Fix typos, batch 2022-04-12

ch_distributions/TeX/ch_distributions.tex

@@ -618,7 +618,7 @@ \subsection{Normal probability examples}
 \end{nexercise}
 \end{exercisewrap}
 \footnotetext{Short answers:
-  (a) $Z_{95} = 1.65 \to 1430$ SAT score.
+  (a) $Z_{95} = 1.6449 \to 1429$ SAT score.
   (b) $Z_{97.5} = 1.96 \to 1492$ SAT score.}
 
 \D{\newpage}
@@ -1099,9 +1099,9 @@ \subsection{Geometric distribution}
   The probability the second person is the first to hit
   her deductible is
   \begin{align*}
-  &P(\text{second person is the first to hit deductible}) \\
+  &P(\text{second person is the first to not hit deductible}) \\
   &\quad
-    = P(\text{the first won't, the second will})
+    = P(\text{the first will, the second won't})
     = (\insureFprob{})(\insureSprob{})
     = \insureDistB{}
   \end{align*}
@@ -1496,7 +1496,7 @@ \subsection{The binomial distribution}
 
   In the outcome of interest, there are $k = 5$ successes
   in $n = 8$ trials (recall that a success is an individual
-  who does \emph{not} exceed the deductible, and the
+  who does \emph{not} exceed the deductible), and the
   probability of a success is $p = \insureSprob{}$.
   So the probability that 5 of 8 will not exceed the
   deductible and 3 will exceed the deductible is given by

ch_distributions/TeX/normal_distribution.tex

@@ -129,7 +129,7 @@
 \begin{parts}
 \item What is the probability of observing an 83\degree F temperature or 
 higher in LA during a randomly chosen day in June?
-\item How cool are the coldest 10\% of the days (days with lowest average 
+\item How cool are the coldest 10\% of the days (days with lowest 
 high temperature) during June in LA?
 \end{parts}
 }{}

ch_foundations_for_inf/TeX/ch_foundations_for_inf.tex

@@ -559,7 +559,7 @@ \subsection{Central Limit Theorem}
 methods.\footnote{For example, we could use what's called the
   \term{finite population correction factor}:
   if the sample is of size $n$ and the population size is $N$,
-  then we can multiple the typical standard error formula by
+  then we can multiply the typical standard error formula by
   $\sqrt{\frac{N-n}{N-1}}$
   to obtain a smaller, more precise estimate of the
   actual standard error.
@@ -865,7 +865,7 @@ \subsection{More details regarding the Central Limit Theorem}
 
 At no point will the distribution of $\hat{p}$ look
 \emph{perfectly} normal, since $\hat{p}$ will always
-be take discrete values ($x / n$).
+take discrete values ($x / n$).
 It is always a matter of degree, and we will use
 the standard success-failure condition with minimums
 of 10 for $np$ and $n (1 - p)$ as our guideline
@@ -886,11 +886,11 @@ \subsection{Extending the framework for other statistics}
 difference in product prices for two websites,
 we might take a random sample of products available
 on both sites, check the prices on each,
-and use then compute the average difference;
+and then compute the average difference;
 this strategy certainly would give us some idea
 of the actual difference through a point estimate.
 
-While this chapter emphases a single proportion
+While this chapter emphasizes a single proportion
 context, we'll encounter many different contexts
 throughout this book where these methods will be
 applied.
@@ -1195,11 +1195,11 @@ \subsection{Changing the confidence level}
   $z^{\star}=1.65$.
   The 90\% confidence interval can then be computed as
   \begin{align*}
-  \hat{p}\ \pm\ 1.65 \times SE_{\hat{p}}
+  \hat{p}\ \pm\ 1.6449 \times SE_{\hat{p}}
       \quad\to\quad 0.887\ \pm\ 1.65 \times 0.0100
-      \quad\to\quad (0.8705, 0.9035)
+      \quad\to\quad (0.8705, 0.9034)
   \end{align*}
-  That is, we are 90\% confident that 87.1\% to 90.4\% of American
+  That is, we are 90\% confident that 87.1\% to 90.3\% of American
   adults supported the expansion of solar power in 2018.
 \end{nexample}
 \end{examplewrap}

