Merge branch 'master' of github.com:DataScienceSpecialization/courses

* 'master' of github.com:DataScienceSpecialization/courses:
  added pdfs for brian in statinf
  Added some lecture files
  Added a bunch of pdfs
  Added some figures
  Fixed some errors
  recompiled 03_04
  Minor corrections to intro page
  Added a hw2
  Added hw2
  Worked on hw2
  Set correct answer for question 2
  Added a hw 2
  Fixed a small error
  Added HW files
  Added some homework
  update

Author: rdpeng

06_StatisticalInference/03_05_MultipleTesting/index.pdf

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+---
+title       : Homework 1 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$set(plot = knitr:::hook_plot_html)
+runif(1)
+```
+
+## About these slides
+- These are some practice problems for Statistical Inference Quiz 1
+- They were created using slidify interactive which you will learn in 
+Creating Data Products
+- Please help improve this with pull requests here
+(https://github.com/bcaffo/courses)
+
+
+--- &radio
+
+Consider influenza epidemics for two parent heterosexual families. Suppose that the probability is 15% that at least one of the parents has contracted the disease. The probability that the father has contracted influenza is 6% while that the mother contracted the disease is 5%. What is the probability that both contracted influenza expressed as a whole number percentage?
+
+1. 15%
+2. 6%
+3. 5%
+4. _2%_
+
+*** .hint
+$A = Father$, $P(A) = .06$, $B = Mother$, $P(B) = .05$ 
+$P(A\cup B) = .15$, 
+
+*** .explanation
+$P(A\cup B) = P(A) + P(B) - 2 P(AB)$ thus
+$$.15 = .06 + .05 - 2 P(AB)$$
+```{r}
+(0.15 - .06 - .05) / 2
+```
+
+---  &radio
+
+A random variable, $X$, is uniform, a box from $0$ to $1$ of height $1$. (So that it's density is $f(x) = 1$ for $0\leq x \leq 1$.) What is it's median expressed to two decimal places? </p>
+
+1. 1.00
+2. 0.75
+3. _0.50_
+4. 0.25
+
+*** .hint
+The median is the point so that 50% of the density lies below it.
+
+*** .explanation
+This density looks like a box. So, notice that $P(X \leq x) = width\times height = x$.
+We want $.5 = P(X\leq x) = x$.
+
+--- &radio
+
+You are playing a game with a friend where you flip a coin and if it comes up heads you give her  $X$ dollars and if it comes up tails she gives you $Y$ dollars. The odds that the coin is heads in $d$. What is your expected earnings?
+
+1. _$-X \frac{d}{1 + d} + Y \frac{1}{1+d} $_
+2. $X \frac{d}{1 + d} + Y \frac{1}{1+d} $
+3. $X \frac{d}{1 + d} - Y \frac{1}{1+d} $
+4. $-X \frac{d}{1 + d} - Y \frac{1}{1+d} $
+
+*** .hint
+The probability that you win on a given round is given by $p / (1 - p) = d$ which implies
+that $p = d / (1 + d)$.
+
+*** .explanation
+You lose $X$ with probability $p = d/(1 +d)$ and you win $Y$ with probability $1-p = 1/(1 + d)$. So your answer is
+$$
+-X \frac{d}{1 + d} + Y \frac{1}{1+d} 
+$$
+
+--- &radio
+A random variable takes the value -4 with probabability .2 and 1 with proabability .8. What
+is the variance of this random variable?
+
+1. 0
+2. _4_
+3. 8
+4. 16
+
+*** .hint
+This random variable has mean 0. The variance would be given by $E[X^2]$ then.
+
+*** .explanation
+$$E[X] = 0$$
+$$
+Var(X) = E[X^2] = (-4)^2 * .2 + (1)^2 * .8
+$$
+```{r}
+-4 * .2 + 1 * .8
+(-4)^2 * .2 + (1)^2 * .8
+```
+
+
+--- &radio
+If $\bar X$ and $\bar Y$ are comprised of $n$ iid random variables arising from distributions
+having  means $\mu_x$ and $\mu_y$, respectively and common variance $\sigma^2$
+what is the variance $\bar X - \bar Y$?
+
+1. 0
+2. _$2\sigma^2/n$_
+3. $\mu_x$ - $\mu_y$
+4. $2\sigma^2$
+
+*** .hint
+Remember that $Var(\bar X) = Var(\bar Y) = \sigma^2 / n$. 
+
+*** .explanation 
+$$
+Var(\bar X - \bar Y) = Var(\bar X) + Var(\bar Y) = \sigma^2 / n + \sigma^2 / n
+$$
+
+--- &radio
+Let $X$ be a random variable having standard deviation $\sigma$. What can
+be said about $X /\sigma$?
+
+1. Nothing
+2. _It must have variance 1._
+3. It must have mean 0.
+4. It must have variance 0.
+
+*** .hint
+$Var(aX) = a^2 Var(X)$
+
+*** .explanation
+$$Var(X / \sigma) = Var(X) / \sigma^2 = 1$$
+
+
+--- &radio
+If a continuous density that never touches the horizontal axis is symmetric about zero, can we say that its associated median is zero?
+
+1. _Yes_
+2. No.
+3. It can not be determined given the information given.
+
+*** .explanation
+This is a surprisingly hard problem. The easy explanation is that 50% of the probability
+is below 0 and 50% is above so yes. However, it is predicated on the density not being
+a flat line at 0 around 0. That's why the caveat that it never touches the horizontal axis
+is important.
+
+
+--- &radio
+
+Consider the following pmf given in R
+```{r}
+p <- c(.1, .2, .3, .4)
+x <- 2 : 5 
+```
+What is the variance expressed to 1 decimal place?
+
+1. _1.0_
+2. 4.0
+3. 6.0
+4. 17.0
+
+*** .hint
+The variance is $E[X^2] - E[X^2]$
+
+*** .explanation 
+```{r}
+sum(x ^ 2 * p) - sum(x * p) ^ 2
+```