Question

4.8 Twitter users and news, Part I: 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 2.6%, 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: Required 99% confidence interval  : (48.30%, 61.7%)

Step-by-step explanation:

Confidence interval for PROPORTION:  $p\pm z^* \times S.E.$, where p = sample proportion , z* = two-tailed critical z-value  and S.E. = standard error.

As per given: p=55%

Critical z-value for 99% confidence = 2.576

S.E. = 2.6%

Then, required confidence interval :

$55\% \pm (2.576)\times 2.6\%\\\\\approx55\% \pm6.70\%\\\\=(55\%-6.70\%,\ 55\%+6.70\%)\\\\=(48.30\%,\ 61.7\%)$

Required 99% confidence interval  : (48.30%, 61.7%)