Question

A statistician calculates that 9% of Americans are vegetarians. If the statistician is correct, what is the probability that the proportion of vegetarians in a sample of 471 Americans would be greater than 8%

Answer 1

The probability that the proportion of vegetarians in a sample of 471 Americans would be greater than 8% is P=0.776.

Explanation:

We have to calculate a probability that a sample of n=471 has a proportion greater than 8%, given that the population proportion is 9%.

First, we have to calculate the parameters of the sampling distribution of the proportions:

`\hat{p}=p=0.09\\\\\\\sigma_{\hat{p}}=\sqrt{(p(1-p))/(n)}=\sqrt{(0.09\cdot 0.91)/(471)}=0.0132`

Now, we can calculate the probability using the z-score:

`z=\frac{p-\hat{p}}{\sigma_{\hat{p}}}=(0.08-0.09)/(0.0132)=(-0.01)/(0.0132)=-0.7576`

Then, the probability is:

`P(p>0.08)=P(z>-0.7576)=0.776`