Question

8.2.19 Question Help 0 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. Click here view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 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 which is not unusual. (Round to four decimal places as needed.) OB. 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.) O C. 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. OD. 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

Answer 1

P=0.10 143 ² = 1300 0-11 Find Probability P(x>0.11) P(Z > P(1-P) n 0.11 -0.10 P(Z > 0.10(1-0.10) 1300 Plz> 1.220 = 0.1151 not option (A) is correct This is necessanly 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 '0.1151 not unsual. cs Scanned with i's which is CamScanner