ch_foundations_for_inf/TeX/hypothesis_testing.tex

@@ -156,7 +156,7 @@
 Do a majority of US adults believe raising
 the minimum wage will help the economy,
 or is there a majority who do not believe this?
-A~Rasmussen Reports survey of 1,000 US adults found
+A~Rasmussen Reports survey of a random sample of 1,000 US adults found
 that 42\% believe it will help the
 economy.\footfullcite{webpage:rasmussen-2019-raise-minimum-wage}
 Conduct an appropriate hypothesis test to help

ch_foundations_for_inf/TeX/review_exercises.tex

@@ -165,7 +165,8 @@
 \item
     Men were, on average, paid more in 19 of those
     21 positions.
-    Complete a hypothesis test using your hypotheses
+    Supposing these 21 positions represent a simple random sample,
+    complete a hypothesis test using your hypotheses
     from part~(\ref{gender_pay_gap_medicine_hypotheses}).
 \end{parts}
 }{}

ch_inference_for_means/TeX/ch_inference_for_means.tex

@@ -2403,7 +2403,7 @@ \subsection{Determining a proper sample size}
 We should solve the problem backwards.
 
 \begin{examplewrap}
-\begin{nexample}{What sample size will lead to a power of 80\%?}
+\begin{nexample}{What sample size will lead to a power of 80\%? Use $\alpha = 0.05$.}
   \label{sample_size_for_80_percent_power}%  This is referenced in EOCE.
   We'll assume we have a large enough sample that the normal
   distribution is a good approximation for the test statistic,

ch_inference_for_means/TeX/power_calculations_for_a_difference_of_means.tex

@@ -8,7 +8,9 @@
 average of 1,215 pounds of corn with a standard deviation of 94 pounds per 
 plot. The owner is interested in detecting any average difference of at 
 least 40 pounds per plot. How many plots of land would be needed for the 
-experiment if the desired power level is 90\%? Assume each plot of land gets 
+experiment if the desired power level is 90\%?
+Use $\alpha = 0.05$.
+Assume each plot of land gets 
 treated with either the current fertilizer or the new fertilizer.
 }{}
 
@@ -26,4 +28,5 @@
 interface or the current interface. How many new enrollees do they need for 
 each interface to detect an effect size of 0.5 surveys per enrollee, if the 
 desired power level is 80\%?
+Use $\alpha = 0.05$.
 }{}

ch_inference_for_means/figures/textbooksF18/textbooksF18HTTails.R

@@ -20,6 +20,6 @@ normTail(L = -abs(m),
          # border = COL[4],
          axes = FALSE)
 at <- c(-100, 0, m, 100)
-labels <- expression(0, mu[0]*' = 0', bar(x)[diff]*" = 2.98", 0)
+labels <- expression(0, mu[0]*' = 0', bar(x)[diff]*" = 3.98", 0)
 axis(1, at, labels, cex.axis = 0.9)
 dev.off()

ch_inference_for_props/TeX/ch_inference_for_props.tex

@@ -463,25 +463,25 @@ \subsection{Choosing a sample size when estimating a proportion}
 and identify three sample sizes to consider.\footnotemark
 \end{nexercise}
 \end{exercisewrap}
-\footnotetext{For a 90\% confidence interval, $z^{\star} = 1.65$,
+\footnotetext{For a 90\% confidence interval, $z^{\star} = 1.6449$,
   and since an estimate of the proportion 0.017 is available,
   we'll use it in the margin of error formula:
   \begin{align*}
-  1.65\times \sqrt{\frac{0.017(1-0.017)}{n}} &\ < \ 0.01
+  1.6449\times \sqrt{\frac{0.017(1-0.017)}{n}} &\ < \ 0.01
     \qquad\to\qquad
       \frac{0.017(1-0.017)}{n} \ < \ 
-          \left(\frac{0.01}{1.65}\right)^2
+          \left(\frac{0.01}{1.6449}\right)^2
     \qquad\to\qquad
-      454.96 \ < \ n
+      452.15 \ < \ n
   \end{align*}
   For sample size calculations, we always round up,
-  so the first tire model suggests 455 tires would
+  so the first tire model suggests 453 tires would
   be sufficient.
 
