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

Answer:

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

Step-by-step 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:

$\mu_{\hat p}=p=0.07$

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

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

$P(|\hat p-p|>0.03)=P(\hat p0.03)$

                           $=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}}\frac{0.03}{0.01132})\\\\=P(Z2.65)\\\\=P(Z2.65)\\\\=P(Z$

*Use the z-table.

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