Nicole Dalzell. July 3, PDF Free Download

UNIT 2: PROBABILITY AND DISTRIBUTIONS LECTURE 1: PROBABILITY AND CONDITIONAL PROBABILITY STATISTICS 101 Nicole Dalzell July 3, 2014

Announcements No team activities today Labs: Individual Write-Ups Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 2 / 32

Project 1 Announcements 2 Project 3 Probability Randomness Defining probability Law of large numbers Disjoint and non-disjoint outcomes Probability distributions Independence Recap 4 Marginal, joint, conditional Statistics 101

Project Project Project instructions posted: http: // stat.duke.edu/ courses/ Summer14/ sta101.001-2/ problem sets/ project.pdf Think about research questions to explore. Decide if you ll be collecting your own observational data, conduct an experiment, or use previously collected data (from a published study or public database). Proposal due Friday, Oct 4. Project due Thursday, Nov 7. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 3 / 32

Project Data resources: Data and GIS Services http:// guides.library.duke.edu/ stat101 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 4 / 32

Project Data resources: Data and GIS Services - office hours For questions related to finding data and getting it into R only, not statistical analysis questions. http:// library.duke.edu/ data/ about/ schedule.html Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 5 / 32

Project Data resources: Hillary Mason & Bit.ly http:// bitly.com/ bundles/ hmason/ 1 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 6 / 32

Project Data resources: Reddit http:// www.reddit.com/ r/ datasets Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 7 / 32

Project Data resources: DASL http:// lib.stat.cmu.edu/ DASL/ Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 8 / 32

Project Data resources: Citizen Statistician http:// citizen-statistician.org/ 2012/ 11/ 07/ data-sets-a-list-in-flux/ Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 9 / 32

Project Data resources and many others, get creative! Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 10 / 32

Project Getting Data in R In lab you have been given nicely formatted data that can be directly loaded into R using either load or source. This will rarely happen outside of this class, and for your project you will need to convert your data into a format R can read. Ideally use a plaintext format: csv, tab delimited, etc. (read.csv, read.table, read.delim) Avoid proprietary formats (usually doable but require extra work) Programs like Excel are useful to convert and clean up data Find your data early, if you run into trouble ask for help Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 11 / 32

1 Announcements 2 Project 3 Probability Randomness Defining probability Law of large numbers Disjoint and non-disjoint outcomes Probability distributions Independence Recap 4 Marginal, joint, conditional Statistics 101

Randomness Random processes A random process is a situation in which we know what outcomes could happen, but we don t know which particular outcome will happen. Examples: coin tosses, die rolls, itunes shuffle, whether the stock market goes up or down tomorrow, etc. It can be helpful to model a process as random even if it is not truly random. http:// www.cnet.com.au/ itunes-just-how-random-is-random-339274094.htm Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 12 / 32

Defining probability Probability There are several possible interpretations of probability but they (almost) completely agree on the mathematical rules probability must follow. P(A) = Probability of event A 0 P(A) 1 Frequentist interpretation: The probability of an outcome is the proportion of times the outcome would occur if we observed the random process an infinite number of times. Single main stream school until recently. Bayesian interpretation: A Bayesian interprets probability as a subjective degree of belief: For the same event, two separate people could have differing probabilities. Largely popularized by revolutionary advance in computational technology and methods during the last twenty years. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 13 / 32

Law of large numbers Participation question Which of the following events would you be most surprised by? (a) 3 heads in 10 coin flips (b) 3 heads in 100 coin flips (c) 3 heads in 1000 coin flips Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 14 / 32

Law of large numbers Law of large numbers Law of large numbers states that as more observations are collected, the proportion of occurrences with a particular outcome, ˆp n, converges to the probability of that outcome, p. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 15 / 32

Law of large numbers Law of large numbers vs. law of averages When tossing a fair coin, if heads comes up on each of the first 10 tosses, what do you think the chance is that another head will come up on the next toss? 0.5, less than 0.5, or more than 0.5? H H H H H H H H H H? The probability is still 0.5, or there is still a 50% chance that another head will come up on the next toss. P(H on 11 th toss) = P(T on 11 th toss) = 0.5 The coin is not due for a tail. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 16 / 32

Law of large numbers Law of large numbers vs. law of averages When tossing a fair coin, if heads comes up on each of the first 10 tosses, what do you think the chance is that another head will come up on the next toss? 0.5, less than 0.5, or more than 0.5? H H H H H H H H H H? The probability is still 0.5, or there is still a 50% chance that another head will come up on the next toss. P(H on 11 th toss) = P(T on 11 th toss) = 0.5 The coin is not due for a tail. The common (mis)understanding of the LLN is that random processes are supposed to compensate for whatever happened in the past; this is just not true and is also called gambler s fallacy (or law of averages). Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 16 / 32

