{"action":"created","comment":{"url":"https://api.github.com/repos/Middlebury-Data-Science/HW-2/pulls/comments/57327092","id":57327092,"diff_hunk":"@@ -0,0 +1,475 @@\n+---\r\n+title: \"MATH 216 Homework 2\"\r\n+author: \"Mohamed Hussein\"\r\n+output: html_document\r\n+---\r\n+\r\n+```{r, echo=FALSE, message=FALSE, warning=FALSE}\r\n+suppressPackageStartupMessages(library(plyr))\r\n+suppressPackageStartupMessages(library(dplyr))\r\n+suppressPackageStartupMessages(library(ggplot2))\r\n+suppressPackageStartupMessages(library(foreign))\r\n+suppressPackageStartupMessages(library(knitr))\r\n+suppressPackageStartupMessages(library(tidyr))\r\n+suppressPackageStartupMessages(library(plotly))\r\n+suppressPackageStartupMessages(library(pander))\r\n+suppressPackageStartupMessages(library(gridExtra))\r\n+```\r\n+\r\n+## Admistrative:\r\n+\r\n+Please indicate\r\n+\r\n+* Who you collaborated with: Christian L. \r\n+* Roughly how much time you spent on this HW: 8 Hrs \r\n+* What gave you the most trouble:\r\n+* Any comments you have: \r\n+\r\n+\r\n+## Question 1:\r\n+\r\n+Question 4 on page 76 from Chapter 4 of Data Analysis Using Regression and\r\n+Multilevel/Hierarchical Models.  The codebook can be found\r\n+[here](http://www.stat.columbia.edu/~gelman/arm/examples/pollution/pollution.txt).\r\n+I've included R code blocks for each question, but use them only if you feel it\r\n+necessary.\r\n+\r\n+```{r, echo=FALSE, cache=TRUE}\r\n+# DO NOT EDIT THIS SECTION!\r\n+url <- \"http://www.stat.columbia.edu/~gelman/arm/examples/pollution/pollution.dta\"\r\n+pollution <- read.dta(url) %>% \r\n+  tbl_df()\r\n+```\r\n+\r\n+###a)\r\n+*Create a scatterplot of mortality rate versus level of nitric oxides. Do you think linear regression will fit these data well? Fit the regression and evaluate residual plot from the regression.\r\n+\r\n+```{r, echo=FALSE, fig.width=12, fig.height=6}\r\n+ggplot(data=pollution, aes(x = nox, y = mort))+\r\n+  geom_point() +\r\n+  ggtitle(\"Mortaility vs. Nitric Oxide Levels\") +\r\n+  xlab(\"Relative Nitric Oxide Pollution Potential\") +\r\n+  ylab(\"Total Age-adjusted Mortality Rate per 100,000\") + \r\n+  theme(plot.title = element_text(lineheight=.8, face=\"bold\")) + \r\n+  theme(axis.title.y=element_text(margin=margin(0,12,0,0)))\r\n+```\r\n+\r\n+At a cursory glance, it appears that a linear model would not fit these data well, since the majority of points are constrained below 100 and display a non-linear relationship in how \r\n+they increase. Adding a linear regression line, we get: \r\n+\r\n+```{r, echo=FALSE, fig.width=12, fig.height=6}\r\n+ggplot(data=pollution, aes(x = nox, y = mort))+\r\n+      geom_point() +\r\n+      ggtitle(\"Mortaility vs. Nitric Oxide Levels\") +\r\n+      xlab(\"Relative Nitric Oxide Pollution Potential\") +\r\n+      ylab(\"Total Age-adjusted Mortality Rate per 100,000\") + \r\n+      theme(plot.title = element_text(lineheight=.8, face=\"bold\")) + \r\n+      theme(axis.title.y=element_text(margin=margin(0,12,0,0))) + \r\n+      geom_smooth(method=\"lm\", fullrange=TRUE) \r\n+```\r\n+\r\n+Indeed, adding a linear regression line through the data shows how poorly a linear\r\n+model can predict the mortality rates given different Nitric Oxide pollution levels.