Question

Suppose the true proportion of students at a college who study abroad is 0.25. I select a random sample of 40 students from the college and record if they have studied abroad. What is the probability that the proportion of students in my sample who have studied abroad is less than 0.2

Answer 1

Answer:

$\mu_{\hat p} = 0.25$

$\sigma_{\hat p}= \sqrt{\frac{0.25*(1-0.25)}{40}}= 0.0685$

$z = \frac{0.2-0.25}{0.0685}= -0.730$

And we can use the normal standard distribution or excel to find this probability and we got:

$P(\hat p$

Step-by-step explanation:

We define the parameter as the proportion of students at a college who study abroad and this value is known $p =0.25$, we select a sample size of n =40 and we are interested in the probability associated to the sample proportion, but we know that the distirbution for the sample proportion is given by:

$\hat p \sim N(p , \sqrt{\frac{p(1-p)}{n}}$

And the paramters for this case are:

$\mu_{\hat p} = 0.25$

$\sigma_{\hat p}= \sqrt{\frac{0.25*(1-0.25)}{40}}= 0.0685$

We want to find the following probability:

$P(\hat p< 0.2)$

For this case since we know the distribution for the sample proportion we can use the z score formula given by:

$z = \frac{\hat p -\mu_{\hat p}}{\sigma_{\hat p}}$

Replacing the info given we got:

$z = \frac{0.2-0.25}{0.0685}= -0.730$

And we can use the normal standard distribution or excel to find this probability and we got:

$P(\hat p$