College officials want to estimate the percentage of students who carry a gun, knife, or other such weapon. How many randomly selected students must be surveyed in order to be 94% confident that the sample percentage has a margin of error of 2 percentage points?
To determine the sample size needed for this survey, we can use the formula:
n = (Z^2 * p * (1-p)) / E^2
Where:
n = required sample size
Z = Z-score for the desired level of confidence (94% confidence corresponds to a Z-score of approximately 1.88)
p = estimated sample proportion (since we don't have an estimate, we can use 0.5 as a conservative estimate)
E = margin of error (2 percentage points)
Plugging in the values, the formula becomes:
n = (1.88^2 * 0.5 * (1-0.5)) / 0.02^2
n = (3.5344 * 0.25) / 0.0004
n = 0.8836 / 0.0004
n ≈ 2209
Therefore, in order to be 94% confident that the sample percentage has a margin of error of 2 percentage points, approximately 2209 randomly selected students must be surveyed.
To determine the sample size needed to estimate the percentage of students who carry a weapon with a certain level of confidence and margin of error, we can use the formula:
n = (Z^2 * p * q) / E^2
Where:
n = sample size
Z = Z-score corresponding to the desired confidence level
p = estimated proportion (0.5 is a commonly used conservative value)
q = 1 - p
E = margin of error
In this case, we want to be 94% confident with a margin of error of 2 percentage points. Considering that we don't have an actual estimate for the proportion of students carrying a weapon, we can use a conservative estimate of p = 0.5 since it yields the maximum sample size. Therefore, q = 1 - p = 1 - 0.5 = 0.5.
Now let's plug in the values into the formula:
n = (Z^2 * p * q) / E^2
Z for 94% confidence level is approximately 1.88 (you can find this value using a standard normal distribution table or a statistical calculator).
n = (1.88^2 * 0.5 * 0.5) / (0.02^2)
n = (3.5344 * 0.25) / 0.0004
n = 0.8836 / 0.0004
n ≈ 2209
Rounded up, you will need to survey at least 2209 randomly selected students to be 94% confident that the sample percentage has a margin of error of 2 percentage points.