\r\n+To assess more rigirously the accuracy of our intuition regarding how bad a linear model \r\n+explains the data, we plot a residuals graph: \r\n+\r\n+```{r, echo=FALSE, fig.width=12, fig.height=6}\r\n+model1a.1  <- \r\n+  lm(mort~nox, data=pollution)\r\n+\r\n+residuals1a.1 <- \r\n+  resid(model1a.1)\r\n+\r\n+ggplot(data=pollution, aes(x = nox, y = residuals1a.1))+\r\n+      geom_point()+\r\n+      theme(plot.title = element_text(lineheight=.8, face=\"bold\")) + \r\n+      ggtitle(\"Residuals vs. Nitric Oxide Levels\") +\r\n+      xlab(\"Relative Nitric Oxide Pollution Potential\") +\r\n+      ylab(\"Residuals\")+ \r\n+      geom_hline(yintercept=0, color=\"red\")\r\n+```\r\n+\r\n+This residuals graph confirms our intuition. Despite showing residuals centered \r\n+around zero, the graph depicts a non-constant variance (demonstrated by seeing a \r\n+pattern in the dispersion of the points), thereby indicating that a linear model \r\n+is a poor fit for the data at hand. \r\n+\r\n+###b)\r\n+*Find an appropriate transformation that will result in data more appropriate for linear regression. Fit a regression to the transformed data and evaluatethe new residual plot: \r\n+\r\n+The transformation we consider is the log of the Nirtic Oxide levels. \r\n+```{r, echo=FALSE, fig.width=12, fig.height=6}\r\n+pollution <- \r\n+            pollution %>% \r\n+            mutate(log_nox=log(nox))\r\n+\r\n+  ggplot(data=pollution, aes(x=log_nox, y=mort)) +\r\n+        geom_jitter() +\r\n+        ggtitle(\"Mortaility vs. Logged Nitric Oxide Levels\") +\r\n+        xlab(\"Log of Relative Nitric Oxide Pollution Potential\") +\r\n+        ylab(\"Total Age-adjusted Mortality Rate per 100,000\") + \r\n+        theme(plot.title = element_text(lineheight=.8, face=\"bold\")) + \r\n+        theme(axis.title.y=element_text(size=9, margin=margin(0,12,0,0)))+\r\n+        theme(axis.title.x=element_text(size=9))+\r\n+        geom_smooth(method=\"lm\", fullrange=TRUE)\r\n+\r\n+model1b.1  <- \r\n+              lm(mort~log_nox, data=pollution)\r\n+residuals1b.1 <- \r\n+                resid(model1b.1)\r\n+\r\n+graph1b.1 <- \r\n+              ggplot(data=pollution, aes(x = log_nox, y = residuals1b.1))+\r\n+              geom_point()+ \r\n+              ggtitle(\"Residuals vs. Logged Nitric Oxide Levels\") +\r\n+              xlab(\"Logged Relative Nitric Oxide Pollution Potential\") +\r\n+              ylab(\"Residuals\")\r\n+graph1b.1 <- \r\n+              graph1b.1 + geom_hline(yintercept=0, color=\"red\")\r\n+graph1b.1\r\n+\r\n+```\r\n+\r\n+The resulting residual graph shows an improvement over the previous one. Not only \r\n+are the residuals centered around zero, as was the case in the previous graph, but \r\n+we also see a random dispersion of residuals, suggesting a constant \r\n+variance.This implies that our second model fits the data relatively better than\r\n+a simple linear model. \r\n+\r\n+\r\n+###c)\r\n+*Interpret the slope coefficient from the model you chose in (b).\r\n+```{r, echo=FALSE, fig.width=12, fig.height=6}\r\n+log_model  <- lm(mort~log_nox, data=pollution)\r\n+panderOptions(\"digits\", 3)\r\n+pander(log_model, covariate.labels = ('Logged Nitrogin Oxide'))\r\n+```\r\n+\r\n+An intuitive way of interpreting the results of this model is to differenciate \r\n+the model with respect to the Nitric Oxide. Thus, a one perecent increase \r\n+in Nitric Oxide levels is associated with a 0.15 increase in the age-adjusted \r\n+total mortality rate per 100,000. We are 95% confident that the effect lies between\r\n+0.213 and 0.285 increases in mortailities per 100,000 adjusted for age (Of course, if we say a one UNIT change in the levels of Nitric Oxide levels, then the associated change in mortality rates is about 15.3). \r\n+\r\n+###d)\r\n+*Now fit a model predicting mortality rate using levels of nitric oxides, sulfur dioxide, and hydrocarbons as inputs. Use appropriate transformations when helpful. Plot the fitted regression model and interpret the coefficients.