added hw2.rmd

Author: bcaffo

06_StatisticalInference/homework/hw2.Rmd

@@ -0,0 +1,240 @@
+---
+title       : Homework 2 for Stat Inference
+subtitle    : Extra problems for Stat Inference
+author      : Brian Caffo
+job         : Johns Hopkins Bloomberg School of Public Health
+framework   : io2012
+highlighter : highlight.js  
+hitheme     : tomorrow       
+#url:
+#    lib: ../../librariesNew #Remove new if using old slidify
+#    assets: ../../assets
+widgets     : [mathjax, quiz, bootstrap]
+mode        : selfcontained # {standalone, draft}
+---
+```{r setup, cache = F, echo = F, message = F, warning = F, tidy = F, results='hide'}
+# make this an external chunk that can be included in any file
+library(knitr)
+options(width = 100)
+opts_chunk$set(message = F, error = F, warning = F, comment = NA, fig.align = 'center', dpi = 100, tidy = F, cache.path = '.cache/', fig.path = 'fig/')
+
+options(xtable.type = 'html')
+knit_hooks$set(inline = function(x) {
+  if(is.numeric(x)) {
+    round(x, getOption('digits'))
+  } else {
+    paste(as.character(x), collapse = ', ')
+  }
+})
+knit_hooks$set(plot = knitr:::hook_plot_html)
+runif(1)
+```
+
+## About these slides
+- These are some practice problems for Statistical Inference Quiz 1
+- They were created using slidify interactive which you will learn in 
+Creating Data Products
+- Please help improve this with pull requests here
+(https://github.com/bcaffo/courses)
+runif(1)
+
+--- &radio
+The probability that a manuscript gets accepted to a journal is 12% (say). However,
+given that a revision is asked for, the probability that it gets accepted
+is 90%. Is it possible that the probability that a manuscript has a revision
+asked for is 20%? 
+
+1. Yeah, that's totally possible.
+2. _No, it's not possible._
+3. It's not possible to answer this question.
+
+*** .hint
+$A = accepted$, $B = revision$. $P(A) = .12$, $P(A | B) = .90$. $P(B) = .20$
+
+*** .explanation
+$P(A \cap B) = P(A | B) * P(B) = .9 \times .2 = .18$ this is larger than
+$P(A) = .12$, which is not possible since $A \cap B \subset A$.
+
+
+--- &radio
+Suppose that the number of web hits to a particular site are approximately normally
+distributed with a mean of 100 hits per day and a standard deviation of 10 hits per day. What's the probability that a given day has fewer than 93 hits per day
+expressed as a percentage to the nearest percentage point?
+
+1. 76%
+2. _24%_
+3. 47%
+4. 94%
+
+*** .hint
+Let $X$ be the number of hits per day. We want $P(X \leq 93)$ given that
+$X$ is $N(100, 10^2)$.
+
+*** .explanation
+```{r}
+round(pnorm(93, mean = 100, sd = 10) * 100)
+```
+
+
+--- &radio
+Suppose 5% of housing projects have issues with asbestos. The sensitivity of a test
+for asbestos is 93% and the specificity is 88%. What is the probability that a 
+housing project has no asbestos given a negative test expressed as a percentage
+to the nearest percentage point?
+
+1. 0%
+2. 5%
+3. 10%
+4. 20%
+5. 50%
+6. _100%_
+
+*** .hint
+$A = asbestos$, $T_+ = tests positive$, $T_- = tests negative$. 
+$P(T_+ | A) = .93$, $P(T_- | A^c) = .88$, $P(A) = .05$.
+
+*** .explanation
+We want
+$$
+P(A^c | T_-) = \frac{P(T_- | A^c) P(A^c)}{P(T_- | A^c) P(A^c) + P(T_- | A) P(A)}
+$$
+```{r}
+(.88 * .95) / (.88 * .95 + .07 * .05)
+```
+
+
+
+---  &multitext
+Suppose that the number of web hits to a particular site are approximately normally
+distributed with a mean of 100 hits per day and a standard deviation of 10 hits per day. 
+
+1. What number of web hits per day represents the number so that only
+5% of days have more hits? Express your answer to 3 decimal places.
+
+
+
+*** .hint
