According to a study conducted by an organization, the proportion of Americans who were afraid to fly in 2006 was 0.10. A random sample of 1 comma 200 Americans results in 132 indicating that they are afraid to fly. Explain why this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased.

Even is not unusual.

Step-by-step explanation:

Given:

P = 0.10

n = 1200

x = 132

1. Sample of proportion ( ˆp ) = x/n

= 132/1200

= 0.11

2. Z = ˆp - p / √p ( 1 - p )/ n

= 0.11 - 0.10 / √0.10 ( 1 - 0.10 ) / 132

= 0.01 / 0.02611

= 0.3829

3. Probability

P ( ≥ 132 ) = P( Z > 1.50 ) = 1 - P( Z > 1.50 )

= 1 - 0.9332

= 0.0668

The probability is higher than (0.05). Which makes the event unusual, therefore there is no enough evidence which suggest that the number of Americans who are afraid to fly has increased.