Statistics

statistics-ai999x

Descriptive statistics, probability distributions, confidence intervals, hypothesis tests, regression and correlation, experimental design, quality control and reliability.

Mathematics & Statistics Question Answering Token Classification Text Classification Tabular Regression Sentence Similarity Feature Extraction

Hugging Face ↗GitHub ↗See it on the capability orb

110capabilities

16topics

110formulas typeset

8worked examples below

1.0.0version

About this model

What Statistics does

Statistics (statistics-ai999x) is an offline Python package for statistics. It covers descriptive statistics, probability distributions, confidence intervals, hypothesis tests, regression and correlation, experimental design, quality control and reliability.

It has 110 capabilities — 110 computational procedures — in 16 topics, the largest being Descriptive statistics, Probability distributions, Hypothesis tests, Regression and correlation, Confidence intervals and Nonparametric tests. Ask a question in words with ask(), or call a capability by name with typed inputs with solve(). Every answer carries its value and unit, the steps that produced it and the formula in LaTeX.

When a question is not determined — a value is missing, an input is out of range, or the request is outside the package — it says so and names what it needs, instead of guessing. It runs on an ordinary CPU without a network connection (Python >=3.10), and the same question always gets the same answer.

Install

pip install statistics-ai999x
hf download AI999X/statistics-ai999x --local-dir statistics-ai999x

What it covers

Analysis of variance1One-way ANOVA Categorical data2Fisher's exact test (2 x 2) · McNemar's test for paired proportions Confidence intervals7Confidence interval for a mean (sigma known, z-interval) · Confidence interval for a mean from data (t-interval) · Confidence interval for a mean from summary statistics (t-interval) …Confidence intervals and tests4Critical value of chi-square · Critical value of F · Critical value of Student's t …Descriptive statistics22Arithmetic mean · Coefficient of variation · Five-number summary and interquartile range …Effect sizes1Cohen's d and Hedges' g Experimental design42^2 factorial design: main effects and interaction · 2^3 factorial design: main effects and interactions · Cronbach's alpha (internal consistency) …Hypothesis tests14Bartlett's test for equal variances · Chi-square goodness-of-fit test · Chi-square test for a variance …Nonparametric tests7Friedman test · Kruskal-Wallis H test · Mann-Whitney U test (exact) …Power and sample size5Power of a one-sample z-test · Power of a two-sample z-test · Sample size per group to compare two means (z-test power) …Probability1Probability of at least one occurrence Probability distributions17Binomial probability: at least k successes · Binomial probability: at most k successes · Binomial probability: exactly k successes …Quality control6c chart limits (defects per unit) · Individuals and moving range (I-MR) chart limits · p chart limits (fraction defective) …Regression and correlation10Confidence interval for a correlation (Fisher z) · Covariance · Durbin-Watson statistic …Reliability and survival7Exponential reliability (constant failure rate) · k-out-of-n system reliability · Kaplan-Meier survival estimate …Sampling distributions2Probability that a sample mean is above a value · Probability that a sample mean is below a value

Try it

Worked examples

Each question below was put to statistics-ai999x and the answer shown is the one it returned, with the steps it took.

Hypothesis testsm.ask(…)

Perform a one-sample t-test of [5.1, 4.9, 5.3, 5.5, 5.0, 5.2] against mu0 = 5, two-sided.

t = 1.889822365

df = 5

p_value = 0.1173867951

mean = 5.166666667

standard_deviation = 0.2160246899

Show the steps

One-sample t-test
t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}},\ \nu = n - 1
with x = [5.1, 4.9, 5.3, 5.5, 5, 5.2], mu0 = 5, alternative = two-sided
→ t = 1.889822365, df = 5, p_value = 0.1173867951, mean = 5.166666667, standard_deviation = 0.2160246899

Capability: One-sample t-test

Probability distributionsm.ask(…)

For a normal distribution with mean = 100 and standard deviation = 15, find P(X < 120).

probability = 0.9087887803

z = 1.333333333

Show the steps

Normal probability below a value
P(X < x) = \Phi\left(\frac{x - \mu}{\sigma}\right)
with mu = 100, sigma = 15, x = 120
→ probability = 0.9087887803, z = 1.333333333

Capability: Normal probability below a value

Confidence intervalsm.ask(…)

Find the 95% confidence interval for the mean of [12, 15, 11, 18, 14, 16].

lower = 11.62370216

upper = 17.0429645

margin_of_error = 2.709631171

critical_value = 2.570581836

df = 5

mean = 14.33333333

Show the steps

Confidence interval for a mean from data (t-interval)
\bar{x} \pm t_{\alpha/2,\,n-1}\frac{s}{\sqrt{n}}
with x = [12, 15, 11, 18, 14, 16], confidence = 0.95
→ lower = 11.62370216, upper = 17.0429645, margin_of_error = 2.709631171, critical_value = 2.570581836, df = 5, mean = 14.33333333

Capability: Confidence interval for a mean from data (t-interval)

Hypothesis testsm.ask(…)

Chi-square test of independence for the table [[20, 30], [25, 25]].

chi_square = 1.01010101

df = 1

p_value = 0.3148786413

expected = [[22.5, 27.5], [22.5, 27.5]]

Show the steps

Chi-square test of independence
\chi^2 = \sum_{i,j}\frac{(O_{ij} - E_{ij})^2}{E_{ij}},\ E_{ij} = \frac{R_i C_j}{n}
with table = [20 30; 25 25]
→ chi_square = 1.01010101, df = 1, p_value = 0.3148786413, expected = [[22.5, 27.5], [22.5, 27.5]]

Capability: Chi-square test of independence

Reliability and survivalm.solve(…)

m.solve("weibull_reliability", beta=2, eta=1000, t=500)

reliability = 0.7788007831

unreliability = 0.2211992169

hazard_rate = 0.001

mttf = 886.2269255

Show the steps

Weibull reliability, hazard rate and MTTF
R(t) = e^{-(t/\eta)^\beta},\quad h(t) = \frac{\beta}{\eta}\left(\frac{t}{\eta}\right)^{\beta-1}
with beta = 2, eta = 1000, t = 500
→ reliability = 0.7788007831, unreliability = 0.2211992169, hazard_rate = 0.001, mttf = 886.2269255

