Statistical Graphics and Inference

The Good, The Bad, and The Misleading

January Academy 2016

Chester Ismay (Office: ETC 223)

cismay@reed.edu

http://blogs.reed.edu/datablog

http://blogs.reed.edu/ed-tech

The Importance of Data Visualization

Describe what you see when analyzing the summary information

• What do you think the plots of x versus y look like for the four datasets?

Visualizing this data

• What have you learned from visualizing the data?

The raw Anscombe’s Quartet data

Data is in pairs (x1 goes with y1, x2 goes with y2, etc.)

Visualizations: The Bad

What’s wrong?

What’s wrong?

2015 State of the Union Twitter heat map for Lower 48 US

Source: Glamour News

2010 Population Distribution in the US and Puerto Rico

Source: Census.gov

The Misleading

Common plots and
their uses

Hopefully, The Good

REVIEW: Variable types

• Quantitative (Numeric)
  • It is sensible to add, subtract, or take averages of the data values
  • Continuous - numerical values in a range (fractions and decimals included)
  • Discrete - numerical values with jumps (usually counts)
• Qualitative (Categorical)
  • Values designated by different groupings/categories (usually not numeric)

How plots can assist with INFERENTIAL STATISTICS

Setting up the hypotheses

• What’s the null hypothesis (\(H_0\))?
  • The population mean leniency score is the same for smile versus neutral faces
  • \(H_0: \mu_s = \mu_n\) or \(H_0: \mu_s - \mu_n = 0\)
• What’s the alternative hypothesis (\(H_a\))?
  • The average leniency score is higher for smiling students than it is for students with a neutral facial expression.
  • \(H_a: \mu_s > \mu_n\) or \(H_a: \mu_s - \mu_n > 0\)

Side-by-side Histograms

• One continuous variable versus one categorical variable
• Shows how many values are in different bins
• Questions to ask:
  • What is the shape? (Symmetric, skewed, etc.)
  • How many peaks are there? (Unimodal, bimodal, etc.)
  • How do the values vary?

Side-by-side Boxplots

• One continuous variable versus one categorical variable
• Vertical bars represent quantiles
• Dots represent outliers

Do we have reason to believe,
based on the distributions of leniency scores
over these two expression groups,
that there is a significant increase
in the mean leniency score for
faces that smile compared to neutral faces?

Source: Statistics4u.info

For the Smile Leniency problem

• What’s the sample?
  • The 68 student pictures showing either smiling or neutral expressions
• What’s the population?
  • Ideally, ALL students with either smiling or neutral expressions assigned punishment after an infraction

• We see that the sample mean leniency for those smiling, \(\bar{x}_s\), is greater than the similar measure for those with neutral faces, \(\bar{x}_n\).

• But is it statistically significantly greater?

Assuming the null hypothesis is true…

• Recall that we have \(H_0: \mu_s - \mu_n = 0\). What does this mean in laymen’s terms?
  • We are assuming that there is no relationship in leniency score and type of facial expression.
• We want to see if the difference in sample means could be explained by chance or if it is highly unlikely that chance is a good explanation for seeing a difference of that magnitude or larger.

The Chance Process

• We can use a tactile point of view to explain what “chance” means here:
  • Use \(n_n = 34\) blue index cards corresponding to neutral faces and \(n_s = 34\) red index cards corresponding to faces that smile. What’s next?
  • Write the values of the corresponding leniency score on each of the index cards. What’s next?
  • Put the two stacks of index cards together, creating a new set of 68 cards.

The Chance Process (continued)

• We can use the index cards to create two new stacks for those smiling and those with neutral faces.
  • First, we must shuffle all the cards thoroughly.
  • After doing so, in this case with equal values of sample sizes, we split the deck in half.
  • We then calculate the new sample mean leniency score of the smiling deck, and also the new sample mean leniency score of the neutral deck.
  • This creates one simulation of the samples.
  • We next want to calculate a statistic from these two samples.

The Chance Process (continued again)

• We could do this actual shuffling and calculating, but it’s simpler to let the computer do mundane tasks.

• Recall that the original sample mean difference is \(4.9117647 - 4.1176471 = 0.7941176\).

• From our simulation we obtain a difference of \(-0.4117647\).

• What do we do next?
  • More simulations! Repeat this shuffling process, say, 10,000 times and then look at the distribution of the resulting sample mean differences.

• What’s next?

The p-value!

