Question

If the statistician is right, what is the probability that the proportion of Rolls Royce owners in a sample of 508 Americans would differ from the population proportion by greater than 3%

Answer 1

The probability that the sample proportion would differ from the population proportion by greater than 3% is 0.008.

Explanation:

The complete question is:

A statistician calculates that 7% of Americans own a Rolls Royce. If the statistician is right, what is the probability that the proportion of Rolls Royce owners in a sample of 508 Americans would differ from the population proportion by greater than 3%? Round your answer to four decimal places.

Solution:

Let the random variable X represent the number of Americans owning a Rolls Royce.

The proportion of Americans who own a Rolls Royce is, p = 0.07.

A random sample of n = 508 Americans were selected.

As the sample size is quite large, i.e. n = 508 > 30, according to the central limit theorem the sampling distribution of sample proportion would follow a normal distribution approximately.

The mean and standard deviation are as follows:

`\mu_(\hat p)=p=0.07`

`\sigma_(\hat p)=\sqrt{(p(1-p))/(n)}=\sqrt{(0.07(1-0.07))/(508)}=0.01132`

Compute the probability that the sample proportion would differ from the population proportion by greater than 3% as follows:

`P(|\hat p-p|>0.03)=P(\hat p<-0.03\ \cup\ \hat p>0.03)`

`=P((\hat p-\mu_(\hat p))/(\sigma_(\hat p))}<(-0.03)/(0.01132)\ \cup\ (\hat p-\mu_(\hat p))/(\sigma_(\hat p))}>(0.03)/(0.01132))\\\\=P(Z<-2.65\ \cup\ Z>2.65)\\\\=P(Z<-2.65)+P(Z>2.65)\\\\=P(Z<-2.65)+1-P(Z<2.65)\\\\=0.00402+1-0.99598\\\\=0.00804\\\\\approx 0.008`

*Use the z-table.

Thus, the probability that the sample proportion would differ from the population proportion by greater than 3% is 0.008.