Question

From driverless cars to a workplace staffed by robots, automation has the potential to reshape many facets of American life. The large majority of Americans (87%) would favor a requirement that all driverless vehicles have a human in the driver's seat who can take control of the vehicle in the event of an emergency, while 56% of U.S. adults say that they would not ride in a driverless vehicle.† If these figures are correct, what is the probability that in a sample of n = 100 U.S. adults, the sample proportion p̂ of adults who would not ride in a driverless vehicle falls between 54% and 64%? (Round your answer to four decimal places.)

Answer 1

Solution:

Population proportion of US adults who would not ride a driverless vehicle is, P = 56% = 0.56

We want to obtain the probability that sample proportion (p̂) of adults who would not ride in a driverless vehicle in a sample of 100 US adults falls between 54% and 64%.

i.e. We have to obtain Pr(0.54 < p̂ < 0.64).

If nP ≥ 10 and nQ ≥ 10 then sampling distribution of sample proportion follows approximately normal distribution with mean P and variance (PQ/n).

i.e. p̂ ~ N(P, PQ/n)

(Where, P is population proportion, Q = 1 - P and n is sample size.

We have, P = 0.56 , Q = 1 - 0.56 = 0.44 and n = 100

nP = 100×0.56 = 56 which is greater than 10.

nQ = 100×0.44 = 44 which is greater than 10.

Hence, we can consider sample proportion p̂ as approximately normally distributed random variable with mean P and variance PQ/n.

We have to obtain Pr(0.54 < p̂ < 0.64).

We know that if p̂ ~ N(P, PQ/n) then,

We have, P = 0.56 , Q = 1 - 0.56 = 0.44 and n = 100

Pr(0.54 < p̂ < 0.64) = Pr(p̂ < 0.64) - Pr(p̂ < 0.54)

Using "pnorm" function of R we get,

Pr(Z < 1.6116) = 0.9465 and Pr(Z < -0.4029) = 0.3435

Hence, the probability that in a sample of n = 100 U.S. adults, the sample proportion p̂ of adults who would not ride in a driverless vehicle falls between 54% and 64% is 0.6030.

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