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 \(X \sim \mathcal{N}(\mu, \sigma)\), equals the mean \(\mu_0\) of a theoretical distribution \(X_0 \sim \mathcal{N}(\mu_0, \sigma)\). Note we assume that standard deviation of the unknown population \(X\) is known and equal to the standard deviation of the theoretical population \(\sigma\). 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 ministats 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 \(\sigma_{X_0} = 5\):
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 \(\sigma_{X_A} = 3\):
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
(np.float64(3.1180664906014357), np.float64(0.0018204172963375287))
The helper function ztest in the ministats module performs
exactly the same sequence of steps to compute the \(p\)-value.
from ministats import ztest
ztest(xAs, mu0=muX0, sigma0=sigmaX0)
np.float64(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 ministats import ztest
ztest(xBs, mu0=muX0, sigma0=sigmaX0)
np.float64(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 \(\mu_{X_0} = 100\).
Effect size estimates#
Discussion#
Links#
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)