• Identify how many simulated mean differences are as extreme or more extreme as what we witnessed in our original sample mean difference.
  • Here this value is 0.0244.
• So only 2.44% of the simulated values are at or greater than what we saw with our original sample. Do we have evidence to reject \(H_0: \mu_s - \mu_n = 0\) in favor of \(H_a: \mu_s > \mu_n\)?
  • Yes, our \(p\)-value of 0.0244 is a small proportion so we have little evidence in favor of the null hypothesis.

The p-value Visualized

But I thought this was a
\(t\)-test problem?

• For the \(t\)-test, the test statistic is \(t = \dfrac{\bar{x}_s - \bar{x}_n}{\sqrt{\dfrac{{s_s}^2}{n_s} + \dfrac{{s_n}^2}{n_n}}}\).

• So if we divided all of our simulated statistics by \({\sqrt{\frac{{s_s}^2}{n_s} + \frac{{s_n}^2}{n_n}}}\), we would be on the same scale as the \(t\).

• What would the histogram of transformed values and corresponding \(t\) curve look like plotted together?

Results of t.test

## 
##  Welch Two Sample t-test
## 
## data:  leniency by expression
## t = 2.0415, df = 65.367, p-value = 0.02262
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
##  0.1451043       Inf
## sample estimates:
##   mean in group smile mean in group neutral 
##              4.911765              4.117647

Tying it together

• The \(t\) distribution and also the normal distribution were developed to approximate the “by chance” distribution.

• When they were developed, scientists/statisticians couldn’t replicate the simulation 10,000 times like I have here.

• That’s why the t.test comes with lots of assumptions:
  • Sample sizes must be larger than 30
  • If sample sizes are not large enough, we need to assume the population distributions are normal.
    • That’s a BIG assumption!

Another example of INFERENTIAL STATISTICS and Plotting

Does spending extra time with non-academic activities hinder academic performance?

Scatterplots

• Two continuous variables
• Questions to ask:
  • Pattern - Positive/No/Negative relation
  • Form - Linear/Non-linear
  • Strength - Closely matches fitted line?

How was this “line of best fit” obtained?

Least squares estimation

• Given paired data \((x_1, y_1)\), \((x_2, y_2)\), \(\cdots\), \((x_n, y_n)\), we model via the equation \(y_j = \beta_0 + \beta_1 x_j + \varepsilon_j\).
• We want to minimize the sum of squared error terms (\(\varepsilon_j\)). Solving for \(\varepsilon_j\), we minimize (using partial derivatives from calculus) \[ \sum_{j=1}^n (y_j - \beta_0 - \beta_1 x_j)^2. \]

Statistical software to the rescue!

The equation of the fitted line is

\[ \hat{gpa} = 3.598 -0.006 \times hours \]

So…Is hours doing non-academic things a SIGNIFICANT NEGATIVE predictor of gpa?

• \(H_0 : \beta_1 = 0\) versus \(H_a: \beta_1 < 0\)

• Our sample slope is -0.006. Is this far enough from zero to conclude that this observed slope is significant?

• We can do the same sort of resampling here by shuffling all of the response variable values and assigning them to each of the values of the explanatory variable.

More shuffling!

• We assume that hours and gpa are not related.
• Any \(x\) value could match up with any \(y\) value.
• We shuffle the values of gpa and assign them to values of hours.
• We calculate the simulated slope statistic.
• We then repeat this process, say, 20,000 times and then see where our observed slope statistic \(\hat{\beta}_1\) falls on that distribution.

• The \(p\)-value is 0.0323 so we reject the null hypothesis.
• Based on this sample of university students, we have evidence to conclude that hours doing non-academic things is a significant negative predictor of gpa.
• A similar transformation shows that these results can be well-approximated by a \(t\) distribution.

Data Visualization Examples
on the Web

Links

• Where Are the Hardest Places to Live in the U.S.?
• Think with Google - Travel Trends
• Palo Alto Maintenance Requests
• The Science of Social Timing
• Data Visualization Catalogue
• Why Most Twitter Maps Can’t Be Trusted
• Social Explorer

What can I help you with?

• Data analysis
• Data wrangling/cleaning
• Data visualization
• Data tidying/manipulating
• Reproducible research

When am I available?

• Email me at cismay@reed.edu or chester.ismay@reed.edu
• Tentative Spring 2016 office (ETC 223) hours
  • Mondays (10 AM to 11 AM)
  • Tuesdays (2 PM to 3 PM)
  • Wednesdays (1:30 PM to 2:30 PM)
  • Fridays (1:30 PM to 2:30 PM)
• Sometimes available for virtual office hours via Google Hangouts (email me for details)

Thanks!

cismay@reed.edu

Slides available at http://rpubs.com/cismay/ja_2016

Code for slide creation on my GitHub page