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.

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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");
    }
}