   A similar computation can be accomplished using 0.062
   and 0.013 for $p$, and you should verify that using these
-  proportions results in minimum sample sizes of 1584
-  and~350 tires, respectively.}
+  proportions results in minimum sample sizes of 1574
+  and~348 tires, respectively.}
 
 \begin{examplewrap}
 \begin{nexample}{The sample sizes vary widely in
@@ -756,14 +756,14 @@ \subsection{Sampling distribution of the difference
       \frac{0.22 (1 - 0.22)}{50}}
     = 0.095
   \end{align*}
-  For a 90\% confidence interval, we use $z^{\star} = 1.65$:
+  For a 90\% confidence interval, we use $z^{\star} = 1.6449$:
   \begin{align*}
   \text{point estimate} \ \pm\ z^{\star} \times SE
-    \quad \to \quad 0.13 \ \pm\ 1.65 \times  0.095
-    \quad \to \quad (-0.027, 0.287)
+    \quad \to \quad 0.13 \ \pm\ 1.6449 \times  0.095
+    \quad \to \quad (-0.026, 0.286)
   \end{align*}
   We are 90\% confident that blood thinners have
-  a difference of -2.7\% to +28.7\% percentage point
+  a difference of -2.6\% to +28.6\% percentage point
   impact on survival rate for patients who are like
   those in the study.
   Because 0\% is contained in the interval,
@@ -1063,7 +1063,7 @@ \subsection{Sampling distribution of the difference
 \begin{itemize}
 \setlength{\itemsep}{0mm}
 \item
-    We do not accept the null hypothesis, which means
+    We do not reject the null hypothesis, which means
     we don't have sufficient evidence to conclude that
     mammograms reduce or increase breast cancer deaths.
 \item

ch_inference_for_props/TeX/difference_of_two_proportions.tex

@@ -64,7 +64,8 @@
 the proportions of males and females whose favorite color is black 
 $(p_{male} - p_{female})$ was calculated to be (0.02, 0.06).
 Based on this 
-information, determine if the following statements are true or false, and 
+information, determine if the following statements about
+undergraduate college students are true or false, and 
 explain your reasoning for each statement you identify as false.
 \footfullcite{Ellis:2001}
 \begin{parts}

ch_intro_to_data/TeX/ch_intro_to_data.tex

@@ -226,7 +226,7 @@ \subsection{Observations, variables, and data matrices}
     in an annual percentage.  \\
 \var{term} & The length of the loan, which is always set
     as a whole number of months. \\
-\var{grade} & Loan grade, which takes a values A through G
+\var{grade} & Loan grade, which takes values A through G
     and represents the quality of the loan and its likelihood
     of being repaid.  \\
 \var{state} & US state where the borrower resides. \\

ch_probability/TeX/conditional_probability.tex

@@ -147,8 +147,7 @@
 phenotypes mate with one another more frequently than what would be expected 
 under a random mating pattern. Researchers studying this topic collected data on 
 eye colors of 204 Scandinavian men and their female partners. The table below 
-summarizes the results. For simplicity, we only include heterosexual 
-relationships in this exercise. \footfullcite{Laeng:2007}
+summarizes the results.\footfullcite{Laeng:2007}
 \begin{center}
 \begin{tabular}{ll  ccc c} 
                                         &           & \multicolumn{3}{c}{\textit{Partner (female)}} \\

ch_regr_mult_and_log/TeX/ch_regr_mult_and_log.tex

@@ -434,7 +434,7 @@ \subsection{Including and assessing many variables in a model}
 such models are a common first step in gaining insights
 or providing some evidence of a causal connection.
 
