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Homework 1 for Stat Inference
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Brian Caffo
Johns Hopkins Bloomberg School of Public Health
io2012
highlight.js
tomorrow
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selfcontained

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 10% while that the mother contracted the disease is 9%. What is the probability that both contracted influenza expressed as a whole number percentage?

  1. 15%
  2. 10%
  3. 9%
  4. 4%

*** .hint $A = Father$, $P(A) = .10$, $B = Mother$, $P(B) = .09$ $P(A\cup B) = .15$,

*** .explanation $P(A\cup B) = P(A) + P(B) - P(AB)$ thus $$.15 = .10 + .09 - P(AB)$$

.10 + .09 - .15
[1] 0.04

--- &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?

  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 odds 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 probability .2 and 1 with probability .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 $$E[X] = 0$$ $$ Var(X) = E[X^2] = (-4)^2 * .2 + (1)^2 * .8 $$

-4 * .2 + 1 * .8
[1] 0

(-4)^2 * .2 + (1)^2 * .8
[1] 4

--- &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

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

sum(x ^ 2 * p) - sum(x * p) ^ 2
[1] 1