Statistical inference: one and two-sample t-tests

Statistical Inference and Science

• Previously: descriptive statistics. “Here are data; what do they say?”.
• May need to take some action based on information in data.
• Or want to generalize beyond data (sample) to larger world (population).
• Science: first guess about how world works.
• Then collect data, by sampling.
• Is guess correct (based on data) for whole world, or not?

Sample data are imperfect

• Sample data never entirely represent what you’re observing.
• There is always random error present.
• Thus you can never be entirely certain about your conclusions.
• The Toronto Blue Jays’ average home attendance in part of 2015 season was 25,070 (up to May 27 2015, from baseball-reference.com).
• Does that mean the attendance at every game was exactly 25,070? Certainly not. Actual attendance depends on many things, eg.:
  • how well the Jays are playing
  • the opposition
  • day of week
  • weather
  • random chance

Packages for this section

library(tidyverse)

Reading the attendances

…as a .csv file:

my_url <- "http://ritsokiguess.site/datafiles/jays15-home.csv"
jays <- read_csv(my_url) 
jays

Another way

• This is a “big” data set: only 25 observations, but a lot of variables.

• To see the first few values in all the variables, can also use glimpse:

glimpse(jays)

Rows: 25
Columns: 21
$ row         <dbl> 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96…
$ game        <dbl> 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 27, 28, 29, 30, 31, 3…
$ date        <chr> "Monday, Apr 13", "Tuesday, Apr 14", "Wednesday, Apr 15", …
$ box         <chr> "boxscore", "boxscore", "boxscore", "boxscore", "boxscore"…
$ team        <chr> "TOR", "TOR", "TOR", "TOR", "TOR", "TOR", "TOR", "TOR", "T…
$ venue       <lgl> NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA…
$ opp         <chr> "TBR", "TBR", "TBR", "TBR", "ATL", "ATL", "ATL", "BAL", "B…
$ result      <chr> "L", "L", "W", "L", "L", "W-wo", "L", "W", "W", "W", "W", …
$ runs        <dbl> 1, 2, 12, 2, 7, 6, 2, 13, 4, 7, 3, 3, 5, 7, 7, 3, 10, 2, 3…
$ Oppruns     <dbl> 2, 3, 7, 4, 8, 5, 5, 6, 2, 6, 1, 6, 1, 0, 1, 6, 6, 3, 4, 4…
$ innings     <dbl> NA, NA, NA, NA, NA, 10, NA, NA, NA, NA, NA, NA, NA, NA, NA…
$ wl          <chr> "4-3", "4-4", "5-4", "5-5", "5-6", "6-6", "6-7", "7-7", "8…
$ position    <dbl> 2, 3, 2, 4, 4, 3, 4, 2, 2, 1, 4, 5, 3, 3, 3, 3, 5, 5, 5, 5…
$ gb          <chr> "1", "2", "1", "1.5", "2.5", "1.5", "1.5", "2", "1", "Tied…
$ winner      <chr> "Odorizzi", "Geltz", "Buehrle", "Archer", "Martin", "Cecil…
$ loser       <chr> "Dickey", "Castro", "Ramirez", "Sanchez", "Cecil", "Marimo…
$ save        <chr> "Boxberger", "Jepsen", NA, "Boxberger", "Grilli", NA, "Gri…
$ `game time` <time> 02:30:00, 03:06:00, 03:02:00, 03:00:00, 03:09:00, 02:41:0…
$ Daynight    <chr> "N", "N", "N", "N", "N", "D", "D", "N", "N", "N", "N", "N"…
$ attendance  <dbl> 48414, 17264, 15086, 14433, 21397, 34743, 44794, 14184, 15…
$ streak      <chr> "-", "--", "+", "-", "--", "+", "-", "+", "++", "+++", "+"…

Attendance histogram

ggplot(jays, aes(x = attendance)) + geom_histogram(bins = 6)

Comments

• Attendances have substantial variability, ranging from just over 10,000 to around 50,000.
• Distribution somewhat skewed to right (but no outliers).
• These are a sample of “all possible games” (or maybe “all possible games played in April and May”). What can we say about mean attendance in all possible games based on this evidence?
• Think about:
  • Confidence interval
  • Hypothesis test.

