
n ≈ 673.099
So, the admissions officers should sample approximately 673 applications to estimate the true percentage of private school applicants within 4%, with 90% confidence.
b. Given that they selected a random sample of 450 applications and found that 46 of those students attend private schools, we can calculate the confidence interval.
First, we calculate the sample proportion (p-hat):
p-hat = x / n
p-hat = 46 / 450
p-hat ≈ 0.102
Next, we calculate the standard error (SE) using the formula:
SE = sqrt((p-hat * (1-p-hat))/n)
SE = sqrt((0.102 * (1-0.102))/450)
SE ≈ 0.015
Now, we can calculate the margin of error (ME):
ME = Z * SE
ME = 1.645 * 0.015
ME ≈ 0.025
Finally, we can calculate the confidence interval:
Lower limit = p-hat - ME
Lower limit = 0.102 - 0.025
Lower limit ≈ 0.077
Upper limit = p-hat + ME
Upper limit = 0.102 + 0.025
Upper limit ≈ 0.127
The 90% confidence interval is approximately 0.077 to 0.127. This means that we are 90% confident that the true proportion of private school students in the applicant pool falls within this range.
c. Since the statewide enrollment rate of private school students is 12% and the lower limit of the confidence interval is 0.077 (7.7%), we can conclude that the percentage of private school students in the applicant pool is lower than the statewide enrollment rate.
a. To estimate the true percentage of private school applicants within a certain margin of error with a certain level of confidence, you can use the formula for sample size calculation:
n = (Z^2 * p * (1-p)) / E^2
Where:
n = sample size
Z = z-score corresponding to the desired confidence level (in this case, 90% confidence corresponds to Z = 1.645)
p = estimated proportion (use 0.12, the statewide enrollment rate)
E = margin of error (in this case, 4% or 0.04)
By plugging in the values, we get:
n = (1.645^2 * 0.12 * (1-0.12)) / 0.04^2
n = (2.706, 0.12 * 0.88) / 0.0016
n = (2.706, 0.1056) / 0.0016
n = (1,691.25, 66,000)
Therefore, the admissions officers should sample at least 1,691 students to estimate the true percentage of private school applicants within 4% with 90% confidence.
b. The 90% confidence interval can be calculated using the formula:
CI = p̂ ± Z * sqrt((p̂ * (1-p̂)) / n)
Where:
CI = confidence interval
p̂ = sample proportion (46 out of 450, or 0.102)
Z = z-score corresponding to the desired confidence level (here, Z = 1.645)
n = sample size (450)
Plugging in the values, we get:
CI = 0.102 ± 1.645 * sqrt((0.102 * (1-0.102)) / 450)
Calculating the expression inside the square root:
= 0.102 ± 1.645 * sqrt(0.092796 / 450)
= 0.102 ± 1.645 * 0.01346
Calculating the upper and lower bounds of the confidence interval:
CI = (0.0807, 0.1233)
Interpretation: The 90% confidence interval for the percentage of students applying for admission who attend private schools is between 8.07% and 12.33%. We can be 90% confident that the true percentage lies within this interval.
c. To determine whether the percentage of private school students in the applicant pool is lower than the statewide enrollment rate of 12%, we need to check if the lower bound of the confidence interval is less than 12%.
From the previous calculation, the lower bound of the confidence interval is 8.07%. Since 8.07% is less than 12%, the admissions officers can conclude that the percentage of private school students in their applicant pool is lower than the statewide enrollment rate of 12%.