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 9898% confident that the sample percentage has a margin of error of 1.51.5 percentage points? (a) assume that there is no available information that could be used as an estimate of p^p^.

When determining the confidence interval, the margin of error is crucial.
The critical z value,the sample size, and proportion can all be used to determine the margin of error in the case of proportion.

The following can be used to represent the formula for calculating the margin of error for a single proportion:

ME = z√p(1 - p)/n

In order to be 91% certain that the sampling percentage seems to have an error margin of 1.5 percentage points, the number of randomly chosen students that must be surveyed is equal to;

0.015 = 1.6954×√0.5(1 - 0.5)/n

0.015/1.6954 = √0.5(0.5)/n

On further solving,

n = 3205.128

n ≈ 3206

As a result, 3206 students are randomly chosen since there is no knowledge of the sample fraction.

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The correct question is-

College officials want to estimate the percentage of students who carry a gun, knife, or other such weapons. How many randomly selected students must be surveyed in order to be 91% confident that the sample percentage has a margin of error of 1.5 percentage points?

a) Assume that there is no available information that could be used as an estimate of p.