\r\n+\r\n+First, let's examine the relationship between the two new covariates and the \r\n+mortality rate: \r\n+```{r echo=FALSE, out.width=c('450px', '450px'), fig.show='hold'}\r\n+ggplot(data=pollution, aes(x=hc, y=mort)) + geom_jitter()\r\n+ggplot(data=pollution, aes(x=so2, y=mort)) + geom_jitter()\r\n+```\r\n+\r\n+\r\n+A quick look at the graphs, which display the relationship between the two additional\r\n+covariates and our outcome variable, suggests that there exists a non-linear \r\n+relationship between our two new dependent variables and our outcome variable that is best approximated by a log transformation. \r\n+\r\n+Next, we construct a new model that takes into account the logged values of Sulfur \r\n+Dioxide and Hydrocarbons. From this new model we see that a one one perecent increase in Nitric Oxide levels is associated with a 0.58 increase in the age-adjusted total mortality rate per 100,000, controlling forlevels of sulphur dioxide and hydrocarbon (Again, if we were changing the Nitric Oxide by one unit, the associated increase in motality would be 58).  \r\n+\r\n+\r\n+```{r, echo=FALSE, fig.width=12, fig.height=6}\r\n+pollution <- \r\n+             pollution %>% \r\n+             mutate(log_hc=log(hc)) %>% \r\n+             mutate(log_so2=log(so2))\r\n+\r\n+log_model2  <- \r\n+              lm(mort~log_nox + log_hc + log_so2, data=pollution)\r\n+\r\n+pander(log_model2, covariate.labels = (c('Logged Nitrogin Oxide', \r\n+                                         'Logged Hydrocarbons', \r\n+                                         'Logged Sulfur Dioxide')))\r\n+```\r\n+\r\n+### e)\r\n+*Cross-validate: fit the model you chose above to the first half of the data and\r\n+then predict for the second half. (You used all the data to construct the mode\r\n+in (d), so this is not really cross-validation, but it gives a sense of how the\r\n+steps of cross-validation can be implemented.)\r\n+\r\n+To answer this question, let's follow thw Monte Carlo cross-validation technique:\r\n+In a nutshell, what we will do imagine that our 'pollution' dataset is the population, \r\n+all the data in the world that relates pollutants to mortality rates. Then, we \r\n+will partition two subsamples from our dataset, train and test. We then compare \r\n+the predicted values obtained from the train dataset (P1) to the predicted values \r\n+obtained form the test dataset (P2). If our model is accurate, we would expect, on \r\n+average for the P1=P2. Why? Because the expected value of P1 should equal P2, since\r\n+both are randomly drawn from a normally-distributed population. Graphically, if our \r\n+model is accurate, we would see our points lying on a 45 degree line. \r\n+\r\n+```{r, echo=FALSE, fig.width=12, fig.height=6}\r\n+#First, extract a fraction of the dataset: \r\n+train <- \r\n+  sample_frac(pollution, 0.5)\r\n+\r\n+#Define the model: \r\n+log_model3  <- \r\n+  lm(mort~log_nox + log_hc + log_so2, data=train)\r\n+\r\n+#Obtain the predicted values: \r\n+P1 <- \r\n+  predict (log_model3, train)\r\n+\r\n+#Second, extract another fraction of the \r\n+test <- \r\n+  sample_frac (pollution, 0.5)\r\n+P2 <- \r\n+  predict(log_model3)\r\n+\r\n+ggplot(test, aes(x=P1, y=P2)) + \r\n+       geom_point() + \r\n+       geom_smooth() + \r\n+       labs(title = \"Train vs. Test Predicted Values\", \r\n+            x = \"Train Predicted Values\", y = \"Test Predicted Values\")\r\n+```\r\n+\r\n+Indeed, it does seem to be the case that our model is able to accurately predict\r\n+'new' data, which we would expect, given how the model was derived by\r\n+using the same data.