Question

A poll found that 58% of U.S. adult Twitter users get at least some news on Twitter. The standard error for this estimate was 2.3%, 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.) We are 99% confident that % to % of U.S. adult Twitter users get some news on Twitter.

Answer 1

Answer: (53.49%, 62.51%)

Step-by-step explanation:

Given : A poll found that 58% of U.S. adult Twitter users get at least some news on Twitter.

i.e. p= 0.58

The standard error for this estimate was 2.3%

i.e.  S.E.=0.023

Critical value for 95% confidence : $z_{\alpha/2}=1.96$

Confidence interval : $p\pm z_{\alpha/2}(S.E.)$

i.e. $0.58\pm (1.96)(0.023)$

$=0.58\pm 0.04508=(0.58-0.04508,\ 0.58+0.04508)\\\\=(0.53492,\ 0.62508)=(53.492\%,\ 62.508\%)=(53.49\%,\ 62.51\%)$

Conclusion : We are 95% confident that the true population proportion of U.S. adult Twitter users who get some news on Twitter falls in interval (53.49%, 62.51%)