Disjoint and non-disjoint outcomes Disjoint and non-disjoint outcomes Disjoint (mutually exclusive) outcomes: Cannot happen at the same time. The outcome of a single coin toss cannot be a head and a tail. A student cannot fail and pass a class. A card drawn from a deck cannot be an ace and a queen. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 17 / 32

Disjoint and non-disjoint outcomes Disjoint and non-disjoint outcomes Disjoint (mutually exclusive) outcomes: Cannot happen at the same time. The outcome of a single coin toss cannot be a head and a tail. A student cannot fail and pass a class. A card drawn from a deck cannot be an ace and a queen. Non-disjoint outcomes: Can happen at the same time. A student can get an A in Stats and A in Econ in the same semester. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 17 / 32

Disjoint and non-disjoint outcomes Participation question What is the probability that a randomly sampled student thinks marijuana should be legalized or they agree with their parents political views? (a) 40+36 78 165 (b) 114+118 78 165 Parent Politics Legalize MJ No Yes Total No 11 40 51 Yes 36 78 114 Total 47 118 165 (c) (d) 78 165 78 188 (e) 11 47 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 18 / 32

Disjoint and non-disjoint outcomes Participation question What is the probability that a randomly sampled student thinks marijuana should be legalized or they agree with their parents political views? (a) 40+36 78 165 (b) 114+118 78 165 Parent Politics Legalize MJ No Yes Total No 11 40 51 Yes 36 78 114 Total 47 118 165 (c) (d) 78 165 78 188 (e) 11 47 General addition rule P(A or B) = P(A) + P(B) P(A and B) For disjoint events P(A and B) = 0, hence the above formula simplifies to P(A or B) = P(A) + P(B). Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 18 / 32

Probability distributions Participation question In a survey, 52% of respondents said they are Democrats. What is the probability that a randomly selected respondent from this sample is a Republican? (a) 0.48 (b) more than 0.48 (c) less than 0.48 (d) cannot calculate using only the information given While (a) and (c) are also possible, (b) is definitely not possible. Probability distributions: 1 The events listed must be disjoint 2 Each probability must be between 0 and 1 3 The probabilities must total 1 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 19 / 32

Probability distributions Disjoint vs. complementary Do the sum of probabilities of two disjoint events always add up to 1? Do the sum of probabilities of two complementary events always add up to 1? Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 20 / 32

Probability distributions Disjoint vs. complementary Do the sum of probabilities of two disjoint events always add up to 1? Not necessarily, there may be more than 2 events in the sample space, e.g. party affiliation. Do the sum of probabilities of two complementary events always add up to 1? Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 20 / 32

Probability distributions Disjoint vs. complementary Do the sum of probabilities of two disjoint events always add up to 1? Not necessarily, there may be more than 2 events in the sample space, e.g. party affiliation. Do the sum of probabilities of two complementary events always add up to 1? Yes, that s the definition of complementary, e.g. heads and tails. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 20 / 32

Independence Independence Two processes are independent if knowing the outcome of one provides no useful information about the outcome of the other. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 21 / 32

Independence Independence Two processes are independent if knowing the outcome of one provides no useful information about the outcome of the other. coin flips are independent card draws (without replacement) are dependent Independence and disjointness do not mean the same thing: Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 21 / 32

Independence Independence Two processes are independent if knowing the outcome of one provides no useful information about the outcome of the other. coin flips are independent card draws (without replacement) are dependent Independence and disjointness do not mean the same thing: independent events do not affect each other Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 21 / 32

Independence Independence Two processes are independent if knowing the outcome of one provides no useful information about the outcome of the other. coin flips are independent card draws (without replacement) are dependent Independence and disjointness do not mean the same thing: independent events do not affect each other disjoint (mutually exclusive) events cannot happen at the same time Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 21 / 32

Independence Participation question Between January 9-12, 2013, SurveyUSA interviewed a random sample of 500 NC residents asking them whether they think widespread gun ownership protects law abiding citizens from crime, or makes society more dangerous. 58% of all respondents said it protects citizens. 67% of White respondents, 28% of Black respondents, and 64% of Hispanic respondents shared this view. Which of the below is true? Opinion on gun ownership and race ethnicity are most likely (a) complementary (b) mutually exclusive (c) independent (d) dependent (e) disjoint http:// www.surveyusa.com/ client/ PollReport.aspx?g=a5f460ef-bba9-484b-8579-1101ea26421b Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 22 / 32

Independence Checking for independence If P(A B) = P(A), then A and B are independent. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 23 / 32

Independence Checking for independence If P(A B) = P(A), then A and B are independent. P(protects citizens) = 0.58 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 23 / 32

