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typos for ch5 and more elaborate tidyverse example code w more verbose

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aaronshaw
2020-10-16 16:32:18 -05:00
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os_exercises/ch5_exercises_solutions.rmd
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@@ -57,7 +57,7 @@ SE = \sqrt{\frac{p(1-p)}{n}}\\
\phantom{SE} = 0.0177
\end{array}$$
# 5.8 Twitter users and news, Part I
# 5.8 Twitter users & news I
The general formula for a confidence interval is $point~estimate~±~z^*\times~SE$. Where $z^*$ corresponds to the z-score for the desired value of $\alpha$.
@@ -70,7 +70,7 @@ $$95\% CI = (45.8\%, 58.2\%)$$
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.