Question

A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter.40. The standard error for this estimate was 2.4%, 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.

Answer 1

Answer:  (0.458,0.582)

Step-by-step explanation:

Given : Sample size : $n=815$

The proportion of U.S. adult Twitter users get at least some news on Twitter. : $p=0.52$

Standard error : $s.=0.024$

Significance level : $\alpha: 1-0.99=0.01$

Critical value : $z_{\alpha/2}=2.576$

The confidence interval for population proportion is given by :-

$p\pm\ z_{\alpha/2}\ s\\\\=0.52\pm(2.576)(0.024)\\\\\approx0.52\pm0.062=(0.52-0.062,0.52+0.062)=(0.458,0.582)$

Hence, the 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= (0.458,0.582)