# One-sample z-test for the mean#

The goal of the one-sample $$z$$-test to check if the mean $$\mu$$ of an unknown population $$X \sim \mathcal{N}(\mu, \sigma_0)$$, equals the mean $$\mu_0$$ of a theoretical distribution $$X_0 \sim \mathcal{N}(\mu_0, \sigma_0)$$. Note we assume that standard deviation of the unknown population $$X$$ is known and equal to the standard deviation of the theoretical population $$\sigma_0$$. We discussed this hypothesis test in notebooks/34_analytical_approx.ipynb.

import matplotlib.pyplot as plt
import numpy as np
import seaborn as sns

%matplotlib inline
%config InlineBackend.figure_format = 'retina'


## Data#

One sample of numerical observations $$\mathbf{x}=[x_1, x_2, \ldots, x_n]$$.

## Modeling assumptions#

We model the unknown population as…

and the theoretical distribution is…

We assume the population is normally distributed $$\textbf{(NORM)}$$, or the sample is large enough $$\textbf{(LARGEn)}$$.

We also assume that the variance of the unknown population is known and equal to the variance of the theoretical population under $$H_0$$

## Hypotheses#

$$H_0: \mu = \mu_0$$ and $$H_A: \mu \neq \mu_0$$, where $$\mu$$ is the unknown population mean, $$\mu_0$$ is the theoretical mean we are comparing to.

## Statistical design#

for $$n=5$$

for $$n=20$$

## Test statistic#

Compute $$z = \frac{\overline{\mathbf{x}} - \mu_0}{\sigma_0/\sqrt{n}}$$, where $$\overline{\mathbf{x}}$$ is the sample mean, $$\mu_0$$ is the theoretical population mean, $$\sigma_0$$ is the known standard deviation.

## Sampling distribution#

Standard normal distribution $$Z \sim \mathcal{N}(0,1)$$.

## P-value calculation#

from stats_helpers import ztest

%psource ztest


To perform the one-sample $$z$$-test on the sample xs, we call ztest(xs, mu0=..., sigma0=...), with the ...s replaced by the mean and standard deviation parameters of the distribution under $$H_0$$.

## Examples#

For all the examples we present below, we assume the theoretical distribution we expect under the null hypothesis, is normally distributed with mean $$\mu_{X_0}=100$$ and standard deviation $$\sigma_{X_0} = 5$$:

$\texttt{rvX0} = X_0 \sim \mathcal{N}(100,5).$
from scipy.stats import norm
muX0 = 100
sigmaX0 = 5
rvX0 = norm(muX0, sigmaX0)


### Example A: population different from $$H_0$$#

Suppose the unknown population is normally distributed with mean $$\mu_{X_A}=104$$ and standard deviation $$\sigma_{X_A} = 3$$:

$\texttt{rvXA} = X_A \sim \mathcal{N}(110,3).$
muXA = 104
sigmaXA = 3
rvXA = norm(muXA, sigmaXA)


Let’s generate a sample xAs of size $$n=20$$ from the random variable $$X = \texttt{rvXA}$$.

np.random.seed(42)

# generate a random sample of size n=20
n = 20
xAs = rvXA.rvs(n)
xAs

array([105.49014246, 103.5852071 , 105.94306561, 108.56908957,
103.29753988, 103.29758913, 108.73763845, 106.30230419,
102.59157684, 105.62768013, 102.60974692, 102.60281074,
104.72588681,  98.26015927,  98.8252465 , 102.31313741,
100.96150664, 104.942742  , 101.27592777,  99.7630889 ])

import seaborn as sns
with plt.rc_context({"figure.figsize":(7,1)}):
sns.stripplot(x=xAs, jitter=0, alpha=0.5)


To obtain the $$p$$-value, we first compute the observed $$z$$-statistic, then calculate the tail probabilities in the two tails of the standard normal distribution $$Z\sim\mathcal{N}(0,1)$$.

obsmean = np.mean(xAs)
se = sigmaX0 / np.sqrt(n)
obsz = (obsmean - muX0) / se
rvZ = norm(0,1)
pvalue = rvZ.cdf(-obsz) + 1-rvZ.cdf(obsz)
obsz, pvalue

(3.1180664906014357, 0.0018204172963375287)


The helper function ztest in the stats_helpers module performs exactly the same sequence of steps to compute the $$p$$-value.

from stats_helpers import ztest
ztest(xAs, mu0=muX0, sigma0=sigmaX0)

0.0018204172963375855


The $$p$$-value we obtain is 0.0018 (0.18%), which is below the cutoff value $$\alpha=0.05$$, so our conclusion is we reject the null hypothesis: the mean of the sample xAs is not statistically significantly different from the theoretically expected mean $$\mu_{X_0} = 100$$.

### Example B: sample from a population as expected under $$H_0$$#

# unknown population X = X0
rvXB = norm(muX0, sigmaX0)


Let’s generate a sample xBs of size $$n=20$$ from the random variable $$X = \texttt{rvX}$$, which has the same distribution as the theoretical distribution we expect under the null hypothesis.

# np.random.seed(32)  produces false positie
np.random.seed(31)

# generate a random sample of size n=20
n = 20
xBs = rvXB.rvs(n)
xBs

array([ 97.92621393,  98.33315666, 100.40545993,  96.04486524,
98.90700164,  96.18401578,  96.11439878, 109.24678261,
96.47199845,  99.56978983, 101.43966651,  99.34306739,
95.08627924,  95.40604357, 105.99717245,  98.29312879,
91.20696344, 100.25558735,  97.14035504,  96.49716979])

import seaborn as sns
with plt.rc_context({"figure.figsize":(7,1)}):
sns.stripplot(x=xBs, jitter=0, alpha=0.5)

from stats_helpers import ztest
ztest(xBs, mu0=muX0, sigma0=sigmaX0)

0.17782115942197962


The $$p$$-value we obtain is 0.18, which is above the cutoff value $$\alpha=0.05$$ so our conclusion is that we’ve failed to reject the null hypothesis: the mean of the sample xBs is not significantly different from the theoretically expected mean $$\mu_{X_0} = 100$$.

## CUT MATERIAL#

# NOT GOOD: because we can't specify sigma0 manually;
#           the statsmodels function uses sample std instead of sigmaX0
# from statsmodels.stats import weightstats
# weightstats.ztest(x1, x2=None, value=0)