w09 r lecture materials

[stats_class_2019.git] / problem_sets / week_04 / ps4-worked_solution.Rmd

---
title: 'Week 4 Problem set: Worked solutions'
subtitle: "Statistics and statistical programming  \nNorthwestern University  \nMTS 525"
author: "Aaron Shaw"
date: "April 25, 2019"
output: html_document
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```

# Programming challenges


## PC2

You may need to edit these first lines to work on your own machine. Note that for working with .Rmd files interactively in Rstudio you may find it easier to do this using the drop down menus: "Session" → "Set Working Directory" → "To Source File Location" 

```{r}
## setwd("~/Documents/Teaching/2019/stats/")
## list.files("data/week_04")

mobile <- read.csv("data/week_04/COS-Statistics-Mobile_Sessions.csv")
total <- read.csv("data/week_04/COS-Statistics-Gov-Domains-Only.csv")
```

I'll write a little function to help inspect the data. Make sure you understand what the last line of the function is doing.
```{r}
summary.df <- function (d) {
    print(nrow(d))
    print(ncol(d))
    print(head(d))
    print(d[sample(seq(1, nrow(d)), 5),])
}
```

Then I can run these two lines a few times to look at some samples
```{r}
summary.df(mobile)

summary.df(total)
```
I can check for missing values and summarize the different columns using `lapply`:

```{r}
lapply(total, summary)

lapply(mobile, summary)
```

## PC3

First let's create a table/array using `tapply` that sums pageviews per month across all the sites:
```{r}    
total.views.bymonth.tbl <- tapply(total$pageviews, total$month, sum)
total.views.bymonth.tbl
```
If you run `class` on `total.views.bymonth.tbl` you'll notice it's not a data frame yet. We can change that:
```{r}
total.views <- data.frame(months=names(total.views.bymonth.tbl),
                          total=total.views.bymonth.tbl)

head(total.views)
```
Let's cleanup the rownames (this would all work the same if i didn't do this part).

```{r}
rownames(total.views) <- NULL

head(total.views)
```
## PC4
Onwards to the mobile dataset!

Here we have a challenge because we have to estimate total pageviews (it's not given in the raw dataset). I'll do this by multiplying sessions by pages-per-session. This assumes that the original pages-per-session calculation is precise, but I'm not sure what else we could do under the circumstances.
```{r}
mobile$total.pages <- mobile$Sessions * mobile$PagesPerSession 
```
Then, making the views-per-month array is more or less copy/pasted from above:
```{r}
mobile.views.bymonth.tbl <- tapply(mobile$total.pages, mobile$Month, sum)
mobile.views.bymonth.tbl

mobile.views <- data.frame(months=names(mobile.views.bymonth.tbl),
                           mobile=mobile.views.bymonth.tbl)
rownames(mobile.views) <- NULL
```
## PC5
Now we merge the two datasets. Notice that I have created the `months` column in both datasets with *exactly* the same name.
```{r}
views <- merge(mobile.views, total.views, all.x=TRUE, all.y=TRUE, by="months")
```

These are sorted in strange ways and will be difficult to graph because the dates are stored as characters. Let's convert them into Date objects. Then I can use `sort.list` to sort everything.

```{r}
views$months <- as.Date(views$months, format="%m/%d/%Y %H:%M:%S")

views <- views[sort.list(views$months),]
```

Take a look at the data. Some rows are missing observations. We can drop those rows using `complete.cases`:
```{r}
lapply(views, summary)

views[rowSums(is.na(views)) > 0,]

views.complete <- views[complete.cases(views),]
```

## PC6 

For my proportion measure, I'll take the mobile views divided by the total views.

```{r}
views.complete$prop.mobile <- views.complete$mobile / views.complete$total
    
```
## PC7. 

```{r}
library(ggplot2)
ggplot(data=views.complete) + aes(x=months, y=prop.mobile) + geom_point() + geom_line() + scale_y_continuous(limits=c(0, 1))

```

(a) For my estimate of the proportion I'll just calculate an average from the monthly numbers:

```{r}
mean(views.complete$prop.mobile)

```
(b) From the graph, this proportion seems quite stable with the exception of a single outlier month in late 2015. 

# Statistical questions

## SQ1 — 4.8

The general formula for a confidence interval is $point~estimate~±~z^*\times~SE$. First, identify the three different values. The point estimate is 45%, $z^* = 2.58$ for a 99% confidence level (that's the number of standard deviations around the mean that ensure that 99% of a Z-score distribution is included), and $SE = 2.4\%$. 

With this we can plug and chug:

$$52\% ± 2.58 \times 2.4\% → (45.8\%, 58.2\%)$$

From this data we are 99% confident that between 45.8% and 58.2% U.S. adult Twitter users get some news through the site.

