Question

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 98% confident that the sample percentage has a margin of error of 2 percentage points?

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

n=

(b)    Assume that another study indicated that 7%7% of college students carry weapons.

n=

Answer 1

Solution,

Given that,

a) =  1 - = 0.5

margin of error = E = 0.02

At 98% confidence level

= 1 - 98%  

= 1 - 0.98 =0.02

/2 = 0.01

Z/2 = Z0.01  = 2.326

sample size = n = (Z / 2 / E )2 * * (1 - )

= (2.326 / 0.02)2 * 0.5 * 0.5

= 3381.42

sample size = n = 3382

b) = 0.07

1 - = 1 - 0.07 = 0.93

sample size = n = (Z / 2 / E )2 * * (1 - )

= (2.326 / 0.02)2 * 0.07 * 0.93

= 880.52

sample size = n = 881