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.

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.)

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.

C.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.)

D.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(1minus−p) is less than 10.

Answer #1

n= | 1300 | p= | 0.1000 |

here mean of distribution=μ=np= | 130.00 | |

and standard deviation σ=sqrt(np(1-p))= | 10.82 | |

for normal distribution z score =(X-μ)/σx |

therefore from normal approximation of binomial distribution and continuity correction: |

probability =P(X>142.5)=P(Z>(142.5-130)/10.817)=P(Z>1.16)=1-P(Z<1.16)=1-0.877=0.1230 |

option A is correct

his 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
**0.1230** which is not unusual.

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