Capability: Weibull reliability, hazard rate and MTTF

Analysis of variancem.ask(…)

Perform a one-way ANOVA on groups [23, 25, 21, 22], [30, 28, 31, 29] and [26, 27, 24, 25].

f = 22.12

p_value = 0.0003357531392

df_between = 2

df_within = 9

ss_between = 92.16666667

ss_within = 18.75

Show the steps

One-way ANOVA
F = \frac{SS_B/(k-1)}{SS_W/(N-k)}
with g1 = [23, 25, 21, 22], g2 = [30, 28, 31, 29], g3 = [26, 27, 24, 25]
→ f = 22.12, p_value = 0.0003357531392, df_between = 2, df_within = 9, ss_between = 92.16666667, ss_within = 18.75, ms_between = 46.08333333, ms_within = 2.083333333, eta_squared = 0.8309541698

Capability: One-way ANOVA

Categorical datam.ask(…)

Fisher's exact test for the table [[8, 2], [1, 5]], two-sided.

p_value = 0.03496503497

odds_ratio = 20

Show the steps

Fisher's exact test (2 x 2)
P(a) = \frac{\binom{a+b}{a}\binom{c+d}{c}}{\binom{n}{a+c}}
with table = [8 2; 1 5], alternative = two-sided
→ p_value = 0.03496503497, odds_ratio = 20

Capability: Fisher's exact test (2 x 2)

Confidence intervals and testsm.ask(…)

Find the chi-square critical value with df = 4 and alpha = 0.05.

critical_value = 9.487729037

left_critical_value = 0.7107230214

Show the steps

Critical value of chi-square
\chi^2_{\alpha,\nu}:\ P(\chi^2_\nu > c) = \alpha
with df = 4, alpha = 0.05
→ critical_value = 9.487729037, left_critical_value = 0.7107230214

Capability: Critical value of chi-square

Use it

From Python or the command line

Every snippet was run as shown; the output under it is what it printed.

import statistics_ai999x as m

a = m.ask("Perform a one-sample t-test of [5.1, 4.9, 5.3, 5.5, 5.0, 5.2] against mu0 = 5, two-sided.")
print(a.outcome)
print(a.values)
for step in a.steps:
    print(step["name"])

Output

SOLVED
{'t': 1.88982236504614, 'df': 5, 'p_value': 0.11738679514629244, 'mean': 5.166666666666667, 'standard_deviation': 0.2160246899469286}
One-sample t-test

import statistics_ai999x as m

a = m.solve("weibull_reliability", beta=2, eta=1000, t=500)
print(a.outcome)
print(a.values)

Output

SOLVED
{'reliability': 0.7788007830714049, 'unreliability': 0.22119921692859512, 'hazard_rate': 0.001, 'mttf': 886.226925452758}

import statistics_ai999x as m

# an input is missing: the answer says what it needs instead of guessing
a = m.solve("weibull_reliability")
print(a.outcome)
print(a.message)

Output

NEEDS_INPUT
weibull_reliability needs beta, eta, t (shape parameter beta; scale parameter eta (characteristic life); time). Example: solve('weibull_reliability', beta=2, eta=1000, t=500).

statistics-ai999x "Perform a one-sample t-test of [5.1, 4.9, 5.3, 5.5, 5.0, 5.2] against mu0 = 5, two-sided."

Output

t = 1.88982, df = 5, p_value = 0.117387, mean = 5.16667, standard_deviation = 0.216025
1. One-sample t-test: t statistic with n - 1 degrees of freedom; tail probability from the incomplete beta function  with x = [5.1, 4.9, 5.3, 5.5, 5, 5.2], mu0 = 5, alternative = two-sided  ->  t = 1.88982, df = 5, p_value = 0.117387, mean = 5.16667, standard_deviation = 0.216025
Assumptions: random sample from a roughly normal population (or large n); the alternative must be stated for a one-sided claim
…

Figures

Computed, not drawn by hand

Every value in these figures was computed by the package, each point a run of solve(); in a notebook, plot() and plot_relation() draw a capability's curves the same way.

One-sample t test: statistic against the rejection region

t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}},\ \nu = n - 1

The one-sample t test of x = 5.1, 4.9, 5.3, 5.5, 5.0, 5.2 against μ₀ = 5: the t density with 5 degrees of freedom (from the package's quantiles at 121 cumulative probabilities, differenced), the two-tailed rejection region beyond ±2.57058 for α = 0.05 shaded; marked: t = 1.88982, p-value = 0.117387.

Poisson distribution and an at-least probability

P(X \ge k) = 1 - \sum_{j=0}^{k-1} \frac{e^{-\lambda}\lambda^j}{j!}

The Poisson distribution with mean λ = 4: P(X = k) for k = 0 … 14, each from the package; the stems with k ≥ 2 carry P(X ≥ 2) = 0.908422; marked: the most probable count k = 4, P = 0.195367.

Three groups as box plots, with their ANOVA

F = \frac{SS_B/(k-1)}{SS_W/(N-k)}

Box plots of 3 groups (g1 = 23, 25, 21, 22; g2 = 30, 28, 31, 29; g3 = 26, 27, 24, 25): each box spans the quartiles and the bar is the median, all from the package's percentiles (g1 median 22.5, quartiles 21.75–23.5; g2 median 29.5, quartiles 28.75–30.25; g3 median 25.5, quartiles 24.75–26.25). One-way ANOVA: F = 22.12 with 2 and 9 degrees of freedom, p-value 0.000335753, η² = 0.830954.

