Question

Binge drinking In a recent National Survey of Drug Use and Health, 2312 of 5914 randomly selected full-time U.S. college students were classified as binge drinkers.13 (a) Calculate and interpret a 99% confidence interval for the population proportion p that are binge drinkers. (b) A newspaper article claims that 45% of full-time U.S. college students are binge drinkers. Use your result from part (a) to comment on this claim.

Answer 1

Answer:

a) (0.374,0.406)

b) There is not enough evidence to support the claim.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 5914

Number of binge drinkers, x = 2312

$\hat{p} = \dfrac{x}{n} = \dfrac{2312}{5914} = 0.39$

a) 99% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.58$

Putting the values, we get:

$0.39\pm 2.58(\sqrt{\frac{0.39(1-0.39)}{5914}}) = 0.39\pm 0.016\\\\=(0.374,0.406)$

b) It is claimed that 45% of students are binge drinkers. But since 0.45 does not lie in the calculated confidence interval, thus, we cannot accept the claim and 45% of full-time U.S. college students are not binge drinkers. There is not enough evidence to support the claim.

Answer 2

Using the z-distribution, we have that:

a) The 99% confidence interval for the population proportion p that are binge drinkers is (0.3785, 0.4033), which means that we are 99% sure that the true proportion of all college students that are binge drinkers in between 0.3785 and 0.4033.

b) The upper bound of the interval is below 45%, hence there is enough evidence to reject the newspaper's claim.

Confidence interval of proportions

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

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

• In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

Item a

• 2312 of 5914 randomly selected full-time U.S. college students were classified as binge drinkers, hence $n = 5914, \pi = \frac{2312}{5914} = 0.3909$.

• 99% confidence level, hence$\alpha = 0.99$, z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$, so $z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3909 - 1.96\sqrt{\frac{0.3909(0.6091)}{5914}} = 0.3785$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3909 + 1.96\sqrt{\frac{0.3909(0.6091)}{5914}} = 0.4033$

The 99% confidence interval for the population proportion p that are binge drinkers is (0.3785, 0.4033), which means that we are 99% sure that the true proportion of all college students that are binge drinkers in between 0.3785 and 0.4033.

Item b:

The upper bound of the interval is below 45%, hence there is enough evidence to reject the newspaper's claim.

You can learn more about the z-distribution at https://brainly.com/question/25730047