T-Test

For this test, suppose the data below. Moreover, suppose that the population variance is unknown and that we would like to test whether the population mean is different from 0.

$$ X = ( 0.1, -0.4, 0.2, -0.4, 0.5 ) $$

So we have:

• 5 observations: $n = 5$
• mean of the sample: $\bar{x} \approx 0.04 $

We want to test if the mean of the sampel is equal to zero or not. We define our test hypothesis:

$$ H_{0}: \mu_{0} = 0 $$ and $$ H_{1}: \mu_{0} \neq 0 $$

Test statistics is defined as: $$ t = \sqrt n \frac{\bar{x} - \mu_{0}}{\sigma} $$

With the given example, we get the following statistics $$ t = \sqrt 5 \frac{0.04 - 0.0}{0.416} \approx 0.215 $$

Critical value: $ \pm t(1.0 - \alpha/2, n-1) = \pm t() \approx \pm 0.225$. The regions which are recjected are thus from $-\inf$ to -0.225 and from 0.225 to $+\inf$. The test statistic is outside the rejection regionsd so we do not reject the null hypothesis $H_{0}$. In terms of the initial question: At the $\alpha = 0.05 $ significance level, we do not reject the hypothesis that the population mean $\mu_{0}$ is equal to 0, or there is no sufficient evidence in the data to conclude that the population mean is different from 0.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33

use mathru::statistics::{
    distrib::{Continuous, Normal, T},
    test::{Test, T as T_Test},
};

/// One sample, two tailed
fn main() {
    let mean: f64 = 0.0;

    let alpha: f64 = 0.05;

    let x: Vec<f64> = vec![0.3, -0.4, 0.2, -0.4, 0.5];

    let sample_size: u32 = x.len() as u32;

    let normal: Normal<f64> = Normal::from_data(&x);

    println!("x_bar: {}", normal.mean());
    println!("standard deviation: {}", normal.variance().sqrt());

    let test: T_Test<f64> = T_Test::one_sample(&x, mean);

    let t: f64 = T::new((sample_size - 1) as f64).pdf(1.0 - alpha);

    println!("test: {}", test.value());
    println!("t(1.0 - alpha, n - 1) = {}", t);

    if -t <= test.value() && test.value() <= t {
        println!("H0 is accepted");
    } else {
        println!("H0 is rejected and H1 is accepted");
    }
}