Getting CI for mean attendance

• t.test function does CI and test. Look at CI first:

t.test(jays$attendance)

    One Sample t-test

data:  jays$attendance
t = 11.389, df = 24, p-value = 3.661e-11
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 20526.82 29613.50
sample estimates:
mean of x 
 25070.16 

• From 20,500 to 29,600.

Or, 90% CI

• by including a value for conf.level:

t.test(jays$attendance, conf.level = 0.90)

    One Sample t-test

data:  jays$attendance
t = 11.389, df = 24, p-value = 3.661e-11
alternative hypothesis: true mean is not equal to 0
90 percent confidence interval:
 21303.93 28836.39
sample estimates:
mean of x 
 25070.16 

• From 21,300 to 28,800. (Shorter, as it should be.)

Comments

• Need to say “column attendance within data frame jays” using $.
• 95% CI from about 20,000 to about 30,000.
• Not estimating mean attendance well at all!
• Generally want confidence interval to be shorter, which happens if:
  • SD smaller
  • sample size bigger
  • confidence level smaller
• Last one is a cheat, really, since reducing confidence level increases chance that interval won’t contain pop. mean at all!

Another way to access data frame columns

with(jays, t.test(attendance))

    One Sample t-test

data:  attendance
t = 11.389, df = 24, p-value = 3.661e-11
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 20526.82 29613.50
sample estimates:
mean of x 
 25070.16 

Hypothesis test

• CI answers question “what is the mean?”
• Might have a value \(\mu\) in mind for the mean, and question “Is the mean equal to \(\mu\), or not?”
• For example, 2014 average attendance was 29,327.
• “Is the mean this?” answered by hypothesis test.
• Value being assessed goes in null hypothesis: here, \(H_0 : \mu = 29327\).
• Alternative hypothesis says how null might be wrong, eg. \(H_a : \mu \ne 29327\).
• Assess evidence against null. If that evidence strong enough, reject null hypothesis; if not, fail to reject null hypothesis (sometimes retain null).
• Note asymmetry between null and alternative, and utter absence of word “accept”.

\(\alpha\) and errors

• Hypothesis test ends with decision:
  • reject null hypothesis
  • do not reject null hypothesis.
• but decision may be wrong:

• Either type of error is bad, but for now focus on controlling Type I error: write \(\alpha\) = P(type I error), and devise test so that \(\alpha\) small, typically 0.05.
• That is, if null hypothesis true, have only small chance to reject it (which would be a mistake).
• Worry about type II errors later (when we consider power of test).

Why 0.05? This man.

Why 0.05? (2)

• From The Arrangement of Field Experiments (1926):

• and

Three steps:

• from data to test statistic
  • how far are data from null hypothesis
• from test statistic to P-value
  • how likely are you to see “data like this” if the null hypothesis is true
• from P-value to decision
  • reject null hypothesis if P-value small enough, fail to reject it otherwise

Using t.test:

t.test(jays$attendance, mu=29327)

    One Sample t-test

data:  jays$attendance
t = -1.9338, df = 24, p-value = 0.06502
alternative hypothesis: true mean is not equal to 29327
95 percent confidence interval:
 20526.82 29613.50
sample estimates:
mean of x 
 25070.16 

• See test statistic \(-1.93\), P-value 0.065.
• Do not reject null at \(\alpha=0.05\): no evidence that mean attendance has changed.

Assumptions

• Theory for \(t\)-test: assumes normally-distributed data.
• What actually matters is sampling distribution of sample mean: if this is approximately normal, \(t\)-test is OK, even if data distribution is not normal.
• Central limit theorem: if sample size large, sampling distribution approx. normal even if data distribution somewhat non-normal.
• So look at shape of data distribution, and make a call about whether it is normal enough, given the sample size.

Blue Jays attendances again:

• You might say that this is not normal enough for a sample size of \(n = 25\), in which case you don’t trust the \(t\)-test result:

ggplot(jays, aes(x = attendance)) + geom_histogram(bins = 6)

Another example: learning to read

• You devised new method for teaching children to read.
• Guess it will be more effective than current methods.
• To support this guess, collect data.
• Want to generalize to “all children in Canada”.
• So take random sample of all children in Canada.
• Or, argue that sample you actually have is “typical” of all children in Canada.
• Randomization (1): whether or not a child in sample or not has nothing to do with anything else about that child.
• Randomization (2): randomly choose whether each child gets new reading method (t) or standard one (c).