+Let $X$ be the number of hits per day. We want $P(X \leq 93)$ given that
+$X$ is $N(100, 10^2)$.
+
+*** .explanation
+<span class="answer">`r round(qnorm(.95, mean = 100, sd = 10), 3)`</span>
+```{r}
+round(qnorm(.95, mean = 100, sd = 10), 3)
+round(qnorm(.05, mean = 100, sd = 10, lower.tail = FALSE), 3)
+```
+
+
+---  &multitext
+Suppose that the number of web hits to a particular site are approximately normally
+distributed with a mean of 100 hits per day and a standard deviation of 10 hits per day. Imagine taking a random sample of 50 days. 
+
+1. What number of web hits would
+be the point so that only 5% of averages of 50 days of web traffic have more hits? 
+Express your answer to 3 decimal places. 
+
+*** .hint
+Let $\bar X$ be the average number of hits per day for 50 randomly sampled days.
+$X$ is $N(100, 10^2 / 50)$.
+
+*** .explanation
+<span class="answer">`r round(qnorm(.95, mean = 100, sd = 10 / sqrt(50) ), 3)`</span>
+ 
+```{r}
+round(qnorm(.95, mean = 100, sd = 10 / sqrt(50) ), 3)
+round(qnorm(.05, mean = 100, sd = 10 / sqrt(50), lower.tail = FALSE), 3)
+```
+
+--- &multitext
+
+You don't believe that your friend can discern good wine from cheap. Assuming
+that you're right, in a blind test where you randomize 6 paired varieties (Merlot,
+Chianti, ...) of cheap and expensive wines
+
+1. what is the change that she gets 5 or 6 right expressed as a percentage
+to one decimal place?
+
+*** .hint
+Let $p=.5$ and $X$ be binomial
+
+*** .explanation
+
+<span class="answer">`r round(pbinom(4, prob = .5, size = 6, lower.tail = TRUE) * 100, 1)`</span>
+
+```{r}
+round(pbinom(4, prob = .5, size = 6, lower.tail = TRUE) * 100, 1)
+```
+
+--- &multitext
+
+Consider a uniform distribution. If we were to sample 100 draws from a 
+a uniform distribution (which has mean 0.5, and variance 1/12) and take their
+mean, $\bar X$
+
+1. what is the approximate probability of getting as large as 0.51 or larger expressed to 3 decimal places?
+
+*** .hint
+Use the central limit theorem that says $\bar X \sim N(\mu, \sigma^2/n)$
+
+*** .explanation
+
+<span class="answer"> `r round(pnorm(.51, mean = 0.5, sd = sqrt(1 / 12 / 100), lower.tail = FALSE), 3)`</span>
+
+```{r}
+round(pnorm(.51, mean = 0.5, sd = sqrt(1 / 12 / 100), lower.tail = FALSE), 3)
+```
+
+
+--- &multitext
+
+If you roll ten standard dice, take their average, then repeat this process over and over and construct a histogram, 
+
+1. what would it be centered at?
+
+
+*** .hint
+$E[X_i] = E[\bar X]$ where $\bar X = \frac{1}{n}\sum_{i=1}^n X_i$
+
+*** .explanation
+
+
+The answer will be  <span class="answer">3.5</span> since the mean of the
+sampling distribution of iid draws will be the population mean that the
+individual draws were taken from.
+
+--- &multitext
+
+If you roll ten standard dice, take their average, then repeat this process over and over and construct a histogram, 
+
+1. what would be its variance expressed to 3 decimal places?
+
+*** .hint
+$$Var(\bar X) = \sigma^2 /n$$
+
+*** .explanation
+The answer will be <span class="answer">`r round( mean(1 : 6 - 3.5) ^2 / 100, 3)`</span> 
+since the variance of the sampling distribution of the mean is $\sigma^2/12$
+and the variance of a die roll is 
+
+```{r}
+mean((1 : 6 - 3.5)^2)
+```
+
+--- &multitext
+The number of web hits to a site is Poisson with mean 16.5 per day. 
+
+1. What is the probability of getting 20 or fewer in 2 days expressed
+as a percentage to one decimal place?
+
+*** .hint
+Let $X$ be the number of hits in 2 days then $X \sim Poisson(2\lambda)$
+
+*** .explanation
+<span class="answer">`r round(ppois(20, lambda = 16.5 * 2) * 100, 1)`</span>
+
+```{r}
+round(ppois(20, lambda = 16.5 * 2) * 100, 1)
+```
+
+
+