Question

A consumer group wishes to estimate the proportion of adult Twitter users that get at least some news on Twitter. A poll is conducted and it is found that from a SRS of 1251 adult Twitter users, 641 of them get at least some news on Twitter.

2b) What is the 90% confidence interval for the proportion of adult Twitter users that get at least some news on Twitter. Round to 4 decimal places.

Answer 1

Applying the steps to find the confidence interval of a proportion, it is found that the 90% confidence interval for the proportion of adult Twitter users that get at least some news on Twitter is (0.4892, 0.5356).

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 $\frac{1+\alpha}{2}$.

From a SRS of 1251 adult Twitter users, 641 of them get at least some news on Twitter, hence:

$n = 1251, \pi = \frac{641}{1251} = 0.5124$

90% confidence level

So $\alpha = 0.9$, z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$, so $z = 1.645$.  

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5124 - 1.645\sqrt{\frac{0.5124(0.4876)}{1251}} = 0.4892$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5124 + 1.645\sqrt{\frac{0.5124(0.4876)}{1251}} = 0.5356$

The 90% confidence interval for the proportion of adult Twitter users that get at least some news on Twitter is (0.4892, 0.5356).

A similar problem is given at https://brainly.com/question/16807970