Independence Checking for independence If P(A B) = P(A), then A and B are independent. P(protects citizens) = 0.58 P(protects citizens White) = 0.67 P(protects citizens Black) = 0.28 P(protects citizens Hispanic) = 0.64 P(protects citizens) varies by race/ethnicity, therefore opinion on gun ownership and race ethnicity are most likely dependent. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 23 / 32

Independence Evaluating dependence based on sample data Conditional probabilities calculated based on sample data suggest dependence conduct a hypothesis test to determine if the observed difference is unlikely to have happened by chance. We have seen that P(protects citizens White) = 0.67 and P(protects citizens Hispanic) = 0.64. Under which condition would you be more convinced of a real difference between the proportions of Whites and Hispanics who think gun widespread gun ownership protects citizens? n = 500 or n = 50, 000 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 24 / 32

Independence Evaluating dependence based on sample data Conditional probabilities calculated based on sample data suggest dependence conduct a hypothesis test to determine if the observed difference is unlikely to have happened by chance. We have seen that P(protects citizens White) = 0.67 and P(protects citizens Hispanic) = 0.64. Under which condition would you be more convinced of a real difference between the proportions of Whites and Hispanics who think gun widespread gun ownership protects citizens? n = 500 or n = 50, 000 n = 50,000 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 24 / 32

Independence Evaluating dependence based on sample data Conditional probabilities calculated based on sample data suggest dependence conduct a hypothesis test to determine if the observed difference is unlikely to have happened by chance. We have seen that P(protects citizens White) = 0.67 and P(protects citizens Hispanic) = 0.64. Under which condition would you be more convinced of a real difference between the proportions of Whites and Hispanics who think gun widespread gun ownership protects citizens? n = 500 or n = 50, 000 n = 50,000 Large observed difference hypothesis test will likely be significant. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 24 / 32

Independence Evaluating dependence based on sample data Conditional probabilities calculated based on sample data suggest dependence conduct a hypothesis test to determine if the observed difference is unlikely to have happened by chance. We have seen that P(protects citizens White) = 0.67 and P(protects citizens Hispanic) = 0.64. Under which condition would you be more convinced of a real difference between the proportions of Whites and Hispanics who think gun widespread gun ownership protects citizens? n = 500 or n = 50, 000 n = 50,000 Large observed difference hypothesis test will likely be significant. Small observed difference, and n large hypothesis test may be significant. n small then hypothesis test will likely not be significant. Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 24 / 32

Independence Participation question A 2012 Gallup poll suggests that 19.4% of North Carolinians don t have health insurance. Assuming that the uninsured rate stayed constant, what is the probability that two randomly selected North Carolinians are both uninsured? (a) 19.4 2 (b) 0.194 2 (c) 0.194 2 (d) (1 0.194) 2 http:// www.gallup.com/ poll/ 125066/ State-States.aspx?ref=interactive Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 25 / 32

Independence Participation question A 2012 Gallup poll suggests that 19.4% of North Carolinians don t have health insurance. Assuming that the uninsured rate stayed constant, what is the probability that two randomly selected North Carolinians are both uninsured? (a) 19.4 2 (b) 0.194 2 (c) 0.194 2 (d) (1 0.194) 2 http:// www.gallup.com/ poll/ 125066/ State-States.aspx?ref=interactive Product rule for independent events P(A and B) = P(A) P(B) Or more generally, P(A 1 and and A k ) = P(A 1 ) P(A k ) Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 25 / 32

Recap Participation question In a NC emergency room, 5 patients are waiting to be seen. Assuming that these patients constitute a random sample, what is the probability that at least one is uninsured? (a) 1 0.194 5 (b) 1 0.194 5 (c) 0.806 5 (d) 1 0.806 5 (e) 1 0.806 5 Sample space of uninsured NC residents in this sample: S = {0, 1, 2, 3, 4, 5} We are interested in instances where at least one person is uninsured: S = {0, 1, 2, 3, 4, 5} So we can divide up the sample space intro two categories: S = {0, at least one} Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 26 / 32

Recap Participation question In a NC emergency room, 5 patients are waiting to be seen. Assuming that these patients constitute a random sample, what is the probability that at least one is uninsured? (a) 1 0.194 5 (b) 1 0.194 5 (c) 0.806 5 (d) 1 0.806 5 (e) 1 0.806 5 Sample space of uninsured NC residents in this sample: S = {0, 1, 2, 3, 4, 5} We are interested in instances where at least one person is uninsured: S = {0, 1, 2, 3, 4, 5} So we can divide up the sample space intro two categories: S = {0, at least one} Since the probability of the sample space adds up to 1: P(at least 1 uninsured) = 1 P(none uninsured) = 1 0.806 5 0.65 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 26 / 32

