Question

I want to take a survey of students at my university to find out what proportion like the new bus service on campus. how many will i need to survey if i want to estimate with 99% confidence the true proportion to within 2% if i believe that 75% of students like the bus service?

Answer 1

Answer:

N = 3,120 (Approx)

Explanation:

Given:

z- score for 99% confidence = 2.58

Proportion (P) = 75% = 75 / 100 = 0.75

E = 2% = 0.02

Q = 1 - P = 1 - 0.75 = 0.25

Computation:

$N = (Z/E)^2 \times pq$

$N = (2.58/0.02)^2 \times (0.75 \ times 0.25)$

$N = (129)^2 \times (0.1875)$

$N = (16,641) \times (0.1875)$

$N = 3,120.1875$

Therefore, N = 3,120 (Approx).