Question

College officials want to estimate the percentage of students who carry a gun, knife, or other such weapon. A 9898% 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^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. Your final answer however should be an integer.

Answer:

(b)    How many randomly selected students must be surveyed if we assume that another study indicated that 66% of college students carry weapons. Answer:

A college student surveyed fellow students in order to determine a 90% confidence interval on the proportion of students who plan on voting in the upcoming student election for Student Government Association President. Of those surveyed, it is determined that 82 plan on voting and 169 do not plan on voting.

(a) Find the 90% confidence interval.
Enter the smaller number in the first box.

Confidence interval: ( , ).

Answer 1

Solution:- Given that E = 0.05,

98% Confidence interval for Z = 2.33

(a) for p = 0.5

n = (Z/E)^2*p*q = (2.33/0.05)^2*0.5*0.5 = 542.89 = 543

(b) for p = 0.06,

n = (Z/E)^2*p*q = (2.33/0.05)^2*0.06*0.94 = 122.47 = 123

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Solution:- Given that Total = 82+169 = 251

p = x/n = 82/251 = 0.327

(a) 90% Confidence interval for population proportion = (0.278,0.375)