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Homework 2 for Stat Inference
Extra problems for Stat Inference
Brian Caffo
Johns Hopkins Bloomberg School of Public Health
io2012
highlight.js
tomorrow
mathjax
quiz
bootstrap
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) 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

round(pnorm(93, mean = 100, sd = 10) * 100)
[1] 24

--- &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 $$ P(A^c | T_-) = \frac{P(T_- | A^c) P(A^c)}{P(T_- | A^c) P(A^c) + P(T_- | A) P(A)} $$

(.88 * .95) / (.88 * .95 + .07 * .05)
[1] 0.9958

--- &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 116.449

round(qnorm(.95, mean = 100, sd = 10), 3)
[1] 116.4

round(qnorm(.05, mean = 100, sd = 10, lower.tail = FALSE), 3)
[1] 116.4

--- &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 102.326

round(qnorm(.95, mean = 100, sd = 10 / sqrt(50) ), 3)
[1] 102.3

round(qnorm(.05, mean = 100, sd = 10 / sqrt(50), lower.tail = FALSE), 3)
[1] 102.3

--- &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

89.1

round(pbinom(4, prob = .5, size = 6, lower.tail = TRUE) * 100, 1)
[1] 89.1

--- &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

0.365

round(pnorm(.51, mean = 0.5, sd = sqrt(1 / 12 / 100), lower.tail = FALSE), 3)
[1] 0.365

--- &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 3.5 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 0 since the variance of the sampling distribution of the mean is $\sigma^2/12$ and the variance of a die roll is

mean((1 : 6 - 3.5)^2)
[1] 2.917

--- &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 1

round(ppois(20, lambda = 16.5 * 2) * 100, 1)
[1] 1