Question

A recent report states that 89% of Americans consider themselves above average drivers. A local newspaper is planning on conducting a survey to investigate whether this is true locally. If the newspaper assumes that the 89% claim is true and plans to use the normal approximation to calculate probabilities associated with their sample proportion, which sample size would be most appropriate?
A) 10

B) 11

C) 30

D) 91

E) The number depends on the size of the population.

Answer 1

Answer:

D) 91

Step-by-step explanation:

For the normal approximation, we need that:

np > 5 and np(1-p) > 5

In this question:

$p = 0.89$

So

A) 10

n = 10, then

10*0.89 = 8.9 > 5

10*0.89*0.11 = 0.979 < 5

Does not satisfy.

B) 11

n = 11, then

11*0.89 = 9.79

11*0.89*0.11 = 1.0769 < 5

Does not satisfy

C) 30

n = 30

30*0.89 = 26.7

30*0.89*0.11 = 2.937 < 5

D) 91

n = 91

90*0.89 = 80.1

90*0.89*0.11 = 8.881 > 5

So this satisfies, and 91 is the answer