In a college student poll, it is of interest to estimate the proportion p of students in favor of changing from a quarter-system to a semester-system. How many students should be polled so that p can be estimated to within 0.09 using a 99% confidence interval?
To estimate the required sample size to estimate the proportion p with a certain degree of accuracy, we can use the formula:
n = (Z^2 * p * (1 - p)) / (E^2)
where:
- n is the required sample size
- Z is the Z-value for the desired confidence level (99% in this case)
- p is the estimated proportion of students in favor of changing systems (we can use 0.5 as a conservative estimate to get the maximum sample size)
- E is the desired margin of error, which is 0.09 in this case
Let's calculate the sample size using the formula:
Z = 2.576 (for 99% confidence level)
p = 0.5 (conservative estimate)
E = 0.09 (margin of error)
n = (2.576^2 * 0.5 * (1 - 0.5)) / (0.09^2)
Simplifying the equation:
n = (6.656 * 0.25) / 0.0081
n = 1664
Therefore, to estimate the proportion p within 0.09 with a confidence level of 99%, you should poll at least 1664 college students.
To determine the sample size required to estimate the proportion with a specific level of confidence and margin of error, we can use the formula:
n = (Z^2 * p * (1-p)) / E^2
Where:
n = sample size
Z = Z-score corresponding to the desired level of confidence (99% confidence corresponds to a Z-score of approximately 2.576)
p = estimated proportion of students in favor of changing systems (if unknown, it is common to use 0.5 as a conservative estimate)
E = margin of error (0.09 in this case)
Substituting the values into the formula:
n = (2.576^2 * 0.5 * (1-0.5)) / 0.09^2
n ≈ (6.656 * 0.25) / 0.0081
n ≈ 166.4 / 0.0081
n ≈ 20543.2
Rounding up to the nearest whole number, we get:
n ≈ 20544
Therefore, approximately 20,544 students should be polled in order to estimate the proportion of students in favor of changing systems to within 0.09 with 99% confidence.