Question

College officials want to estimate the percentage of students who carry a gun, knife, or other such weapon. A 90% confidence interval is desired where the interval is no wider than 5 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^. First Hint: note the question is limiting the "width" of the interval to 5 percentage points. Also, use at least five decimals places of accuracy with all values used in calculations. Answer: (b) How many randomly selected students must be surveyed if we assume that another study indicated that 7% of college students carry weapons. Answer:

Answer 1

A confidence interval for p, the population proportion, is given by :

where is the sample proportion;

E is the margin of error and ;

n is the sample size; and

is upper point of the standard normal distribution.

Now, the width of this interval is

Now, we are asked to find out the value of n such that the width fo the interval is no more than 0.05. Thus :

(a)

We have:

but we don't have the value of . In such cases, we replace in the expression for finding the value of n by 0.25, which is the maximum value of .

Thus, to find n, we solve the following expression:

Thus, to obtain a 90% confidence interval no wider than 5 percentage points, atleast 1083 students (found by rounding up 1082.27138) must be surveyed.

(b)

This part is same as part (a) with only difference that we are given an estimate of = 0.07. Thus, we use the exact formula to find the value of n.

Thus, to obtain a 90% confidence interval no wider than 5 percentage points, atleast 282 students (found by rounding up 281.80941) must be surveyed.

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