Question

A recent survey showed that among 100 randomly selected college seniors, 20 plan to attend graduate school and 80 do not. Determine a 68 % confidence interval for the population proportion of college seniors who plan to attend graduate school. (Enter each answer rounded to three decimal places.)

Answer 1

Answer:

The 68% confidence interval for the population proportion of college seniors who plan to attend graduate school is (0.16, 0.24).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

A recent survey showed that among 100 randomly selected college seniors, 20 plan to attend graduate school and 80 do not.

This means that $n = 100, \pi = \frac{20}{100} = 0.2$

68% confidence level

So $\alpha = 0.32$, z is the value of Z that has a pvalue of $1 - \frac{0.32}{2} = 0.84$, so $Z = 0.995$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 - 0.995\sqrt{\frac{0.2*0.8}{100}} = 0.16$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 + 0.995\sqrt{\frac{0.2*0.8}{100}} = 0.24$

The 68% confidence interval for the population proportion of college seniors who plan to attend graduate school is (0.16, 0.24).