## SQ2 — 4.10

(a) False. See the answer to 4.8 above. With $\alpha = 0.01$, we can consult the 99% confidence interval. It includes 50% but also goes lower.

(b) False. The standard error of the sample does not contain any information about the proportion of the population included in the sample. It measures the variability of the sample distribution.

(c) False. Increasing the sample size will decrease the standard error. Consider the formula: $\frac{\sigma}{\sqrt{n}}$. A smaller $n$ will result in a larger standard error.

(d) False. All else being equal, a lower confidence interval will cover a narrower range. A higher interval will cover a wider range. To confirm this, revisit the formula in SQ1 above. and plug in the corresponding alpha value of .9, resulting in a $z^*$ value of 1.28 (see the Z-score table in the back of *OpenIntro*).

## SQ3 — 4.19

The hypotheses should be about the population mean ($\mu$) and not the sample mean ($\bar{x}$). The null hypothesis should have an equal sign. The alternative hypothesis should be about the critical value, not the sample mean. The following would have been better:

$$H_0: \mu = 10~hours$$
$$H_A: \mu \gt 10~hours$$

## SQ4 — 4.32

(a) True. See part (d) of SQ2 above.
(b) False. A lower alpha value is the probability of Type 1 Error, so reducing the one reduces the other.
(c) False. Failure to reject the null is evidence that we cannot conclude that the true value is different from the null. This is **very** different from evidence that the null hypothesis is true.
(d) True. Consult the section of *OpenIntro* discussing statistical power and Type 2 Error. 
(e) True. We'll revisit this in a moment below, but consider the relationship between statistical test, the standard error, and the sample size. As the sample size increases towards infinity, the standard error approaches zero, resulting in arbitrarily precise point estimates that will result in rejecting the null hypothesis for *any* value of a test statistic for any critical value of $\alpha$.

# Empirical paper questions

## EQ1

In my words (or rather formulas since I think that's less ambiguous), the key pairs of null/alternative hypotheses look something like the following:

Let $\Delta$ be the parameter estimate for the difference in mean percentage of positive ($\mu_{pos}$) and negative ($\mu_{neg}$) words between the experimental and control conditions for the treatments of reduced negative content ($R_{neg}$ and reduced positive content ($R_{pos}$).

For the reduced negative content conditions (the left-hand side of Figure 1), the paper tests:

$$HR_{neg}1_0: \Delta_{\mu_{pos}} = 0$$
$$HR_{neg}1_a: \Delta{\mu_{pos}} \gt 0$$
And:
$$HR_{neg}2_0: \Delta_{\mu_{neg}} = 0$$
$$HR_{neg}2_a: \Delta_{\mu_{neg}} \lt 0$$
Then, for the reduced positive content conditions (the right-hand side of Figure 1), the paper tests:

$$HR_{pos}1_0:~~ \Delta_{\mu_{pos}} = 0$$
$$HR_{pos}1_a:~~ \Delta{\mu_{pos}} \lt 0$$

And:

$$HR_{pos}2_0:~~ \Delta_{\mu_{neg}} = 0$$
$$HR_{pos}2_a:~~ \Delta_{\mu_{neg}} \gt 0$$
Note that the theories the authors used to motivate the study imply directions for the alternative hypotheses, but nothing in the description of the analysis suggests that they used one-tailed tests. I've written these all in terms of specific directions here to correspond with the theories stated in the paper. They could also (arguably more accurately) have been written in terms of inequalities ("$\neq$").

## EQ2

The authors' estimates suggest that reduced negative News Feed content causes an increase in the percentage of positive words and a decrease in the percentage of negative words in subsequent News Feed posts by study participants (supporting $HR_{neg}1_a$ and $HR_{neg}2_a$ respectively).

They also find that reduced positive News Feed content causes a decrease in the percentage of negative words and an increase in the percentage of positive words in susbequent News Feed posts (supporting $HR_{pos}1_a$ and $HR_{pos}2_a$)

## EQ3

Cohen's $d$ puts estimates of experimental effects in standardized units (much like a Z-score!) in order to help understand their size relative to the underlying distribution of the dependent variable(s). The d-values for each of the effects estimated in the paper are 0.02, 0.001, 0.02, and 0.008 respectively (in the order presented in the paper, not in order of the hypotheses above!). These are miniscule effects. However, the treatment itself is also quite narrow in scope, suggesting that the presence of any treatment effect at all is an indication of the underlying phenomenon (emotional contagion). Personally, I find it difficult to attribute much substantive significance to the results because I'm not even convinced that tiny shifts in the percentage of positive/negative words used in News Feed updates accurately index meaningful emotional shifts (maybe we could call it linguistic contagion instead?). Despite these concerns and the ethical considerations that attracted so much public attention, I consider this a clever, well-executed study and I think it's quite compelling. I expect many of you will have different opinions of various kinds!