Recap Participation question In a NC emergency room, 5 patients are waiting to be seen. Assuming that these patients constitute a random sample, what is the probability that at least one is uninsured? (a) 1 0.194 5 (b) 1 0.194 5 (c) 0.806 5 (d) 1 0.806 5 (e) 1 0.806 5 At least 1 Sample space of uninsured NC residents in this sample: S = {0, 1, 2, 3, 4, 5} We are interested in instances where at least one person is uninsured: S = {0, 1, 2, 3, 4, 5} So we can divide up the sample space intro two categories: S = {0, at least one} Since the probability of the sample space adds up to 1: P(at least 1 uninsured) = 1 P(none uninsured) = 1 0.806 5 0.65 P(at least one) = 1 P(none) Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 26 / 32

Marginal, joint, conditional 1 Announcements 2 Project 3 Probability Randomness Defining probability Law of large numbers Disjoint and non-disjoint outcomes Probability distributions Independence Recap 4 Marginal, joint, conditional Statistics 101

Marginal, joint, conditional Relapse Researchers randomly assigned 72 chronic users of cocaine into three groups: desipramine (antidepressant), lithium (standard treatment for cocaine) and placebo. Results of the study are summarized below. no relapse relapse total desipramine 10 14 24 lithium 18 6 24 placebo 20 4 24 total 48 24 72 http:// www.oswego.edu/ srp/ stats/ 2 way tbl 1.htm Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 27 / 32

Marginal, joint, conditional Marginal probability What is the probability that a patient relapsed? no relapse relapse total desipramine 10 14 24 lithium 18 6 24 placebo 20 4 24 total 48 24 72 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 28 / 32

Marginal, joint, conditional Marginal probability What is the probability that a patient relapsed? P(relapsed) = 48 72 0.67 no relapse relapse total desipramine 10 14 24 lithium 18 6 24 placebo 20 4 24 total 48 24 72 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 28 / 32

Marginal, joint, conditional Joint probability What is the probability that a patient received the the antidepressant (desipramine) and relapsed? no relapse relapse total desipramine 10 14 24 lithium 18 6 24 placebo 20 4 24 total 48 24 72 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 29 / 32

Marginal, joint, conditional Joint probability What is the probability that a patient received the the antidepressant (desipramine) and relapsed? no relapse relapse total desipramine 10 14 24 lithium 18 6 24 placebo 20 4 24 total 48 24 72 P(relapsed and desipramine) = 10 72 0.14 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 29 / 32

Marginal, joint, conditional Conditional probability If we know that a patient received the antidepressant (desipramine), what is the probability that they relapsed? no relapse relapse total desipramine 10 14 24 lithium 18 6 24 placebo 20 4 24 total 48 24 72 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 30 / 32

Marginal, joint, conditional Conditional probability If we know that a patient received the antidepressant (desipramine), what is the probability that they relapsed? no relapse relapse total desipramine 10 14 24 lithium 18 6 24 placebo 20 4 24 total 48 24 72 P(relapsed desipramine) = 10 24 0.42 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 30 / 32

Marginal, joint, conditional Conditional probability If we know that a patient received the antidepressant (desipramine), what is the probability that they relapsed? no relapse relapse total desipramine 10 14 24 lithium 18 6 24 placebo 20 4 24 total 48 24 72 P(relapsed desipramine) = 10 24 0.42 P(relapsed lithium) = 18 24 0.75 P(relapsed placebo) = 20 24 0.83 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 30 / 32

Marginal, joint, conditional Conditional probability If we know that a patient relapsed, what is the probability that they received the antidepressant (desipramine)? no relapse relapse total desipramine 10 14 24 lithium 18 6 24 placebo 20 4 24 total 48 24 72 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 31 / 32

Marginal, joint, conditional Conditional probability If we know that a patient relapsed, what is the probability that they received the antidepressant (desipramine)? no relapse relapse total desipramine 10 14 24 lithium 18 6 24 placebo 20 4 24 total 48 24 72 P(desipramine relapsed) = 10 48 0.21 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 31 / 32

Marginal, joint, conditional Conditional probability If we know that a patient relapsed, what is the probability that they received the antidepressant (desipramine)? no relapse relapse total desipramine 10 14 24 lithium 18 6 24 placebo 20 4 24 total 48 24 72 P(desipramine relapsed) = 10 48 0.21 P(lithium relapsed) = 18 48 0.375 P(placebo relapsed) = 20 48 0.42 Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 31 / 32

Marginal, joint, conditional What if we don t have counts to create a contingency table with counts? Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 32 / 32

Marginal, joint, conditional What if we don t have counts to create a contingency table with counts?... next time Statistics 101 ( Nicole Dalzell ) U2 - L1: Probability July 3, 2014 32 / 32

Nicole Dalzell. July 3, 2014