Simple linear regression: data and fitted line

\hat{y} = b_0 + b_1 x,\quad b_1 = \frac{S_{xy}}{S_{xx}},\quad b_0 = \bar{y} - b_1\bar{x}

The least-squares line through x = 1, 2, 3, 4, 5, y = 2.2, 4.1, 6.1, 7.9, 10.2: intercept 0.16 (standard error 0.143295), slope 1.98 (standard error 0.0432049), r = 0.999287, R² = 0.998574, p-value of the slope 2.2873e-05, as returned by the package.

Kaplan–Meier survival curve

\hat{S}(t) = \prod_{t_i \le t}\left(1 - \frac{d_i}{n_i}\right)

The Kaplan–Meier survival curve returned by the package for times 6, 6, 6, 7, 10, 13, 16, 22, 23 with event indicators 1, 1, 1, 0, 1, 1, 1, 1, 1 (0 = censored): it steps down at each event time 6, 10, 13, 16, 22, 23; marked: t = 12, S = 0.533333.

Confidence intervals for a mean at five levels

\bar{x} \pm t_{\alpha/2,\,n-1}\frac{s}{\sqrt{n}}

t confidence intervals for the mean of x = 12, 15, 11, 18, 14, 16 at 5 confidence levels, each from the package: 50 % [13.5673, 15.0993]; 80 % [12.7776, 15.8891]; 90 % [12.2093, 16.4574]; 95 % [11.6237, 17.043]; 99 % [10.0831, 18.5836] (sample mean 14.3333, 5 degrees of freedom); higher confidence, wider interval.

Formulas

The mathematics inside

12 of its 110 formulas; every one is in the capability browser below.

One-sample t-test

t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}},\ \nu = n - 1

Normal probability below a value

P(X < x) = \Phi\left(\frac{x - \mu}{\sigma}\right)

Confidence interval for a mean from data (t-interval)

\bar{x} \pm t_{\alpha/2,\,n-1}\frac{s}{\sqrt{n}}

Chi-square test of independence

\chi^2 = \sum_{i,j}\frac{(O_{ij} - E_{ij})^2}{E_{ij}},\ E_{ij} = \frac{R_i C_j}{n}

Weibull reliability, hazard rate and MTTF

R(t) = e^{-(t/\eta)^\beta},\quad h(t) = \frac{\beta}{\eta}\left(\frac{t}{\eta}\right)^{\beta-1}

One-way ANOVA

F = \frac{SS_B/(k-1)}{SS_W/(N-k)}

Fisher's exact test (2 x 2)

P(a) = \frac{\binom{a+b}{a}\binom{c+d}{c}}{\binom{n}{a+c}}

Critical value of chi-square

\chi^2_{\alpha,\nu}:\ P(\chi^2_\nu > c) = \alpha

McNemar's test for paired proportions

\chi^2 = \frac{(|b - c| - 1)^2}{b + c}

Confidence interval for a mean (sigma known, z-interval)

\bar{x} \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

Critical value of F

F_{\alpha;\,\nu_1,\nu_2}:\ P(F_{\nu_1,\nu_2} > c) = \alpha

Arithmetic mean

\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i

Capabilities

Every capability, with an example

Search by name, formula, input or topic. Each capability opens to show what it takes, what it returns, when it applies, and a question it answered.

110 capabilities

Analysis of variance 1

One-way ANOVAone_way_anova

One-way analysis of variance for 2 to 5 independent groups: F statistic, p-value, sums of squares and eta squared.

F = \frac{SS_B/(k-1)}{SS_W/(N-k)}

Inputs

g1 group 1 vector
g2 group 2 vector
g3 group 3 vector optional
g4 group 4 vector optional
g5 group 5 vector optional

Returns

f MS between / MS within
p_value upper-tail p-value
df_between k - 1
df_within N - k
ss_between between-group sum of squares
ss_within within-group sum of squares
ms_between mean square between
ms_within mean square within
eta_squared SSB / SST

Applies when

independent groups
roughly normal populations with equal variances

Example

m.ask("Perform a one-way ANOVA on groups [23, 25, 21, 22], [30, 28, 31, 29] and [26, 27, 24, 25].")

→ f = 22.12, p_value = 0.000335753, df_between = 2, df_within = 9, ss_between = 92.1667, ss_within = 18.75, ms_between = 46.0833, ms_within = 2.08333, eta_squared = 0.830954

Categorical data 2

Fisher's exact test (2 x 2)fisher_exact_test

Fisher's exact test of association in a 2 x 2 table, with the sample odds ratio.

P(a) = \frac{\binom{a+b}{a}\binom{c+d}{c}}{\binom{n}{a+c}}

Inputs

table 2 x 2 table of counts [[a, b], [c, d]] matrix
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

p_value exact p-value (two-sided: tables no more likely than the observed one)
odds_ratio ad/(bc); empty when b c = 0

Applies when

fixed margins (conditional test)
whole-number counts

Example

m.ask("Fisher's exact test for the table [[8, 2], [1, 5]], two-sided.")

→ p_value = 0.034965, odds_ratio = 20

McNemar's test for paired proportionsmcnemar_test

McNemar's test for a 2 x 2 table of paired outcomes: continuity-corrected chi-square and the exact binomial p-value.

\chi^2 = \frac{(|b - c| - 1)^2}{b + c}

Inputs

table 2 x 2 table of paired counts [[both yes, yes/no], [no/yes, both no]] matrix

Returns

chi_square (|b - c| - 1)^2 / (b + c)
p_value chi-square p-value (1 df)
exact_p_value exact binomial p-value
discordant b + c

Applies when

paired (matched) binary outcomes
b + c > 0

Example

m.ask("McNemar's test for the paired table [[30, 12], [5, 53]].")

→ chi_square = 2.11765, p_value = 0.14561, exact_p_value = 0.143463, discordant = 17

Confidence intervals 7

Confidence interval for a mean (sigma known, z-interval)ci_mean_known_sigma

The z confidence interval for a population mean when the population standard deviation sigma is known.

