Alternatives and two-sample t-test

Linguistic Data: Quantitative Analysis and Visualisation: linguistic theory

Alternatives

my_sample <- c(80, 90, 100, 100, 130)
mu0 <- 125
mean(my_sample)

[1] 100

\[H_0\colon \mu > \mu_0\]

t.test(my_sample, mu = mu0, alternative = 'greater')

    One Sample t-test

data:  my_sample
t = -2.9881, df = 4, p-value = 0.9798
alternative hypothesis: true mean is greater than 125
95 percent confidence interval:
 82.16369      Inf
sample estimates:
mean of x 
      100 

x <- seq(-2.9881, 5, 0.01)
curve(dt(t, df=4), from = -5, to=5, xname = 't')

# add shading
polygon(c(x[1], x, x[length(x)]), 
        c(0, dt(x, df=4), 0), col="steelblue")

With this alternative, p-value is “probability to get sample mean that is equal to sample mean we actually obtained (100) or more (i.e. ”more favoring alternative”), provided that null hypothesis holds (i.e. population mean = 125)“.

Exercise

Find p-value for alternative = ‘less’

\[H_1\colon \mu < \mu_0\]

We can do it without function t.test. It should be equal to the opposite area:

x <- seq(-5, -2.9881, 0.01)
curve(dt(t, df=4), from = -5, to=5, xname='t')

# add shading
polygon(c(x[1], x, x[length(x)]), 
        c(0, dt(x, df=4), 0), col="steelblue")

The first area (0.9798) is probability that \(\bar x\) is larger than 100 provided that null hypothesis holds. The second area (we are trying te find) is probability that \(\bar x\) is less than 100 provided than null hypothesis holds. As \(\bar x\) either larger or less than 100 (it is almost impossible that it’s equal to 100 exactly), then sum of probabilities should be 1. So, p-value for alternative “less” is

1 - 0.9798

Let’s check it:

t.test(my_sample, mu = mu0, alternative = 'less')

    One Sample t-test

data:  my_sample
t = -2.9881, df = 4, p-value = 0.02021
alternative hypothesis: true mean is less than 125
95 percent confidence interval:
     -Inf 117.8363
sample estimates:
mean of x 
      100 

Exact hit!

Exercise

Find p-value for alternative = ‘two.sided’

Now we have:

\[H_1\colon \mu \ne \mu_0\] Again, we can do it without t.test, then check.

x <- seq(-5, -2.9881, 0.01)
curve(dt(t, df=4), from = -5, to=5, xname='t')

# add shading
polygon(c(x[1], x, x[length(x)]), 
        c(0, dt(x, df=4), 0), col="steelblue")

# add shading symmetrical
polygon(-c(x[1], x, x[length(x)]), 
        c(0, dt(-x, df=4), 0), col="steelblue")

Now we can reject null hypothesis either if \(\bar x\) is too small or if \(\bar x\) is too large. Thus p-value is an area of symmetric figure shown on the picture. And it’s twice as large as p-value for alternative “less”.

0.02021 * 2

[1] 0.04042

Let’s check

t.test(my_sample, mu = mu0, alternative = 'two.sided')

    One Sample t-test

data:  my_sample
t = -2.9881, df = 4, p-value = 0.04042
alternative hypothesis: true mean is not equal to 125
95 percent confidence interval:
  76.77059 123.22941
sample estimates:
mean of x 
      100 

Two-sample t-test

x <- c(3, 2, 5, 1, 7) # treatment group
y <- c(4, 5, 3, 5, 7, 8) # control group
t.test(x, y, alternative = 'less')

    Welch Two Sample t-test

data:  x and y
t = -1.3149, df = 7.4904, p-value = 0.1137
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
      -Inf 0.7398083
sample estimates:
mean of x mean of y 
 3.600000  5.333333 

t.test(x, y, alternative = 'two.sided')

    Welch Two Sample t-test

data:  x and y
t = -1.3149, df = 7.4904, p-value = 0.2274
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -4.809596  1.342930
sample estimates:
mean of x mean of y 
 3.600000  5.333333 

t.test(rep(x, 3), rep(y, 3), alternative = 'less')

    Welch Two Sample t-test

data:  rep(x, 3) and rep(y, 3)
t = -2.448, df = 26.335, p-value = 0.01067
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
      -Inf -0.526204
sample estimates:
mean of x mean of y 
 3.600000  5.333333 

Let’s use data from thesis “Vowel duration and aspiration effects in Icelandic” by Stefano Coretta, University of York, 2016.

dat <- read.csv("https://bit.ly/39Fr0gD")
dat

Is it true that time for round sounds is greater or less than time for unrounded sounds?

Get values of time variable where roundness == “round”

round_times <- dat[dat$roundness == "round", "time"]
unrounded_times <- dat[dat$roundness == "unrounded", "time"]

round_times[1:20]

 [1] 438.02832 440.61933 443.07138 132.67144 137.56550
 [6] 143.38381  45.93431  49.32262  51.80259 309.72030
[11] 312.01393 314.33173 119.19929 122.31784 125.55692
[16] 336.11847 338.59150 341.18800  64.25130  69.46999

mean(round_times)

[1] 295.8432

sd(round_times)

[1] 196.1767

length(round_times)

[1] 300

mean(unrounded_times)

[1] 309.2343

sd(unrounded_times)

[1] 181.4159

length(unrounded_times)

[1] 506

Exercise

Use t.test to test a hypothesis and give an answer: can we say that there is a statistically significant difference between round and unrounded times?

t.test(round_times, unrounded_times)

    Welch Two Sample t-test

data:  round_times and unrounded_times
t = -0.9631, df = 589.36, p-value = 0.3359
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -40.69893  13.91676
sample estimates:
mean of x mean of y 
 295.8432  309.2343 

p-value = 0.33 > 0.05 => do not reject H0

This can be made easier:

t.test(time ~ roundness, data=dat)

    Welch Two Sample t-test

data:  time by roundness
t = -0.9631, df = 589.36, p-value = 0.3359
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -40.69893  13.91676
sample estimates:
    mean in group round mean in group unrounded 
               295.8432                309.2343 

Let’s repeat with different variable:

t.test(vowel.dur ~ roundness, data=dat)

    Welch Two Sample t-test

data:  vowel.dur by roundness
t = 4.3548, df = 554.3, p-value = 1.587e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
  4.463876 11.799644
sample estimates:
    mean in group round mean in group unrounded 
               91.79000                83.65824 

Now we have very small p-value (\(\texttt{1.587e-05} = 1.587\cdot 10^{-5}=0.00001587\), that’s smaller than 0.05 — therefore, reject null hypothesis, i.e. we have statistically significant difference.

Visualization

boxplot(vowel.dur ~ roundness, data=dat)