Question

A poll found that 52% of U.S. adult Twitter users get at least some news on Twitter. The standard error for this estimate was 2.4%, and a normal distribution may be used to model the sample proportion. Construct a 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter, and interpret the confidence interval in context. (Round your answers to two decimal places.)

Answer 1

Answer:

The 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter is (0.46, 0.58). This means that we are 99% sure that the true proportion of all U.S. adult Twitter users who get some news on Twitter is between these two values.

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$.

A poll found that 52% of U.S. adult Twitter users get at least some news on Twitter. The standard error for this estimate was 2.4%

This means that:

$\pi = 0.52, \sqrt{\frac{\pi(1-\pi)}{n}} = 0.024$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 - 2.575(0.024) = 0.46$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 + 2.575(0.024) = 0.58$

The 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter is (0.46, 0.58). This means that we are 99% sure that the true proportion of all U.S. adult Twitter users who get some news on Twitter is between these two values.