\bar{x} \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

Inputs

xbar sample mean
sigma population standard deviation (known)
n sample size integer
confidence confidence level as a fraction (0.95 for 95%)

Returns

lower lower limit
upper upper limit
margin_of_error z sigma / sqrt(n)
critical_value z critical value

Applies when

sigma known
normal population or large n
random sample

Example

m.ask("Find the 95% confidence interval for the mean with sample mean = 52, population standard deviation = 8 and n = 36.")

→ lower = 49.3867, upper = 54.6133, margin_of_error = 2.61329, critical_value = 1.95996

Confidence interval for a mean from data (t-interval)ci_mean_t

The t confidence interval for a population mean from sample data (sigma unknown).

\bar{x} \pm t_{\alpha/2,\,n-1}\frac{s}{\sqrt{n}}

Inputs

x the data values (at least two) vector
confidence confidence level as a fraction (0.95 for 95%)

Returns

lower lower limit
upper upper limit
mean sample mean
margin_of_error t s / sqrt(n)
critical_value t critical value
df n - 1

Applies when

random sample from a roughly normal population (or large n)
sigma unknown

Example

m.ask("Find the 95% confidence interval for the mean of [12, 15, 11, 18, 14, 16].")

→ lower = 11.6237, upper = 17.043, margin_of_error = 2.70963, critical_value = 2.57058, df = 5, mean = 14.3333

Confidence interval for a mean from summary statistics (t-interval)ci_mean_t_summary

The t confidence interval for a mean from the sample mean, sample standard deviation and n.

\bar{x} \pm t_{\alpha/2,\,n-1}\frac{s}{\sqrt{n}}

Inputs

xbar sample mean
s sample standard deviation
n sample size (>= 2) integer
confidence confidence level as a fraction (0.95 for 95%)

Returns

lower lower limit
upper upper limit
margin_of_error t s / sqrt(n)
critical_value t critical value
df n - 1

Applies when

random sample from a roughly normal population (or large n)
sigma unknown

Example

m.ask("Find the 95% confidence interval for the mean with sample mean = 25, sample standard deviation = 4 and n = 16.")

→ lower = 22.8686, upper = 27.1314, margin_of_error = 2.13145, critical_value = 2.13145, df = 15

Confidence interval for a proportion (Wald)ci_proportion

The large-sample (Wald) confidence interval for a population proportion from x successes in n trials.

\hat{p} \pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Inputs

x number of successes integer
n sample size integer
confidence confidence level as a fraction (0.95 for 95%)

Returns

proportion sample proportion x/n
lower lower limit (clipped at 0)
upper upper limit (clipped at 1)
margin_of_error z sqrt(p(1-p)/n)

Applies when

0 < x < n
large sample: n p and n(1 - p) at least about 10
limits clipped to [0, 1]

Example

m.ask("Find the 95% confidence interval for a proportion with x = 48 successes out of n = 120.")

→ proportion = 0.4, lower = 0.312348, upper = 0.487652, margin_of_error = 0.0876523

Confidence interval for a variance and standard deviationci_variance

The chi-square confidence interval for a population variance (and standard deviation) from sample data.

\left[\frac{(n-1)s^2}{\chi^2_{1-\alpha/2,\,n-1}},\ \frac{(n-1)s^2}{\chi^2_{\alpha/2,\,n-1}}\right]

Inputs

x the data values (at least two) vector
confidence confidence level as a fraction (0.95 for 95%)

Returns

variance sample variance
lower lower limit for the variance
upper upper limit for the variance
sd_lower lower limit for the standard deviation
sd_upper upper limit for the standard deviation

Applies when

normal population (the interval is sensitive to non-normality)

Example

m.ask("Find the 95% confidence interval for the population variance of [12, 15, 11, 18, 14, 16].")

→ variance = 6.66667, lower = 2.59757, upper = 40.1021, sd_lower = 1.6117, sd_upper = 6.33262

Confidence interval for the difference of two meansci_two_means

The t confidence interval for mu1 - mu2 from two independent samples (Welch or pooled variances).

(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2,\,\nu}\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}

Inputs

x first sample vector
y second sample vector
confidence confidence level as a fraction (0.95 for 95%)
variance welch (unequal variances) or pooled (equal variances) choice

Returns

difference mean(x) - mean(y)
lower lower limit
upper upper limit
df degrees of freedom (Welch-Satterthwaite or n1 + n2 - 2)
standard_error standard error of the difference

Applies when

independent random samples
roughly normal populations or large samples

Example

m.ask("Find the 95% confidence interval for the difference in means of [12, 15, 11, 18, 14] and [10, 9, 13, 11, 8] using Welch.")

→ difference = 3.8, lower = 0.278295, upper = 7.3217, df = 7.17415, standard_error = 1.49666

Wilson score interval for a proportionwilson_interval

The Wilson score confidence interval for a proportion (good coverage for small n and for p near 0 or 1).

\frac{\hat{p} + \frac{z^2}{2n} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n} + \frac{z^2}{4n^2}}}{1 + \frac{z^2}{n}}

Inputs

x number of successes integer
n sample size integer
confidence confidence level as a fraction (0.95 for 95%)

Returns

proportion x/n
lower lower limit
upper upper limit
center interval centre

Applies when

0 <= x <= n
independent trials

Example

m.ask("Find the 95% Wilson interval for x = 3 successes out of n = 20.")

→ proportion = 0.15, lower = 0.0523687, upper = 0.360419, center = 0.206394

Confidence intervals and tests 4

Critical value of chi-squarechi_square_critical_value

The right-tail chi-square critical value (P(X > c) = alpha) for given degrees of freedom, and the left-tail value (P(X < c) = alpha).

\chi^2_{\alpha,\nu}:\ P(\chi^2_\nu > c) = \alpha

Inputs

df degrees of freedom (> 0)
alpha tail area

Returns

critical_value right-tail critical value
left_critical_value left-tail critical value

Applies when

df > 0
0 < alpha < 1

Example

m.ask("Find the chi-square critical value with df = 4 and alpha = 0.05.")

