College officials want to estimate the percentage of students who carry a gun, k

Question

College officials want to estimate the percentage of students who carry a gun, knife, or other such weapon. A 96% confidence interval is desired where the interval is no wider than 7 percentage points? (a) How many randomly selected students must be surveyed if we assume that there is no available information that could be used as an estimate of p. Answer: (b) How many randomly selected students must be surveyed if we assume that another study indicated that 4% of college students carry weapons. Answer:

Explanation / Answer

Answer to part a)

length of confidence interval is 0.07 (7 %)

This implies margin of error (E) = 0.035 .[error = 0.5 * Length of interval]

For 96% confidence level , z = 2.05

Since there is no other information p = 0.5

.

Sample size n = p * (1-p) * (z / E)^2

.

On plugging the values we get

n = 0.5 *0.5 * (2.05 / 0.035)^2

n = 857.653 ~ 858

.

Answer to part b)

if p = 0.04

We use the same formula

n = 0.04 * 0.96 * (2.05 /0.035)^2

n = 131.736 ~ 132

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