\r\n+\r\n+### f) \r\n+*What do you think are the reasons for using cross-validation?\r\n+\r\n+Cross-validation tests the efficacy of a model by using a subset of the\r\n+available dataset to fit the model, then testing its ability to predict using the\r\n+the remainder of the dataset. This technique is preferable to evaluating \r\n+residuals as it demonstrates how well a model will do when facing 'new data.' In short, having a model without knowing how 'accurately' it can predict new data is not very useful, cross-validation gives us an idea of how well (or not) our model will be able to predict new data. \r\n+\r\n+\r\n+## Question 2:\r\n+\r\n+Perform an Exploratory Data Analysis (EDA) of the OkCupid data, keeping in mind \r\n+in HW-3, you will be fitting a logistic regression to predict gender. What do I mean by\r\n+EDA?\r\n+\r\n+* Visualizations\r\n+* Tables\r\n+* Numerical summaries\r\n+\r\n+```{r, echo=FALSE, cache=TRUE}\r\n+# DO NOT EDIT THIS SECTION!\r\n+profiles <- read.csv(\"profiles.csv\", header=TRUE) %>% tbl_df()\r\n+set.seed(76)\r\n+profiles <- profiles %>% sample_frac(0.1)\r\n+essays <- select(profiles, contains(\"essay\"))\r\n+profiles <- select(profiles, -contains(\"essay\"))\r\n+\r\n+```\r\n+\r\n+The big question we are trying to answer in this section is: Who is in our sample? First, let's explore our sample's basic demographics: \r\n+\r\n+```{r, echo=FALSE, fig.width=12, fig.height=6}\r\n+con_var <- \r\n+           profiles %>% \r\n+           group_by(sex) %>% \r\n+           mutate (\r\n+                    age_mean=mean(age, na.rm=TRUE), \r\n+                    age_sd=sd(age), \r\n+                    height_mean=mean(height, na.rm=TRUE), \r\n+                    height_sd=sd(height, na.rm=TRUE),  \r\n+                    income_mean=mean(ifelse(income>0, income,NA),na.rm=TRUE),\r\n+                    income_sd=sd(ifelse(income>0, income,NA),na.rm=TRUE)\r\n+           ) %>%  \r\n+          select(sex, age_mean:income_sd) %>% \r\n+          slice(which.max(age_mean)) %>%  #We only need one entry per variable per sex\r\n+          gather(\"Stat1\", \"Value\", 2:7) %>%\r\n+          separate (Stat1, c(\"demographic\", \"stat\"), sep=\"_\") %>% \r\n+          spread(demographic, Value)\r\n+                   \r\n+kable(con_var[,c(1, 3:5)], \r\n+      format = \"markdown\",\r\n+      caption=\"Basic Demographics of Okcupid Bay Area Users\", \r\n+      col.names=c(\"Sex\", \"Age\", \"Height\", \"Income\"),  \r\n+      digits=c(2, 2, 2, 2, 2))\r\n+```\r\n+\r\n+\r\n+This table reports the mean and standard deviation of Okcupid users' age, height, and income by sex. In our sample, the average age of male Okcupid user in the Bay Area is 32, while the female average age is 33. As expected, male users are taller on average by about five inches. Men also tend to have higher reported income, $110,000 compared to $81,000 reported by females. It is important to stress that we are unable to verify the