Z-Test
-------

The z-test is used to assess if the mean :math:`\overline{x}` of sample :math:`X` differs from the one of a known population.
The *significance level* of this difference is determined by a *p-value* threshold chosen prior doing the test.

Conditions for using a z-test:

#. Population is normally distributed
#. Population :math:`\mu` and :math:`\sigma` is known
#. Sample size is greater than 30 (see note below)

.. note::
   According to central limit theorem, a distribution is well approximated when reaching 30 samples.
   See `here <https://statisticsbyjim.com/basics/central-limit-theorem/>`__ for more infos.

To perform a z-test with a sample :math:`X` of size :math:`n`, you should compute the sample *standard score* (or *z-score*).
The *z-score*, noted :math:`Z`, characterizes how far from the population mean :math:`\mu` your sample mean :math:`\overline{x}` is, in unit of standard deviation :math:`\sigma`.
It is computed as follow:

.. math::
   Z=\frac{\overline{x}-\mu}{\sigma_\overline{x}}=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}

.. note::
   The SEM is used in the denominator to account for inaccuracies of :math:`\overline{x}`.
   The more samples are collected, the more the denominator tends toward :math:`\sigma`.
   See :ref:`SEM <SEM>` for more details.

From :math:`Z`, a *p-value* can be derived using the :math:`\mathcal{N}(0,1)` :ref:`CDF <CDF>` noted :math:`\Phi_{0,1}(x)`:

* Left "tail" of the :math:`\mathcal{N}(0,1)` distribution:

  .. math::
     \alpha &= P(\mathcal{N}(0,1)<Z\sigma)

     &=P(\mathcal{N}(0,1)<Z\times 1)
     
     &=P(\mathcal{N}(0,1)<Z)=\Phi_{0,1}(Z)

* Right "tail" of the :math:`\mathcal{N}(0,1)` distribution:

  .. math::
     \alpha &= 1-P(\mathcal{N}(0,1)<Z\sigma)

     &=1-P(\mathcal{N}(0,1)<Z\times 1)

     &=1-P(\mathcal{N}(0,1)<Z)=1-\Phi_{0,1}(Z)

.. image:: ../../figures/normal_law_tails.svg
   :align: center
          
|       
| If the test is done over one tail (left OR right) it is called a **one-tailed** z-test.
| If the test is done over both tails (left AND right) it is called a **two-tailed** z-test.

| A *one-tailed* z-test checks whether :math:`\overline{x} < \mu` or :math:`\overline{x} > \mu`.
| A *two-tailed* z-test checks whether :math:`\overline{x} \ne \mu`.

The following code shows you how to obtain the *p-value* in R:

.. literalinclude:: ../../code/ztest_pvalue.R
   :language: R

Output example:

.. code-block:: console
                
   Alpha approximated is 0.0359588035958804
   Alpha from built-in CDF 0.0359303191129258

If the :math:`\alpha` value given by the test is lower or equal to the *p-value* threshold chosen initially,
:math:`H_0` is rejected and :math:`H_1` is considered accepted.

An alternative way of doing the z-test is to build a **rejection region** from the *p-value*.
This is done by using the reverse :ref:`CDF <CDF>` function :math:`\Phi_{0,1}^{-1}(x)` as in the following code:

.. literalinclude:: ../../code/ztest_rejection_region.R
   :language: R

Output:

.. code-block:: console

   Rejection region for left tail: Z in ]-inf,-3.43161440362327]
   Rejection region for right tail: Z in [3.4316144036233,+inf[
   
Thus, if the z-score if part of one of the rejection regions, :math:`H_0` is rejected and :math:`H_1` is considered accepted.

   
Examples
========

One-tailed
^^^^^^^^^^^

This exercice is inpired from `this video <https://www.youtube.com/watch?v=bB-J6_wcGgE>`__.

A complain was registered stating that the boys in the municipal school are underfed.
The average weight of boys of age 10 is 32kg with a standard deviation of 9kg.
A sample of 25 boys of age 10 from the school is selected. Their average weight is 29.5kg.
We want to check whether the complain is true or not with a confidence level of :math:`\alpha=0.05`.

**--- Solution ---**

Hypothesis:

* :math:`H_0` : No significant difference (:math:`\overline{x} \ge 32`), the boys from the are not underfed
* :math:`H_1` : There is significant difference (:math:`\overline{x} < 32`), the boys from the are underfed

.. math::
   Z=\frac{29.5-32}{\frac{9}{\sqrt{25}}}=-1.388889

From this z-score, the *p-value* is 0.08243327. As it is greater than 0.05, we cannot reject :math:`H_0`.
Thus, the boys from the are not underfed.
  
Two-tailed
^^^^^^^^^^^

This exercice is inpired from `this website <https://www.mathandstatistics.com/learn-stats/hypothesis-testing/two-tailed-z-test-hypothesis-test-by-hand>`__.

The USA mean public school yearly funding is $6800 per student per year, with a standard deviation of $400.
We want to assess if a certain state in the USA, Michigan, receives a significantly different amount of public school funding (per student) than the USA average,
with :math:`\alpha=0.05`. A sample of 100 students reveals that in average, they received $6873.

.. note::
   
   Notice, we are not saying **"significantly lower amount"** or **"significantly higher amount"** but **"significantly different amount"**.
   This is a sign that a two-tailed z-test is required since we should check for both, lower and higher.

**--- Solution ---**

Hypothesis:

* :math:`H_0` : No significant difference (:math:`\overline{x} = 6800`), Michigan receives the same amount of public school funding per student
* :math:`H_1` : There is significant difference (:math:`\overline{x} \ne 6800`), Michigan do not receives the same amount of public school funding per student

.. math::
   Z=\frac{6873-6800}{\frac{400}{\sqrt{100}}}=1.825
   
| The *p-value* associated with the left tail (using :math:`-Z` with the CDF) is 0.03400051.
| Thus, as we are doing a *two-tailed* z-test the *p-value* is :math:`2\times 0.03400051 = 0.06800103`.
| We multiply by two has the two tails of the normal law are symetric.

Since :math:`0.06800103 >> 0.05` we cannot reject the null hypothesis :math:`H_0`.
Thus, Michigan receives the same amount of public school funding per student.