Question

(6 points) A poll conducted in 2013 found that the proportion of of U.S. adult Twitter users who get at least some news on Twitter was 0.52. The standard error for this estimate was 0.024, and a normal distribution may be used to model the sample proportion. 1. Construct a 99% confidence interval for the proportion of U.S. adult Twitter users who got some news on Twitter. Round to four decimal places.

Answer 1

Answer:

$0.52 - 2.58 *0.024=0.4581$

$0.52 + 2.58 *0.024=0.5819$

The 99% confidence interval would be given by (0.4581;0.5819)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$. And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The standard error for this case is given:

$SE =\sqrt{\frac{\hat p (1-\hat p)}{n}}=0.024$

If we replace the values obtained we got:

$0.52 - 2.58 *0.024=0.4581$

$0.52 + 2.58 *0.024=0.5819$

The 99% confidence interval would be given by (0.4581;0.5819)