Now we are going to cover how to perform a variety of basic statistical tests in R.

• Correlation
• T-tests/Rank-sum tests
• Linear Regression
• Logistic Regression
• Proportion tests
• Chi-squared
• Fisher’s Exact Test

Note: We will be glossing over the statistical theory and “formulas” for these tests. There are plenty of resources online for learning more about these tests, as well as dedicated Biostatistics series at the School of Public Health

cor() performs correlation in R

cor(x, y = NULL, use = "everything",
    method = c("pearson", "kendall", "spearman"))

Like other functions, if there are NAs, you get NA as the result. But if you specify use only the complete observations, then it will give you correlation on the non-missing data.

library(readr)
circ = read_csv("http://johnmuschelli.com/intro_to_r/data/Charm_City_Circulator_Ridership.csv")
cor(circ$orangeAverage, circ$purpleAverage, use="complete.obs")

[1] 0.9195356

The corrr package allows you to do correlations easily:

library(corrr); library(dplyr)
circ %>% 
  select(orangeAverage, purpleAverage) %>% 
  correlate()

Correlation method: 'pearson'
Missing treated using: 'pairwise.complete.obs'

# A tibble: 2 x 3
  rowname       orangeAverage purpleAverage
  <chr>                 <dbl>         <dbl>
1 orangeAverage        NA             0.920
2 purpleAverage         0.920        NA    

correlate is usually better with more than 2 columns:

avgs = circ %>% select(ends_with("Average"))
cobj = avgs %>% correlate(use = "complete.obs", diagonal = 1)
cobj %>% fashion(decimals = 3)

        rowname orangeAverage purpleAverage greenAverage bannerAverage
1 orangeAverage         1.000          .908         .840          .545
2 purpleAverage          .908         1.000         .867          .521
3  greenAverage          .840          .867        1.000          .453
4 bannerAverage          .545          .521         .453         1.000

cobj %>% rplot()

You can also use cor.test() to test for whether correlation is significant (ie non-zero). Note that linear regression may be better, especially if you want to regress out other confounders.

ct = cor.test(circ$orangeAverage, circ$purpleAverage, 
              use = "complete.obs")
ct

    Pearson's product-moment correlation

data:  circ$orangeAverage and circ$purpleAverage
t = 73.656, df = 991, p-value < 2.2e-16
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 0.9093438 0.9286245
sample estimates:
      cor 
0.9195356 

For many of these testing result objects, you can extract specific slots/results as numbers, as the ct object is just a list.

# str(ct)
names(ct)

[1] "statistic"   "parameter"   "p.value"     "estimate"    "null.value" 
[6] "alternative" "method"      "data.name"   "conf.int"   

ct$statistic

       t 
73.65553 

ct$p.value

[1] 0

The broom package has a tidy function that puts most objects into data.frames so that they are easily manipulated:

library(broom)
tidy_ct = tidy(ct)
tidy_ct

# A tibble: 1 x 8
  estimate statistic p.value parameter conf.low conf.high method     alternative
     <dbl>     <dbl>   <dbl>     <int>    <dbl>     <dbl> <chr>      <chr>      
1    0.920      73.7       0       991    0.909     0.929 Pearson's… two.sided  

Note that you can add the correlation to a plot, via the annotate

library(ggplot2)
txt = paste0("r=", signif(ct$estimate,3))
q = qplot(data = circ, x = orangeAverage, y = purpleAverage)
q + annotate("text", x = 4000, y = 8000, label = txt, size = 5)

The T-test is performed using the t.test() function, which essentially tests for the difference in means of a variable between two groups.

In this syntax, x and y are the column of data for each group.

tt = t.test(circ$orangeAverage, circ$purpleAverage)
tt

    Welch Two Sample t-test

data:  circ$orangeAverage and circ$purpleAverage
t = -17.076, df = 1984, p-value < 2.2e-16
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -1096.7602  -870.7867
sample estimates:
mean of x mean of y 
 3033.161  4016.935 

Using t.test treats the data as independent. Realistically, this data should be treated as a paired t-test. The paired = TRUE argument to do a paired test

t.test(circ$orangeAverage, circ$purpleAverage, paired = TRUE)

    Paired t-test

data:  circ$orangeAverage and circ$purpleAverage
t = -42.075, df = 992, p-value < 2.2e-16
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -799.783 -728.505
sample estimates:
mean of the differences 
               -764.144 

t.test saves a lot of information: the difference in means estimate, confidence interval for the difference conf.int, the p-value p.value, etc.

