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 1,400 Americans results in 154 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. , which O 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 is not unusual. (Round to four decimal places as needed.) O B. This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the sample size n is more than 5% of the population. OC. This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the sample proportion. is very close to 0.10. (Type an integer or a decimal.) OD. This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the value of np(1-P) is less than 10. Click to select and enter your answer(s).

Answer 1

Answer)

Null hypothesis Ho : P = 0.1

Alternate hypothesis Ha : P > 0.1

N = 1400

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 140

N*(1-p) = 1260

Both the conditions are met so we can use standard normal z table to estimate the probability

z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Claimed P = 0.1

Observed P = 154/1400

Z = 1.25

From z table, P(z>1.25) = 0.1056

As we know that an event is unusal when probability is less than 0.05

As here 0.1056 is > 0.05

It is not unusual

So answer here is

Option A is correct

Answer is 0.1056