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

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.

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 - (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.