names(tt)

 [1] "statistic"   "parameter"   "p.value"     "conf.int"    "estimate"   
 [6] "null.value"  "stderr"      "alternative" "method"      "data.name"  

tidy(tt)

# A tibble: 1 x 10
  estimate estimate1 estimate2 statistic  p.value parameter conf.low conf.high
     <dbl>     <dbl>     <dbl>     <dbl>    <dbl>     <dbl>    <dbl>     <dbl>
1    -984.     3033.     4017.     -17.1 4.20e-61     1984.   -1097.     -871.
# … with 2 more variables: method <chr>, alternative <chr>

You can also use the ‘formula’ notation. In this syntax, it is y ~ x, where x is a factor with 2 levels or a binary variable and y is a vector of the same length.

library(tidyr)
long = circ %>% 
  select(date, orangeAverage, purpleAverage) %>% 
  gather(key = line, value = avg, -date)
tt = t.test(avg ~ line, data = long)
tidy(tt)

# A tibble: 1 x 10
  estimate estimate1 estimate2 statistic  p.value parameter conf.low conf.high
     <dbl>     <dbl>     <dbl>     <dbl>    <dbl>     <dbl>    <dbl>     <dbl>
1    -984.     3033.     4017.     -17.1 4.20e-61     1984.   -1097.     -871.
# … with 2 more variables: method <chr>, alternative <chr>

Nonparametric analog to t-test (testing medians):

tidy(wilcox.test(avg ~ line, data = long))

# A tibble: 1 x 4
  statistic  p.value method                                          alternative
      <dbl>    <dbl> <chr>                                           <chr>      
1   336714. 4.56e-58 Wilcoxon rank sum test with continuity correct… two.sided  

Website

The aov function exists for ANOVA, but we’d recommend lm (Linear models):

long3 = circ %>% 
  select(date, orangeAverage, purpleAverage, bannerAverage) %>% 
  gather(key = line, value = avg, -date)
anova_res = aov(avg ~ line, data = long3)
anova_res

Call:
   aov(formula = avg ~ line, data = long3)

Terms:
                      line  Residuals
Sum of Squares  2214242251 3724379811
Deg. of Freedom          2       2396

Residual standard error: 1246.762
Estimated effects may be unbalanced
1039 observations deleted due to missingness

Now we will briefly cover linear regression. I will use a little notation here so some of the commands are easier to put in the proper context. \[ y_i = \alpha + \beta x_i + \varepsilon_i \] where:

• \(y_i\) is the outcome for person i
• \(\alpha\) is the intercept
• \(\beta\) is the slope
• \(x_i\) is the predictor for person i
• \(\varepsilon_i\) is the residual variation for person i

The R version of the regression model is:

y ~ x

where:

• y is your outcome
• x is/are your predictor(s)

For a linear regression, when the predictor is binary this is the same as a t-test:

fit = lm(avg ~ line, data = long)
fit

Call:
lm(formula = avg ~ line, data = long)

Coefficients:
      (Intercept)  linepurpleAverage  
           3033.2              983.8  

‘(Intercept)’ is \(\alpha\)

‘linepurpleAverage’ is \(\beta\)

fit = lm(avg ~ line, data = long3)
fit

Call:
lm(formula = avg ~ line, data = long3)

Coefficients:
      (Intercept)  lineorangeAverage  linepurpleAverage  
            827.3             2205.9             3189.7  

The summary command gets all the additional information (p-values, t-statistics, r-square) that you usually want from a regression.

sfit = summary(fit)
print(sfit)

Call:
lm(formula = avg ~ line, data = long3)

Residuals:
    Min      1Q  Median      3Q     Max 
-4016.9 -1008.4     5.6   973.1  4072.6 

Coefficients:
                  Estimate Std. Error t value Pr(>|t|)    
(Intercept)         827.27      75.88   10.90   <2e-16 ***
lineorangeAverage  2205.89      84.41   26.13   <2e-16 ***
linepurpleAverage  3189.67      85.57   37.27   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1247 on 2396 degrees of freedom
  (1039 observations deleted due to missingness)
Multiple R-squared:  0.3729,    Adjusted R-squared:  0.3723 
F-statistic: 712.2 on 2 and 2396 DF,  p-value: < 2.2e-16

