Decoding the Student’s t-Distribution

Decoding the Student’s \(t\)-Distribution

Mastering Statistical Analysis with Small Sample Sizes

Hypothesis Testing (\(z\)-Test)

Assumptions

  • Independent random samples
  • Sample size is large enough (> 30) and/or Population variance is known

\[z = \frac{\bar x - \mu}{\sigma / \sqrt{n}} \sim N(0, 1)\]

where \(\sigma\) is the population standard deviation.

Student’s \(t\)-Distribution

\[t = \frac{\bar x - \mu}{{s}/{\sqrt{n}}} \sim t(n-1)\]

where \(s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar x)^2}\) is the sample standard deviation, and \(n-1\) is the degrees of freedom (df).

Why \(n-1\) degrees of freedom?

\[\bar x = \frac{x_1 + x_2 + ... + x_n}{n}\]

\[x_1 + x_2 + ... + \color{red}{x_n} = n \cdot \bar x\]

Degrees of freedom also refers to the number of independent observations in a sample of data.

Student’s \(t\)-Distribution

The \(t\)-distribution is a continuous probability distribution that generalizes the standard normal distribution with a heavier tail.

ggplot(data.frame(x = c(-5, 5)), aes(x = x)) +
  stat_function(fun = dt, args = list(df = 10)) +
  labs(x = "Value", y = "Density", title="t-distribution with 10 degrees of freedom") +
  theme_minimal()

\(t\)-Distribution vs. Normal Distribution

ggplot(data.frame(x = c(-5, 5)), aes(x = x)) +
  stat_function(fun = dt, args = list(df = 1), color = "orange") +
  stat_function(fun = dt, args = list(df = 2), color = "purple") +
  stat_function(fun = dt, args = list(df = 5), color = "blue") +
  stat_function(fun = dt, args = list(df = Inf), color = "black") +
  stat_function(fun = dnorm, color = "red", linewidth = 1, linetype="dashed") +
  labs(x = "Value", y = "Density", title="t-distribution (solid) vs. z-distribution (dashed)") +
  scale_color_manual(name = "Distribution", 
                     values = c("df = 1" = "orange", "df = 2" = "purple", "df = 5" = "blue",
                                "df = Inf" = "black", "Normal (z)" = "red")) + 
  theme_minimal()

sd Function in R

Steps for Hypothesis Testing

  • State the null (\(H_0\)) and alternative (\(H_a\)) hypotheses
  • Specify the significance level, \(\alpha\)
  • Compute the test statistic \(t\) (critical value method)
  • Compute the \(p\)-value (\(p\)-value method)
  • State the conclusion

Types of \(t\)-Tests

  • One-sample \(t\)-test: verify a claim about the population mean
  • Two-sample \(t\)-test: compare two population means
  • Paired \(t\)-test: compare two population means with paired data

One-Sample \(t\)-Test

A computer software company claims that a new version of its operating system will crash fewer than six times per year on average. A system administrator installs the operating system on a random sample of 41 computers. At the end of year, the sample mean number of crashes is 7.1, with a standard deviation of 3.6. Can you conclude that the vendor’s claim is false? Use the \(\alpha = 0.05\) significance level.

  • \(H_0\): \(\mu = 6\)
  • \(H_a\): \(\mu > 6\)

One-Sample \(t\)-Test

x_bar = 7.1 
s = 3.6

mu = 6
n = 41

t = (x_bar - mu) / (s / sqrt(n))
t_cv = qt(0.95, n - 1)
p_value = pt(t, n - 1, lower.tail = FALSE) # default is lower.tail = TRUE

lower_ci = x_bar - t_cv * s / sqrt(n)
upper_ci = x_bar + t_cv * s / sqrt(n)

print(str_glue("t = {round(t, 3)}, t_cv = {round(t_cv, 3)}"))
t = 1.957, t_cv = 1.684
print(str_glue("p-value = {round(p_value, 3)}"))
p-value = 0.029
print(str_glue("CI = ({round(lower_ci, 3)}, {round(upper_ci, 3)})"))
CI = (6.153, 8.047)

Since 1.957 > 1.684 and 0.029 < \(0.05\), we reject \(H_0\) and conclude that the vendor’s claim is false.