→ critical_value = 9.48773, left_critical_value = 0.710723

Critical value of Ff_critical_value

The right-tail F critical value (P(F > c) = alpha) for numerator and denominator degrees of freedom.

F_{\alpha;\,\nu_1,\nu_2}:\ P(F_{\nu_1,\nu_2} > c) = \alpha

Inputs

df1 numerator degrees of freedom
df2 denominator degrees of freedom
alpha tail area

Returns

critical_value right-tail F critical value

Applies when

df1, df2 > 0
0 < alpha < 1

Example

m.ask("Find the F critical value with df1 = 3, df2 = 20 and alpha = 0.05.")

→ critical_value = 3.09839

Critical value of Student's tt_critical_value

The Student t critical value for given degrees of freedom and significance (or confidence) level.

t_{\alpha/2,\,\nu}:\ P(T_\nu > t) = \alpha/2

Inputs

df degrees of freedom (> 0)
alpha significance level (1 - confidence)
tails two-tailed, right-tailed or left-tailed choice

Returns

critical_value t critical value (negative for a left-tailed test)

Applies when

df > 0
0 < alpha < 1
a confidence level is read as two-tailed

Example

m.ask("Find the critical t value for a 95% confidence interval with df = 10.")

→ critical_value = 2.22814

Critical value of zz_critical_value

The standard normal critical value for a significance level (or confidence level), two-tailed or one-tailed.

z_{\alpha/2} = \Phi^{-1}(1 - \alpha/2)

Inputs

alpha significance level (1 - confidence)
tails two-tailed, right-tailed or left-tailed choice

Returns

critical_value z critical value (negative for a left-tailed test)

Applies when

0 < alpha < 1
a confidence level is read as two-tailed

Example

m.ask("Find the critical z value for a 95% confidence interval.")

→ critical_value = 1.95996

Descriptive statistics 22

Arithmetic meanmean

The arithmetic mean (average) of a list of numbers.

\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i

Inputs

x the data values vector

Returns

mean arithmetic mean
n number of values
sum sum of the values

Example

m.ask("Find the mean of [2, 4, 9].")

→ mean = 5, n = 3, sum = 15

Coefficient of variationcoefficient_of_variation

The coefficient of variation (sample standard deviation divided by the mean), as a ratio and a percentage.

CV = \frac{s}{\bar{x}}

Inputs

x the data values (at least two) vector

Returns

cv s / mean
cv_percent 100 s / mean
mean mean
standard_deviation sample standard deviation

Applies when

positive mean
sample standard deviation (n - 1)

Example

m.ask("Find the coefficient of variation of [10, 12, 14, 16, 18].")

→ cv = 0.225877, cv_percent = 22.5877, mean = 14, standard_deviation = 3.16228

Five-number summary and interquartile rangefive_number_summary

Minimum, first quartile, median, third quartile, maximum and the interquartile range of a data set.

Q_p = x_{(\lfloor h \rfloor)} + (h - \lfloor h \rfloor)\,(x_{(\lfloor h \rfloor + 1)} - x_{(\lfloor h \rfloor)}),\ h = 1 + (n-1)p

Inputs

x the data values (at least two) vector
method quantile method: linear (type 7, spreadsheet/numpy default) or weibull (type 6, p(n+1)) choice optional

Returns

minimum smallest value
q1 first quartile
median median
q3 third quartile
maximum largest value
iqr interquartile range Q3 - Q1

Applies when

quartiles by linear interpolation between order statistics (Hyndman-Fan type 7, as in spreadsheets and numpy) unless method = weibull (type 6, as in Minitab and SPSS); textbooks that take medians of the halves can differ

Example

m.ask("Find the five-number summary of [1, 3, 5, 7, 9, 11, 13, 15].")

→ minimum = 1, q1 = 4.5, median = 8, q3 = 11.5, maximum = 15, iqr = 7

Geometric meangeometric_mean

The geometric mean of positive numbers (average growth factors, ratios).

G = \left(\prod x_i\right)^{1/n}

Inputs

x positive values vector

Returns

geometric_mean n-th root of the product

Applies when

all values positive

Example

m.ask("Find the geometric mean of [2, 8].")

→ geometric_mean = 4

Harmonic meanharmonic_mean

The harmonic mean of positive numbers (average speeds over equal distances, average rates).

H = \frac{n}{\sum 1/x_i}

Inputs

x positive values vector

Returns

harmonic_mean n divided by the sum of reciprocals

Applies when

all values positive

Example

m.ask("Find the harmonic mean of [40, 60].")

→ harmonic_mean = 48

Mean absolute deviationmean_absolute_deviation

The mean absolute deviation about the mean of a list of numbers.

MAD = \frac{1}{n}\sum |x_i - \bar{x}|

Inputs

x the data values vector

Returns

mean_absolute_deviation average distance from the mean
mean mean

Example

m.ask("Find the mean absolute deviation of [2, 4, 6, 8].")

→ mean_absolute_deviation = 2, mean = 5

Mean and standard deviation of a frequency tablefrequency_table_statistics

The mean and the sample and population standard deviations of data given as values with frequencies.

\bar{x} = \frac{\sum f_i x_i}{\sum f_i},\quad s = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{n - 1}}

Inputs

x the distinct values vector
f their frequencies (whole numbers) vector

Returns

mean sum(f x) / n
sample_standard_deviation divisor n - 1
population_standard_deviation divisor n
n total frequency

Applies when

frequencies are whole numbers
total frequency >= 2

Example

m.ask("Find the mean and standard deviation of the frequency table with values [1, 2, 3, 4] and frequencies [3, 5, 8, 4].")

→ mean = 2.65, sample_standard_deviation = 0.988087, population_standard_deviation = 0.963068, n = 20

Medianmedian

The median (middle value) of a list of numbers; the mean of the two middle values when n is even.

\tilde{x} = x_{((n+1)/2)} \ \text{or}\ \tfrac12\left(x_{(n/2)} + x_{(n/2+1)}\right)

Inputs

x the data values vector

Returns

median median
n number of values

Example

m.ask("Find the median of [7, 1, 5, 3, 9].")

