# One SampleÂ¶

This notebook demonstrates the one-sample t-test and checking normality with a Q-Q plot, using the ZARR13.DAT file from the NIST e-Handbook of Statistical Methods.

## SetupÂ¶

Import our Python modules:

import pandas as pd
import numpy as np
import scipy.stats as sps
import statsmodels.api as sm
import matplotlib.pyplot as plt


data = pd.read_table('ZARR13.DAT', skiprows=25, names=['X'])
data

X
0 9.206343
1 9.299992
2 9.277895
3 9.305795
4 9.275351
... ...
190 9.232963
191 9.234956
192 9.216746
193 9.274107
194 9.273776

195 rows Ã— 1 columns

## Showing the DistributionÂ¶

Letâ€™s look at histogram:

plt.hist(data.X, bins=20)
plt.show()


And the distribution statistics:

data.X.describe()

count    195.000000
mean       9.261461
std        0.022789
min        9.196848
25%        9.246496
50%        9.261952
75%        9.275530
max        9.327973
Name: X, dtype: float64


### Q-Q PlotÂ¶

The data looks normal-ish, but a histogram isnâ€™t a very reliable way to assess normality. A Q-Q plot against the normal distribution lets us be much more precise:

sm.qqplot(data.X, fit=True, line='45')
plt.show()


That data looks normal! Itâ€™s a straight line right through the bulk of the data. Itâ€™s common for the first and last few points to deviate from normal just a little bit more than the central mass of data points.

### Drawing Q-Q OurselvesÂ¶

Letâ€™s try to draw our own. The way we draw a Q-Q plot is this:

1. Sort the data values in ascending order. These will be plotted on the y axis.

2. Compute the percentile for each data point - where is it in the range of data points? We can do this with its position or count. We donâ€™t label any point 0 or 1; instead, for point $$i \in [1,n]$$, we compute $$v = i / (n + 1)$$.

3. Compute the quantiles in the reference distribution (in our case, normal) for each data point position. These will be plotted on the x axis.

Letâ€™s sort:

observed = data['X'].sort_values()


Now we need to compute the percentiles for each position. The arange function is useful for this - it can generate $$i = 1 \dots n$$, and we can divide to rescale the points:

nobs = len(observed)
pred_ps = np.arange(1, nobs + 1) / (nobs + 1)


And we need to convert these percentiles into quantile values from the standard normal:

norm_dist = sps.norm()
pred_vs = norm_dist.ppf(pred_ps)


Finally, we can plot them against each other:

plt.scatter(pred_vs, observed)
plt.xlabel('Theoretical Quantiles')
plt.ylabel('Sample Quantiles')
plt.show()


We can see the straight line, but two issues remain:

1. Our sample quantiles are on the original scale, but theoretical are standardized. Letâ€™s standardize the sample quantiles so the values, not just relative shape, are comparable.

2. We donâ€™t have the refernce line. This will be easier to draw with standardized sample quantiles.

Standardization, for normally-distributed data, is transforming it to have a mean of 0 and standard deviation of 1. We do this by subtracting the mean and dividing by the sample standard deviation:

std_observed = (observed - observed.mean()) / observed.std()
# reference values; once standardized, the line will be y=x
ref_xs = np.linspace(np.min(pred_vs), np.max(pred_vs), 1000)
plt.plot(ref_xs, ref_xs, color='red')

plt.scatter(pred_vs, std_observed)

# and labels
plt.xlabel('Theoretical Quantiles')
plt.ylabel('Sample Quantiles')
plt.show()


## T-testsÂ¶

Letâ€™s start with a 1-sample T-test for $$H_0: \mu = 9$$:

sps.ttest_1samp(data.X, 9)

Ttest_1sampResult(statistic=160.21472007838156, pvalue=4.430052598624794e-208)


If $$\mu$$ were nine, it would be extremely unlikely to find a sample with this observed mean.

These values are very tightly distributed. A 1-sample T-test for $$H_0: \mu = 9.25$$:

sps.ttest_1samp(data.X, 9.25)

Ttest_1sampResult(statistic=7.022783450050035, pvalue=3.5755854129627046e-11)


Our data would also be very unlikely, but not as unlikely, under this null hypothesis.

And just to see it accept the null, letâ€™s try $$H_0: \mu = 9.26$$:

sps.ttest_1samp(data.X, 9.26)

Ttest_1sampResult(statistic=0.895105984916905, pvalue=0.37183966110536093)


Our data is consistent with $$\mu=9.26$$.

Note: What I just did here â€” try several different null hypotheses in a row â€” is not a valid statistical procedure. I am only doing it to demonstrate the results of a t-test both when the null hypothesis holds, and when it does not.