Question

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 1,300 Americans results in 143 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. Select the correct choice below and. if necessary, fill in the answer box to complete your choice. This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the probability of obtaining a value equal to or more extreme than the sample proportion is . which is not unusual.

Answer 1

Answer:

The correct option is A) This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the probability of obtaining a value equal to or more extreme than the sample proportion is . which is not unusual.

Step-by-step explanation:

Consider the provided information.

The formula for testing a proportion is based on the z statistic.

$z=\frac{\hat p-p_0}{\sqrt{p_0\frac{1-p_0}{n}}}$

Were $\hat p$ is sample proportion.

$p_0$ hypothesized proportion and n is the sample space,

The proportion of Americans who were afraid to fly in 2006 was 0.10. A random sample of 1,300 Americans results in 143 indicating that they are afraid to fly.

Therefore, n = 100  $\hat p = \frac{143}{1300}=0.11$ , $p_0 = 0.10$ and $1 - p_0 = 1 - 0.10 = 0.90$

Substitute the respective values as shown:

$z=\frac{0.11-0.10}{\sqrt{\frac{0.10\times 0.90}{1300}}}$

$z=1.201850$

P(x>sample proportion)=p(z>1.20)=0.11507

Hence, the correct option is A) This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the probability of obtaining a value equal to or more extreme than the sample proportion is . which is not unusual.