accuracy of these reported incomes. Thus, it may very well be that males, on average, tend to inflate their incomes. \r","path":"HW-2 - Draft2.Rmd","position":290,"original_position":290,"commit_id":"e90a85bb593f0e26bfdf445e9b8122a193225e4e","original_commit_id":"e90a85bb593f0e26bfdf445e9b8122a193225e4e","user":{"login":"mhusseinmidd","id":17278733,"avatar_url":"https://avatars.githubusercontent.com/u/17278733?v=3","gravatar_id":"","url":"https://api.github.com/users/mhusseinmidd","html_url":"https://github.com/mhusseinmidd","followers_url":"https://api.github.com/users/mhusseinmidd/followers","following_url":"https://api.github.com/users/mhusseinmidd/following{/other_user}","gists_url":"https://api.github.com/users/mhusseinmidd/gists{/gist_id}","starred_url":"https://api.github.com/users/mhusseinmidd/starred{/owner}{/repo}","subscriptions_url":"https://api.github.com/users/mhusseinmidd/subscriptions","organizations_url":"https://api.github.com/users/mhusseinmidd/orgs","repos_url":"https://api.github.com/users/mhusseinmidd/repos","events_url":"https://api.github.com/users/mhusseinmidd/events{/privacy}","received_events_url":"https://api.github.com/users/mhusseinmidd/received_events","type":"User","site_admin":false},"body":"First one is mean, second is standard deviation. Should have either included a variable *stat* to indicate or referenced this in my report. My bad. ","created_at":"2016-03-24T14:45:33Z","updated_at":"2016-03-24T14:45:33Z","html_url":"https://github.com/Middlebury-Data-Science/HW-2/pull/6#discussion_r57327092","pull_request_url":"https://api.github.com/repos/Middlebury-Data-Science/HW-2/pulls/6","_links":{"self":{"href":"https://api.github.com/repos/Middlebury-Data-Science/HW-2/pulls/comments/57327092"},"html":{"href":"https://github.com/Middlebury-Data-Science/HW-2/pull/6#discussion_r57327092"},"pull_request":{"href":"https://api.github.com/repos/Middlebury-Data-Science/HW-2/pulls/6"}}},"pull_request":{"url":"https://api.github.com/repos/Middlebury-Data-Science/HW-2/pulls/6","id":63827934,"html_url":"https://github.com/Middlebury-Data-Science/HW-2/pull/6","diff_url":"https://github.com/Middlebury-Data-Science/HW-2/pull/6.diff","patch_url":"https://github.com/Middlebury-Data-Science/HW-2/pull/6.patch","issue_url":"https://api.github.com/repos/Middlebury-Data-Science/HW-2/issues/6","number":6,"state":"open","locked":false,"title":"3/22 - HW02-Draft02","user":{"login":"mhusseinmidd","id":17278733,"avatar_url":"https://avatars.githubusercontent.com/u/17278733?v=3","gravatar_id":"","url":"https://api.github.com/users/mhusseinmidd","html_url":"https://github.com/mhusseinmidd","followers_url":"https://api.github.com/users/mhusseinmidd/followers","following_url":"https://api.github.com/users/mhusseinmidd/following{/other_user}","gists_url":"https://api.github.com/users/mhusseinmidd/gists{/gist_id}","starred_url":"https://api.github.com/users/mhusseinmidd/starred{/owner}{/repo}","subscriptions_url":"https://api.github.com/users/mhusseinmidd/subscriptions","organizations_url":"https://api.github.com/users/mhusseinmidd/orgs","repos_url":"https://api.github.com/users/mhusseinmidd/repos","events_url":"https://api.github.com/users/mhusseinmidd/events{/privacy}","received_events_url":"https://api.github.com/users/mhusseinmidd/received_events","type":"User","site_admin":false},"body":"This is the second draft of the second homework assignment. 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