→ median = 5, n = 5

Median absolute deviationmedian_absolute_deviation

The median absolute deviation (MAD) about the median, a robust measure of spread (unscaled).

MAD = \operatorname{median}\left(|x_i - \tilde{x}|\right)

Inputs

x the data values vector

Returns

median_absolute_deviation median of |x - median|
median median

Applies when

unscaled (multiply by 1.4826 to estimate a normal sigma)

Example

m.ask("Find the median absolute deviation of [1, 1, 2, 2, 4, 6, 9].")

→ median_absolute_deviation = 1, median = 2

Modemode

The mode(s) of a list of numbers: every value that occurs most often (no mode when all values occur equally often).

\operatorname{mode} = \arg\max_v \#\{i : x_i = v\}

Inputs

x the data values vector

Returns

modes the most frequent value(s), ascending; empty if there is no mode
frequency how often the mode occurs
has_mode yes or no

Applies when

no mode is reported when every distinct value occurs equally often

Example

m.ask("Find the mode of [3, 7, 7, 2, 9, 7, 3].")

→ modes = [7.0], frequency = 3, has_mode = yes

Outliers by the 1.5 IQR ruleiqr_outliers

The lower and upper fences Q1 - 1.5 IQR and Q3 + 1.5 IQR and the values outside them.

[Q_1 - 1.5\,IQR,\ Q_3 + 1.5\,IQR]

Inputs

x the data values (at least four) vector

Returns

lower_fence Q1 - 1.5 IQR
upper_fence Q3 + 1.5 IQR
outliers values outside the fences
q1 first quartile
q3 third quartile
iqr interquartile range

Applies when

quartiles by linear interpolation (type 7)

Example

m.ask("Find the outliers in [10, 12, 12, 13, 12, 11, 14, 13, 15, 102].")

→ lower_fence = 9.375, upper_fence = 16.375, outliers = [102.0], q1 = 12, q3 = 13.75, iqr = 1.75

Percentile of a data setpercentile

The p-th percentile of a data set (linear interpolation, or the weibull/p(n+1) method).

P_p = x_{(\lfloor h \rfloor)} + (h - \lfloor h \rfloor)(x_{(\lfloor h \rfloor+1)} - x_{(\lfloor h \rfloor)})

Inputs

x the data values (at least two) vector
p the percentile, 0 to 100
method quantile method: linear (type 7, spreadsheet/numpy default) or weibull (type 6, p(n+1)) choice optional

Returns

percentile the p-th percentile

Applies when

0 <= p <= 100
linear interpolation (type 7) unless method = weibull (type 6)

Example

m.ask("Find the 90th percentile of [15, 20, 35, 40, 50].")

→ percentile = 46

Population standard deviationpopulation_standard_deviation

The population standard deviation of a list of numbers (divisor n).

\sigma = \sqrt{\frac{1}{n}\sum (x_i - \mu)^2}

Inputs

x the data values vector

Returns

standard_deviation population standard deviation
mean arithmetic mean
n number of values
variance the companion measure

Applies when

the data are the whole population (divisor n)

Example

m.ask("Find the population standard deviation of [2, 4, 4, 4, 5, 5, 7, 9].")

→ mean = 5, n = 8, standard_deviation = 2, variance = 4

Population variancepopulation_variance

The population variance of a list of numbers (divisor n).

\sigma^2 = \frac{1}{n}\sum (x_i - \mu)^2

Inputs

x the data values vector

Returns

variance population variance
mean arithmetic mean
n number of values
standard_deviation the companion measure

Applies when

the data are the whole population (divisor n)

Example

m.ask("Find the population variance of [2, 4, 4, 4, 5, 5, 7, 9].")

→ mean = 5, n = 8, variance = 4, standard_deviation = 2

Rangerange

The range (largest minus smallest value) of a list of numbers.

R = x_{\max} - x_{\min}

Inputs

x the data values vector

Returns

range maximum minus minimum
minimum smallest value
maximum largest value

Example

m.ask("Find the range of [12, 4, 19, 7].")

→ range = 15, minimum = 4, maximum = 19

Sample standard deviationsample_standard_deviation

The sample standard deviation of a list of numbers (divisor n - 1).

s = \sqrt{\frac{1}{n-1}\sum (x_i - \bar{x})^2}

Inputs

x the data values (at least two) vector

Returns

standard_deviation sample standard deviation
mean arithmetic mean
n number of values
variance the companion measure

Applies when

the data are a sample (divisor n - 1, Bessel's correction); at least two values

Example

m.ask("Find the sample standard deviation of [2, 4, 4, 4, 5, 5, 7, 9].")

→ mean = 5, n = 8, standard_deviation = 2.13809, variance = 4.57143

Sample variancesample_variance

The sample variance of a list of numbers (divisor n - 1).

s^2 = \frac{1}{n-1}\sum (x_i - \bar{x})^2

Inputs

x the data values (at least two) vector

Returns

variance sample variance
mean arithmetic mean
n number of values
standard_deviation the companion measure

Applies when

the data are a sample (divisor n - 1, Bessel's correction); at least two values

Example

m.ask("Find the sample variance of [2, 4, 4, 4, 5, 5, 7, 9].")

→ mean = 5, n = 8, variance = 4.57143, standard_deviation = 2.13809

Skewness and kurtosisskewness_kurtosis

Sample skewness and excess kurtosis: the adjusted Fisher-Pearson estimates (G1, G2, as in spreadsheets and most software) and the moment estimates (g1, g2).

G_1 = \frac{\sqrt{n(n-1)}}{n-2}\,\frac{m_3}{m_2^{3/2}},\quad G_2 = \frac{(n-1)\left((n+1)g_2 + 6\right)}{(n-2)(n-3)}

Inputs

x the data values (at least four) vector

Returns

skewness adjusted skewness G1
excess_kurtosis adjusted excess kurtosis G2
skewness_biased moment skewness g1
excess_kurtosis_biased moment excess kurtosis g2

Applies when

at least four values, not all equal

Example

m.ask("Find the skewness and kurtosis of [2, 3, 3, 4, 8, 12, 20].")

