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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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8
os_exercises/ch5_exercises_solutions.html
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@@ -1569,15 +1569,15 @@ SE = \sqrt{\frac{p(1-p)}{n}}\\
\end{array}\]</span></p></li>
</ol>
</div>
<div id="twitter-users-and-news-part-i" class="section level1">
<h1>5.8 Twitter users and news, Part I</h1>
<div id="twitter-users-news-i" class="section level1">
<h1>5.8 Twitter users &amp; news I</h1>
<p>The general formula for a confidence interval is <span class="math inline">\(point~estimate~±~z^*\times~SE\)</span>. Where <span class="math inline">\(z^*\)</span> corresponds to the z-score for the desired value of <span class="math inline">\(\alpha\)</span>.</p>
<p>To estimate the interval from the data described in the question, identify the three different values. The point estimate is 45%, <span class="math inline">\(z^* = 2.58\)</span> for a 99% confidence level (that’s the number of standard deviations around the mean that ensure that 99% of a Z-score distribution is included), and <span class="math inline">\(SE = 2.4\%\)</span>. With this we can plug and chug:</p>
<p><span class="math display">\[52\% ± 2.58 \times 2.4\%\]</span> And that yields: <span class="math display">\[95\% CI = (45.8\%, 58.2\%)\]</span></p>
<p>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.</p>
</div>
<div id="twitter-users-and-news-part-ii" class="section level1">
<h1>5.10 Twitter users and news, Part II</h1>
<div id="twitter-users-news-ii" class="section level1">
<h1>5.10 Twitter users &amp; news II</h1>
<ol style="list-style-type: lower-alpha">
<li><p>False. See the answer to exercise 5.8 above. With <span class="math inline">\(\alpha = 0.01\)</span>, we can consult the 99% confidence interval. It includes 50% but also goes lower. A null hypothesis of <span class="math inline">\(p=0.50\)</span> would not be rejected at this level.</p></li>
<li><p>False. The standard error of the sample proportion does not contain any information about the proportion of the population included in the sample. It estimates the variability of the sample proportion.</p></li>

4
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.