According to a study conducted by an organization, the proportion of Americans w
ID: 3217868 • Letter: A
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 increase. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. 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. B. This is not necessarily evidence 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. c. This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because he 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.) D. This is not 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.)Explanation / Answer
Here p0 = 0.10 and ps = 143/1300 = 0.11
Test statistic => Z = (ps- p0) / sqrt [p0 ( 1- p0)/n]
= ( 0.11 - 0.10) / sqrt [ 0.1 * 0.9 /1300]
= 0.01 / 0.00832 = 1.20
and P value = P ( p0 > 0.11 if p0 = 0.10 ) = 1 - 0.8849 = 0.1151 [ Here for z = 1.20 , p -value = 0.8849 ]
so p- value is 0.1151 which is less than significance level alpha = 0.05 or 0.1
Here we will look into options
(a) this optionis wrong because value of np(1-p) = 117 and not less than 10
(b) THis is some absurd option because we dont know about population size
(c) here this option seems correct , because the probability of obtaining a value equal to or more exteme than the sample proportion is 0.1151, which is not unusual.
(d) This options may be correct because the sample proportion , 0.11 is very close to 0.10 but that seems not a practical approach. It is just a hit and error approach.
so option (c) is correct.
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