→ skewness = 1.39985, excess_kurtosis = 1.37206, skewness_biased = 1.08001, excess_kurtosis_biased = -0.178307

Standard error of the meanstandard_error_of_mean

The standard error of the mean of a sample: s / sqrt(n).

SE_{\bar{x}} = \frac{s}{\sqrt{n}}

Inputs

x the data values (at least two) vector

Returns

standard_error s / sqrt(n)
standard_deviation sample standard deviation
n sample size

Example

m.ask("Find the standard error of the mean of [12, 15, 11, 18, 14].")

→ standard_error = 1.22474, standard_deviation = 2.73861, n = 5

Trimmed meantrimmed_mean

The trimmed mean: the mean after removing the given proportion of values from EACH end.

\bar{x}_{\alpha} = \frac{1}{n - 2k}\sum_{i=k+1}^{n-k} x_{(i)},\ k = \lfloor \alpha n \rfloor

Inputs

x the data values vector
proportion proportion cut from each end (0.1 for a 10% trimmed mean)

Returns

trimmed_mean mean of the remaining values
removed_each_end values removed at each end

Applies when

0 <= proportion < 0.5
floor(proportion * n) values removed from each end

Example

m.ask("Find the 10% trimmed mean of [2, 4, 5, 6, 7, 8, 9, 10, 12, 50].")

→ trimmed_mean = 7.625, removed_each_end = 1

Weighted meanweighted_mean

The weighted mean of values x with weights w (grade-point averages, portfolio averages).

\bar{x}_w = \frac{\sum w_i x_i}{\sum w_i}

Inputs

x the values vector
w the weights (non-negative, same length) vector

Returns

weighted_mean sum(w x) / sum(w)
total_weight sum of the weights

Applies when

same number of values and weights
weights >= 0 with a positive sum

Example

m.ask("Find the weighted mean of values [90, 80, 70] with weights [3, 2, 1].")

→ weighted_mean = 83.3333, total_weight = 6

z-score of a valuez_score

The z-score (standard score) of a value x in a distribution with mean mu and standard deviation sigma.

z = \frac{x - \mu}{\sigma}

Inputs

x the value
mu the mean
sigma the standard deviation (> 0)

Returns

z number of standard deviations x lies above (+) or below (-) the mean

Applies when

sigma > 0

Example

m.ask("Find the z-score of x = 85 when the mean = 70 and the standard deviation = 10.")

→ z = 1.5

Effect sizes 1

Cohen's d and Hedges' gcohens_d

The standardized mean difference of two independent samples: Cohen's d (pooled SD) and Hedges' g (bias-corrected).

d = \frac{\bar{x}_1 - \bar{x}_2}{s_p},\ s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}

Inputs

x first sample vector
y second sample vector

Returns

d (mean x - mean y) / pooled SD
hedges_g d (1 - 3/(4(n1 + n2) - 9))
pooled_sd pooled standard deviation

Applies when

independent samples

Example

m.ask("Find Cohen's d for [12, 15, 11, 18, 14] and [10, 9, 13, 11, 8].")

→ d = 1.60579, hedges_g = 1.45039, pooled_sd = 2.36643

Experimental design 4

2^2 factorial design: main effects and interactionfactorial_2x2_effects

Effects of a two-level 4-run factorial experiment from one response per run in standard (Yates) order ((1), a, b, ab).

\text{Effect} = \frac{\text{contrast}}{2^{k-1}}

Inputs

y 4 responses in standard order vector

Returns

grand_mean average response
A effect A
B effect B
AB effect AB

Applies when

responses in standard order
one response (or cell mean) per run

Example

m.ask("Find the effects of the 2^2 factorial design with responses in standard order [28, 36, 18, 31].")

→ grand_mean = 28.25, A = 10.5, B = -7.5, AB = 2.5

2^3 factorial design: main effects and interactionsfactorial_2x2x2_effects

Effects of a two-level 8-run factorial experiment from one response per run in standard (Yates) order ((1), a, b, ab, c, ac, bc, abc).

\text{Effect} = \frac{\text{contrast}}{2^{k-1}}

Inputs

y 8 responses in standard order vector

Returns

grand_mean average response
A effect A
B effect B
C effect C
AB effect AB
AC effect AC
BC effect BC
ABC effect ABC

Applies when

responses in standard order
one response (or cell mean) per run

Example

m.ask("Find the effects of the 2^3 factorial design with responses in standard order [22, 32, 35, 55, 44, 40, 60, 39].")

→ grand_mean = 40.875, A = 1.25, B = 12.75, C = 9.75, AB = -1.75, AC = -13.75, BC = -5.25, ABC = -6.75

Cronbach's alpha (internal consistency)cronbach_alpha

Cronbach's alpha for a questionnaire or test: rows are respondents, columns are items.

\alpha = \frac{k}{k-1}\left(1 - \frac{\sum s_i^2}{s_T^2}\right)

Inputs

data scores: one row per respondent, one column per item matrix

Returns

alpha k/(k-1) (1 - sum of item variances / variance of totals)
items k
respondents n

Applies when

at least 2 items and 2 respondents

Example

m.ask("Find Cronbach's alpha for [[4, 5, 4], [3, 3, 4], [5, 5, 5], [2, 3, 2], [4, 4, 3]].")

→ alpha = 0.913043, items = 3, respondents = 5

Randomized complete block design (two-way ANOVA without replication)randomized_block_anova

Two-way ANOVA for a randomized complete block design: rows are blocks, columns are treatments, one observation per cell; F-tests for treatments and blocks.

