Question

A poll conducted in 2013 found that 55% of U.S. adult Twitter users get at least some news on Twitter (Pew, 2013). The standard error for this estimate was 2.2%, 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. (please round all percentages to 2 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 between 49.34% and 60.66%, which means that we are 99% sure that the true percentage of US adult Twitter users who get some news is in this interval.

Step-by-step explanation:

Confidence interval:

A confidence interval has the following format:

$M \pm zs$

In which M is the sample mean, z is related to the confidence level and s is the standard error.

55% of U.S. adult Twitter users get at least some news on Twitter (Pew, 2013). The standard error for this estimate was 2.2%.

This means that $M = 55, s = 2.2$

99% confidence level

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

Lower bound of the interval:

$M - zs = 55 - 2.575*2.2 = 49.34$

Upper bound:

$M + zs = 55 + 2.575*2.2 = 60.66$

The 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter is between 49.34% and 60.66%, which means that we are 99% sure that the true percentage of US adult Twitter users who get some news is in this interval.