updating explanation of part I to address reasons for dropped observations

[stats_class_2020.git] / os_exercises / ch5_exercises_solutions.rmd

diff --git a/os_exercises/ch5_exercises_solutions.rmd b/os_exercises/ch5_exercises_solutions.rmd
index 5965223686a3a28694f80bcd30cae31921ebb949..334cb9a4cdf8a148e75e817994e46d17f220d72a 100644 (file)
--- a/os_exercises/ch5_exercises_solutions.rmd
+++ b/os_exercises/ch5_exercises_solutions.rmd
@@ -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.