We can tidy linear models as well and it gives us all of this in a tibble:

tidy(fit)

# A tibble: 3 x 5
  term              estimate std.error statistic   p.value
  <chr>                <dbl>     <dbl>     <dbl>     <dbl>
1 (Intercept)           827.      75.9      10.9 4.76e- 27
2 lineorangeAverage    2206.      84.4      26.1 1.16e-132
3 linepurpleAverage    3190.      85.6      37.3 2.98e-240

The confint argument allows for confidence intervals

tidy(fit, conf.int = TRUE)

# A tibble: 3 x 7
  term              estimate std.error statistic   p.value conf.low conf.high
  <chr>                <dbl>     <dbl>     <dbl>     <dbl>    <dbl>     <dbl>
1 (Intercept)           827.      75.9      10.9 4.76e- 27     678.      976.
2 lineorangeAverage    2206.      84.4      26.1 1.16e-132    2040.     2371.
3 linepurpleAverage    3190.      85.6      37.3 2.98e-240    3022.     3357.

http_data_dir = "http://johnmuschelli.com/intro_to_r/data/"
cars = read_csv(
  paste0(http_data_dir, "kaggleCarAuction.csv"),   
  col_types = cols(VehBCost = col_double()))
head(cars)

# A tibble: 6 x 34
  RefId IsBadBuy PurchDate Auction VehYear VehicleAge Make  Model Trim  SubModel
  <dbl>    <dbl> <chr>     <chr>     <dbl>      <dbl> <chr> <chr> <chr> <chr>   
1     1        0 12/7/2009 ADESA      2006          3 MAZDA MAZD… i     4D SEDA…
2     2        0 12/7/2009 ADESA      2004          5 DODGE 1500… ST    QUAD CA…
3     3        0 12/7/2009 ADESA      2005          4 DODGE STRA… SXT   4D SEDA…
4     4        0 12/7/2009 ADESA      2004          5 DODGE NEON  SXT   4D SEDAN
5     5        0 12/7/2009 ADESA      2005          4 FORD  FOCUS ZX3   2D COUP…
6     6        0 12/7/2009 ADESA      2004          5 MITS… GALA… ES    4D SEDA…
# … with 24 more variables: Color <chr>, Transmission <chr>, WheelTypeID <chr>,
#   WheelType <chr>, VehOdo <dbl>, Nationality <chr>, Size <chr>,
#   TopThreeAmericanName <chr>, MMRAcquisitionAuctionAveragePrice <chr>,
#   MMRAcquisitionAuctionCleanPrice <chr>,
#   MMRAcquisitionRetailAveragePrice <chr>,
#   MMRAcquisitonRetailCleanPrice <chr>, MMRCurrentAuctionAveragePrice <chr>,
#   MMRCurrentAuctionCleanPrice <chr>, MMRCurrentRetailAveragePrice <chr>,
#   MMRCurrentRetailCleanPrice <chr>, PRIMEUNIT <chr>, AUCGUART <chr>,
#   BYRNO <dbl>, VNZIP1 <dbl>, VNST <chr>, VehBCost <dbl>, IsOnlineSale <dbl>,
#   WarrantyCost <dbl>

We’ll look at vehicle odometer value by vehicle age:

fit = lm(VehOdo~VehicleAge, data = cars)
print(fit)

Call:
lm(formula = VehOdo ~ VehicleAge, data = cars)

Coefficients:
(Intercept)   VehicleAge  
      60127         2723  

Note that you can have more than 1 predictor in regression models.The interpretation for each slope is change in the predictor corresponding to a one-unit change in the outcome, holding all other predictors constant.

fit2 = lm(VehOdo ~ IsBadBuy + VehicleAge, data = cars)
tidy(fit2)  

# A tibble: 3 x 5
  term        estimate std.error statistic  p.value
  <chr>          <dbl>     <dbl>     <dbl>    <dbl>
1 (Intercept)   60142.     135.     446.   0.      
2 IsBadBuy       1329.     158.       8.42 3.83e-17
3 VehicleAge     2680.      30.3     88.5  0.      