Reading in data

• File at http://ritsokiguess.site/datafiles/drp.txt.
• Proper reading-in function is read_delim (check file to see)
• Read in thus:

my_url <- "http://ritsokiguess.site/datafiles/drp.txt"
kids <- read_delim(my_url," ")

The data

kids

In group, t is “treatment” (the new reading method) and c is “control” (the old one).

Boxplots

ggplot(kids, aes(x = group, y = score)) + geom_boxplot()

Two kinds of two-sample t-test

• pooled (derived in B57): \(t = { \bar{x}_1 - \bar{x}_2 \over s_p \sqrt{(1 / n_1) + (1 / n_2)}}\),
  • where \(s_p^2 = {(n_1 - 1) s_1^2 + (n_2 - 1)s_2^2 \over n_1 + n_2 -2}\)
• Welch-Satterthwaite: \(t = {\bar{x}_1 - \bar{x}_2 \over \sqrt {{s_1^2 / n_1} + {s_2^2 / n_2}}}\)
  • this \(t\) does not have exact \(t\)-distribution, but is approx \(t\) with non-integer df.

Two kinds of two-sample t-test

• Do the two groups have same spread (SD, variance)?
  • If yes (shaky assumption here), can use pooled t-test.
  • If not, use Welch-Satterthwaite t-test (safe).
• Pooled test derived in STAB57 (easier to derive).
• Welch-Satterthwaite is test used in STAB22 and is generally safe.
• Assess (approx) equality of spreads using boxplot.

The (Welch-Satterthwaite) t-test

• c (control) before t (treatment) alphabetically, so proper alternative is “less”.
• R does Welch-Satterthwaite test by default
• Answer to “does the new reading program really help?”
• (in a moment) how to get R to do pooled test?

Welch-Satterthwaite

t.test(score ~ group, data = kids, alternative = "less")

    Welch Two Sample t-test

data:  score by group
t = -2.3109, df = 37.855, p-value = 0.01319
alternative hypothesis: true difference in means between group c and group t is less than 0
95 percent confidence interval:
      -Inf -2.691293
sample estimates:
mean in group c mean in group t 
       41.52174        51.47619 

The pooled t-test

t.test(score ~ group, data = kids, 
       alternative = "less", var.equal = TRUE)

    Two Sample t-test

data:  score by group
t = -2.2666, df = 42, p-value = 0.01431
alternative hypothesis: true difference in means between group c and group t is less than 0
95 percent confidence interval:
      -Inf -2.567497
sample estimates:
mean in group c mean in group t 
       41.52174        51.47619 

Two-sided test; CI

• To do 2-sided test, leave out alternative:

t.test(score ~ group, data = kids)

    Welch Two Sample t-test

data:  score by group
t = -2.3109, df = 37.855, p-value = 0.02638
alternative hypothesis: true difference in means between group c and group t is not equal to 0
95 percent confidence interval:
 -18.67588  -1.23302
sample estimates:
mean in group c mean in group t 
       41.52174        51.47619 

Comments:

• P-values for pooled and Welch-Satterthwaite tests very similar (even though the pooled test seemed inferior): 0.013 vs. 0.014.
• Two-sided test also gives CI: new reading program increases average scores by somewhere between about 1 and 19 points.
• Confidence intervals inherently two-sided, so do 2-sided test to get them.

Jargon for testing

• Alternative hypothesis: what we are trying to prove (new reading program is effective).
• Null hypothesis: “there is no difference” (new reading program no better than current program). Must contain “equals”.
• One-sided alternative: trying to prove better (as with reading program).
• Two-sided alternative: trying to prove different.
• Test statistic: something expressing difference between data and null (eg. difference in sample means, \(t\) statistic).
• P-value: probability of observing test statistic value as extreme or more extreme, if null is true.
• Decision: either reject null hypothesis or do not reject null hypothesis. Never “accept”.

Logic of testing

• Work out what would happen if null hypothesis were true.
• Compare to what actually did happen.
• If these are too far apart, conclude that null hypothesis is not true after all. (Be guided by P-value.)
• As applied to our reading programs:
  • If reading programs equally good, expect to see a difference in means close to 0.
  • Mean reading score was 10 higher for new program.
  • Difference of 10 was unusually big (P-value small from t-test). So conclude that new reading program is effective.
• Nothing here about what happens if null hypothesis is false. This is power and type II error probability.