\phantom{SE} = 0.0177
\end{array}$$
-# 5.8 Twitter users and news, Part I
+# 5.8 Twitter users & news I
The general formula for a confidence interval is $point~estimate~±~z^*\times~SE$. Where $z^*$ corresponds to the z-score for the desired value of $\alpha$.
Which means that from this data we are 99% confident that between 45.8% and 58.2% U.S. adult Twitter users get some news through the site.
-# 5.10 Twitter users and news, Part II
+# 5.10 Twitter users & news II
(a) False. See the answer to exercise 5.8 above. With $\alpha = 0.01$, we can consult the 99% confidence interval. It includes 50% but also goes lower. A null hypothesis of $p=0.50$ would not be rejected at this level.