One-Sample \(t\)-Test

Suppose you have a batch of batteries, and the manufacturer claims that these batteries last an average of 300 hours. You want to test this claim by taking a sample of batteries and testing their lifespan. \[310, 298, 305, 280, 315, 290, 300, 305, 295, 320\]

  • \(H_0\): \(\mu = 300\)
  • \(H_a\): \(\mu \neq 300\)
batteries = c(310, 298, 305, 280, 315, 290, 300, 305, 295, 320)
print(str_glue("Sample mean: {mean(batteries)}, Sample standard deviation: {round(sd(batteries), 3)}"))
Sample mean: 301.8, Sample standard deviation: 11.887

One-Sample \(t\)-Test

result = t.test(batteries, mu = 300, alternative = "two.sided", conf.level = 0.95)
print(result)

    One Sample t-test

data:  batteries
t = 0.47887, df = 9, p-value = 0.6435
alternative hypothesis: true mean is not equal to 300
95 percent confidence interval:
 293.2969 310.3031
sample estimates:
mean of x 
    301.8
  • We fail to reject \(H_0\) at \(\alpha = 0.05\) and conclude that there is no enough evidence to conclude that the mean lifespan of the batteries is different from 300 hours.

Pooled Variance

  • When the population variances are equal, we can pool the sample variances.
  • The pooled variance is a weighted average of the sample variances. \[s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}\]
  • leveneTest from car package can be used to test the equality of variances.
  • A common rule of thumb is that if the largest sample standard deviation is not more than twice the smallest standard deviation, then the variances are considered equal.
  • var.equal = TRUE in t.test() function.

\[t = \frac{\bar x_1 - \bar x_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\]

Two-Sample \(t\)-Test

Suppose you want to compare the average lifespan of two different brands of batteries. \[\begin{aligned} \text{Brand A: } & 310, 298, 305, 280, 315, 290, 300, 305, 295, 320 \\ \text{Brand B: } & 295, 290, 285, 305, 290, 285, 280, 300, 295, 305, 292, 285 \end{aligned}\]

  • \(H_0\): \(\mu_A = \mu_B\) or \(\mu_A - \mu_B = 0\)
  • \(H_a\): \(\mu_A \neq \mu_B\) or \(\mu_A - \mu_B \neq 0\)
batteries_A = c(310, 298, 305, 280, 315, 290, 300, 305, 295, 320)
batteries_B = c(295, 290, 285, 305, 290, 285, 280, 300, 295, 305, 292, 285)

print(str_glue("Sample mean of Brand A: {mean(batteries_A)}, Sample mean of Brand B: {mean(batteries_B)}"))
Sample mean of Brand A: 301.8, Sample mean of Brand B: 292.25
print(str_glue("Sample standard deviation of Brand A: {round(sd(batteries_A), 3)}, Sample standard deviation of Brand B: {round(sd(batteries_B), 3)}"))
Sample standard deviation of Brand A: 11.887, Sample standard deviation of Brand B: 8.081

Two-Sample \(t\)-Test

result = t.test(batteries_A, batteries_B, alternative = "two.sided", var.equal = TRUE)
print(result)

    Two Sample t-test

data:  batteries_A and batteries_B
t = 2.2361, df = 20, p-value = 0.0369
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
  0.6411308 18.4588692
sample estimates:
mean of x mean of y 
   301.80    292.25
  • We reject \(H_0\) at \(\alpha = 0.05\) and conclude that the mean lifespan of the batteries is different between the two brands.

Paired \(t\)-Test

You are testing the effectiveness of a new diet program. A group of individuals’ weights are recorded before starting the program and after completing the program.

\[\begin{aligned} \text{Before: } & 181, 183, 175, 163, 192, 201, 196, 181, 207, 189 \\ \text{After: } & 166, 163, 178, 180, 175, 166, 172, 176, 172, 172 \end{aligned}\]

  • \(H_0\): \(\mu_{\text{before}} - \mu_{\text{after}} = 0\)
  • \(H_a\): \(\mu_{\text{before}} - \mu_{\text{after}} > 0 \text{ or } \mu_{\text{after}} - \mu_{\text{before}} < 0\)
before = c(181, 183, 175, 163, 192, 201, 196, 181, 207, 189)
after = c(166, 163, 178, 180, 175, 166, 172, 176, 172, 172)
print(str_glue("Sample mean of before: {mean(before)}, Sample mean of after: {mean(after)}"))
Sample mean of before: 186.8, Sample mean of after: 172

Paired \(t\)-Test

# right-tailed test
result = t.test(before, after, alternative = "greater", paired = TRUE)

# left-tailed test
# result = t.test(after, before, alternative = "less", paired = TRUE)
print(result)

    Paired t-test

data:  before and after
t = 2.8892, df = 9, p-value = 0.008954
alternative hypothesis: true mean difference is greater than 0
95 percent confidence interval:
 5.40988     Inf
sample estimates:
mean difference 
           14.8
  • We reject \(H_0\) at \(\alpha = 0.05\) and conclude that the mean weight before and after the diet program is different.

Questions

  • What if we have more than two groups?
  • How do we detect if there is a difference between the groups?
  • ANOVA (Analysis of Variance)