The * does interactions:

fit3 = lm(VehOdo ~ IsBadBuy * VehicleAge, data = cars)
tidy(fit3)  

# A tibble: 4 x 5
  term                estimate std.error statistic p.value
  <chr>                  <dbl>     <dbl>     <dbl>   <dbl>
1 (Intercept)         60140.       143.   420.     0      
2 IsBadBuy             1347.       456.     2.95   0.00315
3 VehicleAge           2681.        32.5   82.4    0      
4 IsBadBuy:VehicleAge    -3.79      88.7   -0.0427 0.966  

You can take out main effects with minus

fit4 = lm(VehOdo ~ IsBadBuy * VehicleAge -IsBadBuy , data = cars)
tidy(fit4)  

# A tibble: 3 x 5
  term                estimate std.error statistic  p.value
  <chr>                  <dbl>     <dbl>     <dbl>    <dbl>
1 (Intercept)           60272.     136.     443.   0.      
2 VehicleAge             2653.      31.1     85.2  0.      
3 IsBadBuy:VehicleAge     242.      30.7      7.89 3.19e-15

tidy(glmfit, conf.int = TRUE)

# A tibble: 3 x 7
  term           estimate   std.error statistic  p.value    conf.low  conf.high
  <chr>             <dbl>       <dbl>     <dbl>    <dbl>       <dbl>      <dbl>
1 (Intercept) -3.78       0.0638         -59.2  0.       -3.90       -3.65     
2 VehOdo       0.00000834 0.000000853      9.78 1.33e-22  0.00000667  0.0000100
3 VehicleAge   0.268      0.00677         39.6  0.        0.255       0.281    

tidy(glmfit, conf.int = TRUE, exponentiate = TRUE)

# A tibble: 3 x 7
  term        estimate   std.error statistic  p.value conf.low conf.high
  <chr>          <dbl>       <dbl>     <dbl>    <dbl>    <dbl>     <dbl>
1 (Intercept)   0.0229 0.0638         -59.2  0.         0.0202    0.0259
2 VehOdo        1.00   0.000000853      9.78 1.33e-22   1.00      1.00  
3 VehicleAge    1.31   0.00677         39.6  0.         1.29      1.32  

Note the coefficients are on the original scale, we must exponentiate them for odds ratios:

exp(coef(glmfit))

(Intercept)      VehOdo  VehicleAge 
 0.02286316  1.00000834  1.30748911 

chisq.test() performs chi-squared contingency table tests and goodness-of-fit tests.

chisq.test(x, y = NULL, correct = TRUE,
           p = rep(1/length(x), length(x)), rescale.p = FALSE,
           simulate.p.value = FALSE, B = 2000)

tab = table(cars$IsBadBuy, cars$IsOnlineSale)
tab

   
        0     1
  0 62375  1632
  1  8763   213

You can also pass in a table object (such as tab here)

cq = chisq.test(tab)
cq

    Pearson's Chi-squared test with Yates' continuity correction

data:  tab
X-squared = 0.92735, df = 1, p-value = 0.3356

names(cq)

[1] "statistic" "parameter" "p.value"   "method"    "data.name" "observed" 
[7] "expected"  "residuals" "stdres"   

cq$p.value

[1] 0.3355516

Note that does the same test as prop.test, for a 2x2 table (prop.test not relevant for greater than 2x2).

chisq.test(tab)

    Pearson's Chi-squared test with Yates' continuity correction

data:  tab
X-squared = 0.92735, df = 1, p-value = 0.3356

prop.test(tab)

    2-sample test for equality of proportions with continuity correction

data:  tab
X-squared = 0.92735, df = 1, p-value = 0.3356
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.005208049  0.001673519
sample estimates:
   prop 1    prop 2 
0.9745028 0.9762701 

fisher.test() performs contingency table test using the hypogeometric distribution (used for small sample sizes).

fisher.test(x, y = NULL, workspace = 200000, hybrid = FALSE,
            control = list(), or = 1, alternative = "two.sided",
            conf.int = TRUE, conf.level = 0.95,
            simulate.p.value = FALSE, B = 2000)

fisher.test(tab)

    Fisher's Exact Test for Count Data

data:  tab
p-value = 0.3324
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.8001727 1.0742114
sample estimates:
odds ratio 
 0.9289923 

Also, if you want to only plot a subset of the data (for speed/time or overplotting)

samp.cars = sample_n(cars, size = 10000)
samp.cars = sample_frac(cars, size = 0.2)
ggplot(aes(x = VehBCost, y = VehOdo), 
       data = samp.cars) + geom_point() 

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