-We want to construct a model that accounts for not only
+We want to construct a model that accounts not only
 for any past bankruptcy or whether the borrower had
 their income source or amount verified,
 but simultaneously accounts for all the variables

ch_summarizing_data/TeX/ch_summarizing_data.tex

@@ -2157,7 +2157,7 @@ \subsection{Variability within data}
 In this study, volunteer patients were randomized
 into one of two experiment groups:
 14 patients received an experimental vaccine
-or 6 patients received a placebo vaccine.
+and 6 patients received a placebo vaccine.
 Nineteen weeks later, all 20 patients were exposed
 to a drug-sensitive malaria virus strain;
 the motivation of using a drug-sensitive strain

extraTeX/eoceSolutions/eoceSolutions.tex

@@ -690,7 +690,7 @@ \chapter{Exercise solutions}
 simplified.
 (b)~No. There are six events under consideration. The Bernoulli distribution 
 allows for only two events or categories. Note that rolling a die could be a 
-Bernoulli trial if we simply to two events, e.g. rolling a 6 and not rolling 
+Bernoulli trial if we simplify to two events, e.g. rolling a 6 and not rolling 
 a 6, though specifying such details would be necessary.}
 
 % 13
@@ -749,9 +749,9 @@ \chapter{Exercise solutions}
 
 % 27
 
-\eocesol{(a)~0.0804.
-(b)~0.0322.
-(c)~0.0193.}
+\eocesol{(a)~Geometric, 0.0804.
+(b)~Binomial, 0.0322.
+(c)~Negative binomial, 0.0193.}
 
 % 29
 
@@ -994,11 +994,11 @@ \chapter{Exercise solutions}
       = 0.010
     \end{align*}}%
 (c)~For a 90\% confidence interval,
-    use $z^{\star} = 1.65$.
+    use $z^{\star} = 1.6449$.
     The confidence interval is
-    $0.085 \pm 1.65 \times 0.010 \to
-    (0.0685, 0.1015)$.
-    We are 90\% confident that 6.85\% to 10.15\%
+    $0.085 \pm 1.6449 \times 0.010 \to
+    (0.0683, 0.1017)$.
+    We are 90\% confident that 6.83\% to 10.17\%
     of first-time site visitors will register using
     the new design.}
 
@@ -1346,17 +1346,17 @@ \chapter{Exercise solutions}
 
 % 15
 
-\eocesol{Since a sample proportion ($\hat{p} = 0.55$) is available,
+\eocesol{Because a sample proportion ($\hat{p} = 0.55$) is available,
 we use this for the sample size calculations.
 The margin of error for a 90\% confidence interval is
-$1.65 \times SE = 1.65 \times \sqrt{\frac{p(1 - p)}{n}}$.
+$1.6449 \times SE = 1.6449 \times \sqrt{\frac{p(1 - p)}{n}}$.
 We want this to be less than 0.01, where we use
 $\hat{p}$ in place of $p$:
 \begin{align*}
-1.65 \times \sqrt{\frac{0.55(1 - 0.55)}{n}} \leq 0.01 \\
-1.65^2 \frac{0.55(1 - 0.55)}{0.01^2} \leq n
+1.6449 \times \sqrt{\frac{0.55(1 - 0.55)}{n}} \leq 0.01 \\
+1.6449^2 \frac{0.55(1 - 0.55)}{0.01^2} \leq n
 \end{align*}
-From this, we get that $n$ must be at least 6739.}
+From this, we get that $n$ must be at least 6697.}
 
 % 17
 

extraTeX/tables/TeX/tTable.tex

@@ -272,7 +272,7 @@ \section{$\pmb{t}$-Probability Table}
 500  &  {\normalsize  1.28} & {\normalsize  1.65} & {\normalsize  1.96} & {\normalsize  2.33} & {\normalsize  2.59}  \\ 
 \hline
 \hline
-$\infty$  &  {\normalsize  1.28} & {\normalsize  1.65} & {\normalsize  1.96} & {\normalsize  2.33} & {\normalsize  2.58}  \\ 
+$\infty$  &  {\normalsize  1.28} & {\normalsize  1.645} & {\normalsize  1.96} & {\normalsize  2.33} & {\normalsize  2.58}  \\ 
 \hline
 \end{tabular}
 \end{center}

main.tex

@@ -1,6 +1,6 @@
 \documentclass[10pt,openany]%,oneside]
 {book}
-\newcommand{\versiondate}[0]{November 13th, 2019}
+\newcommand{\versiondate}[0]{April 12th, 2022}
 
 \usepackage{
   amsmath, calc,