Question

5.8 Twitter users and news, Part I. A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter.12. 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:   $(0.458176,\ 0.581824)$

We are 99% sure that the true population mean falls in interval $(0.458176,\ 0.581824)$.

Step-by-step explanation:

Let $\hat{p}$ be the sample proportion.

As per given , we have

$\hat{p}=0.52$

Standard error = 0.024

Critical value for 99% confidence interval :$z_{\alpha/2}=2.576$

Confidence interval is given by :-

$\hat{p}\pm z_{\alpha/2}\times(S.E.)$

Then, the 99% confidence interval will be :-

$0.52\pm (2.576)\times(0.024)\\\\=0.52\pm0.061824\\\\=(0.52-0.061824,\ 0.52+0.061824)\\\= (0.458176,\ 0.581824)$

Hence, a 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter : $(0.458176,\ 0.581824)$

Interpretation : We are 99% sure that the true population mean falls in interval $(0.458176,\ 0.581824)$.