Hypothesis Tests

One-Sample t Statistic

Compute the one-sample t statistic: t = (mean − mu0) / SE, where SE = s/√n with s = sample std (ddof=1). Two loops: first accumulates the mean, second accumulates the squared deviations for s. Library: scipy.stats.ttest_1samp(sample, mu0) — snapshot shows both statistic and pvalue; RESULT is the statistic only, matching .statistic.

By hand

Sample [4,7,2,9,5,9], n=6, mu0=4. Mean=6.0, s=√8≈2.8284, SE≈1.1547. t = (6.0−4.0)/1.1547 ≈ 1.7321 (= √3).

naive.py
Replay: real traced execution (multi-file project)
import math
sample = [4, 7, 2, 9, 5, 9]
mu0 = 4
n = len(sample)
total = 0.0
for v in sample:
    total = total + v
mean = total / n
sq_diff = 0.0
for v in sample:
    sq_diff = sq_diff + (v - mean) ** 2
s = math.sqrt(sq_diff / (n - 1))
se = s / math.sqrt(n)
t = (mean - mu0) / se
print('RESULT:', round(t, 4))

  1. import math

    1import math2sample = [4, 7, 2, 9, 5, 9]

  2. sample ← [4, 7, 2, 9, 5, 9]

    1import math2sample = [4, 7, 2, 9, 5, 9]3mu0 = 4
    values this step[4, 7, 2, 9, 5, 9]sample

  3. mu0 ← 4

    2sample = [4, 7, 2, 9, 5, 9]3mu0 = 44n = len(sample)
    values this step4mu0

  4. n ← 6

    3mu0 = 44n = len(sample)5total = 0.0
    values this step6n

  5. total ← 0.0

    4n = len(sample)5total = 0.06for v in sample:
    values this step0.0total

  6. v ← 4, total ← 4.0

    pass 1 of 6
    5total = 0.06for v in sample:7    total = total + v8mean = total / n
    values this step4v0.0 → 4.0total
    All 6 passes — pass 1 is the card above
    passvtotal
    140.0 → 4.0
    24 → 74.0 → 11.0
    37 → 211.0 → 13.0
    42 → 913.0 → 22.0
    59 → 522.0 → 27.0
    65 → 927.0 → 36.0

  7. for v in sample:

    5total = 0.06for v in sample:7    total = total + v

  8. mean ← 6.0

    7    total = total + v8mean = total / n9sq_diff = 0.0
    values this step6.0mean

  9. sq_diff ← 0.0

    8mean = total / n9sq_diff = 0.010for v in sample:
    values this step0.0sq_diff

  10. v ← 4, sq_diff ← 4.0

    pass 1 of 6
    9sq_diff = 0.010for v in sample:11    sq_diff = sq_diff + (v - mean) ** 212s = math.sqrt(sq_diff / (n - 1))
    values this step9 → 4v0.0 → 4.0sq_diff
    All 6 passes — pass 1 is the card above
    passvsq_diff
    19 → 40.0 → 4.0
    24 → 74.0 → 5.0
    37 → 25.0 → 21.0
    42 → 921.0 → 30.0
    59 → 530.0 → 31.0
    65 → 931.0 → 40.0

  11. for v in sample:

    9sq_diff = 0.010for v in sample:11    sq_diff = sq_diff + (v - mean) ** 2

  12. s ← 2.8284271247461903

    11    sq_diff = sq_diff + (v - mean) ** 212s = math.sqrt(sq_diff / (n - 1))13se = s / math.sqrt(n)
    values this step2.8284271247461903s

  13. se ← 1.1547005383792517

    12s = math.sqrt(sq_diff / (n - 1))13se = s / math.sqrt(n)14t = (mean - mu0) / se
    values this step1.1547005383792517se

  14. t ← 1.732050807568877

    13se = s / math.sqrt(n)14t = (mean - mu0) / se15print('RESULT:', round(t, 4))
    values this step1.732050807568877t

  15. stdout ← RESULT: 1.7321

    14t = (mean - mu0) / se15print('RESULT:', round(t, 4))
    values this stepRESULT: 1.7321stdout

With the library

scipy.stats.ttest_1samp(sample, mu0) returns a result object; .statistic is the t value and .pvalue is the two-tailed p-value (snapshot only).

library.py 
import numpy as np
from scipy import stats
from dalib.display import set_display
set_display()

sample = [4, 7, 2, 9, 5, 9]
mu0 = 4
n = len(sample)
mean = float(np.mean(sample))
s = float(np.std(sample, ddof=1))
se = float(s / np.sqrt(n))
result = stats.ttest_1samp(sample, mu0)
print('mean:', round(mean, 4))
print('se:', round(se, 6))
print('t_stat:', round(float(result.statistic), 4))
print('pvalue:', round(float(result.pvalue), 4))
print('RESULT:', round(float(result.statistic), 4))

mean: 6.0
se: 1.154701
t_stat: 1.7321
pvalue: 0.1438
RESULT: 1.7321

Honesty

This lesson shows the computation exactly, on a tiny pinned sample. The arithmetic is correct and reproducible, but with a sample this small the result is not a valid statistical finding — it demonstrates the mechanism, not evidence. Real inference needs an adequate sample size and assumption checks (e.g. independence and approximate normality); the p-value / interval here should be read as "how the formula is computed," not as a conclusion about a population.

Implementation notes

  • Design rule §7: RESULT = statistic only; p-value appears in the snapshot for context but is not the comparable between naive and library.
  • df = n−1 = 5. Under H0: mu = mu0, t follows a t distribution with 5 df.
  • Two-tailed p-value: probability of |T| ≥ |t| under H0. Here p≈0.1438, so no rejection at α=0.05.
  • s uses ddof=1 (sample std); using ddof=0 would understate variability and inflate the statistic.
  • Cross-reference: standard-error (ch06) for the SE computation; z-scores (ch02) for the standardization idea the t statistic extends.