You want to estimate the grades of st2%atistics students who gets grades of C or better. How many students should you survey to be 95% confident that the sample percentage is off by no more than 2 percentage points.
To answer this question, we need to use the formula for sample size determination in estimating proportions.
The formula to calculate the necessary sample size is:
n = (Z^2 * p * (1-p)) / (E^2)
Where:
n = sample size needed
Z = Z-score corresponding to the desired confidence level
p = estimated proportion (in decimal form)
E = the desired margin of error (in decimal form)
In this case, we have:
Z = Z-score corresponding to a 95% confidence level. From a standard normal distribution table, a 95% confidence level corresponds to a Z-score of approximately 1.96.
p = We want to estimate the proportion of students who get grades of C or better, so we don't have an estimated proportion yet. However, we can assume a conservative estimate that p = 0.5 (50%).
E = the desired margin of error, which is 2 percentage points. To convert this to decimal form, divide it by 100: E = 0.02.
Plugging these values into the formula, we have:
n = (1.96^2 * 0.5 * (1 - 0.5)) / (0.02^2)
Calculating this formula, we get:
n ≈ (3.8416 * 0.25) / 0.0004
n ≈ 9604
Therefore, you would need to survey at least 9604 statistics students to be 95% confident that the sample percentage is off by no more than 2 percentage points.