F_{\text{trt}} = \frac{SS_{\text{trt}}/(t-1)}{SS_E/((t-1)(b-1))}

Inputs

data table with one row per block and one column per treatment matrix

Returns

f_treatment F for treatments
p_treatment p-value for treatments
f_block F for blocks
p_block p-value for blocks
ss_treatment treatment sum of squares
ss_block block sum of squares
ss_error error sum of squares
df_treatment t - 1
df_block b - 1
df_error (t - 1)(b - 1)

Applies when

one observation per block and treatment
additive model (no block x treatment interaction)
normal errors with constant variance

Example

m.ask("Analyse the randomized block design with data [[12, 14, 15], [10, 13, 13], [11, 15, 16], [9, 12, 14]].")

→ f_treatment = 39, p_treatment = 0.000364431, f_block = 9.25, p_block = 0.0114399, ss_treatment = 34.6667, ss_block = 12.3333, ss_error = 2.66667, df_treatment = 2, df_block = 3, df_error = 6

Hypothesis tests 14

Bartlett's test for equal variancesbartlett_test

Bartlett's test of equal variances across 2 to 5 groups (normal populations).

T = \frac{(N-k)\ln s_p^2 - \sum (n_i - 1)\ln s_i^2}{1 + \frac{1}{3(k-1)}\left(\sum \frac{1}{n_i - 1} - \frac{1}{N-k}\right)}

Inputs

g1 group 1 vector
g2 group 2 vector
g3 group 3 vector optional
g4 group 4 vector optional
g5 group 5 vector optional

Returns

statistic Bartlett's chi-square statistic
df k - 1
p_value p-value

Applies when

normal populations (Bartlett's test is sensitive to non-normality)
at least 2 values per group

Example

m.ask("Bartlett's test for [21, 24, 19, 26, 23], [22, 23, 22, 24, 23] and [18, 28, 20, 30, 24].")

→ statistic = 8.61822, df = 2, p_value = 0.0134455

Chi-square goodness-of-fit testchi_square_goodness_of_fit

The chi-square goodness-of-fit test of observed counts against expected proportions (equal if omitted).

\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i},\ \nu = k - 1

Inputs

observed observed counts per category vector
proportions expected proportions or relative weights per category (equal if omitted) vector optional

Returns

chi_square sum (O - E)^2 / E
df categories - 1
p_value upper-tail p-value
expected expected counts

Applies when

independent observations
expected counts at least about 5
proportions are rescaled to sum to 1

Example

m.ask("Chi-square goodness-of-fit test for observed [18, 22, 20, 25, 15] with equal expected proportions.")

→ chi_square = 2.9, df = 4, p_value = 0.574697, expected = [20.0, 20.0, 20.0, 20.0, 20.0]

Chi-square test for a variancechi_square_variance_test

The chi-square test of H0: sigma = sigma0 for a normal population from sample data.

\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}

Inputs

x the data values (at least two) vector
sigma0 hypothesised standard deviation
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

chi_square (n - 1) s^2 / sigma0^2
df n - 1
p_value p-value
sample_variance s^2

Applies when

normal population (the test is sensitive to non-normality)

Example

m.ask("Chi-square test for a variance: data [10.2, 9.8, 10.5, 10.1, 9.6, 10.4], sigma0 = 0.3, two-sided.")

→ chi_square = 6.66667, df = 5, p_value = 0.493268, sample_variance = 0.12

Chi-square test of independencechi_square_independence

The chi-square test of independence (or homogeneity) for an r x c contingency table of counts.

\chi^2 = \sum_{i,j}\frac{(O_{ij} - E_{ij})^2}{E_{ij}},\ E_{ij} = \frac{R_i C_j}{n}

Inputs

table contingency table of counts (rows x columns) matrix

Returns

chi_square Pearson chi-square statistic
df (r - 1)(c - 1)
p_value upper-tail p-value
expected expected counts under independence

Applies when

independent observations
expected counts at least about 5
no continuity correction

Example

m.ask("Chi-square test of independence for the table [[20, 30], [25, 25]].")

→ chi_square = 1.0101, df = 1, p_value = 0.314879, expected = [[22.5, 27.5], [22.5, 27.5]]

F-test for equality of two variancesf_test_variances

The F-test of H0: sigma1^2 = sigma2^2 from two independent normal samples.

F = \frac{s_1^2}{s_2^2}

Inputs

x first sample vector
y second sample vector
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

f s1^2 / s2^2
df1 n1 - 1
df2 n2 - 1
p_value p-value

Applies when

independent samples from normal populations (sensitive to non-normality)

Example

m.ask("F-test for equality of variances of [21, 24, 19, 26, 23, 25] and [22, 23, 22, 24, 23], two-sided.")

→ f = 9.71429, df1 = 5, df2 = 4, p_value = 0.046857

Levene's test for equal varianceslevene_test

Levene's test of equal variances across 2 to 5 groups, centred at the median (Brown-Forsythe) or the mean.

W = \frac{(N-k)\sum n_i(\bar{z}_{i\cdot} - \bar{z})^2}{(k-1)\sum\sum (z_{ij} - \bar{z}_{i\cdot})^2},\ z_{ij} = |x_{ij} - \tilde{x}_i|

Inputs

g1 group 1 vector
g2 group 2 vector
g3 group 3 vector optional
g4 group 4 vector optional
g5 group 5 vector optional
center median (Brown-Forsythe) or mean (Levene's original) choice

Returns

w Levene statistic
p_value p-value
df1 k - 1
df2 N - k

Applies when

independent groups, at least 2 values each

Example

m.ask("Levene's test (median centred) for [21, 24, 19, 26, 23], [22, 23, 22, 24, 23] and [18, 28, 20, 30, 24].")

→ w = 4.97727, p_value = 0.026665, df1 = 2, df2 = 12

One-proportion z-testproportion_z_test

The z-test of H0: p = p0 for a population proportion from x successes in n trials.

z = \frac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}}

Inputs

x number of successes integer
n sample size integer
p0 hypothesised proportion
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

z (x/n - p0)/sqrt(p0(1 - p0)/n)
p_value p-value
proportion x/n

Applies when

n p0 and n(1 - p0) at least about 10
independent trials

Example

m.ask("Test H0: p = 0.5 for a proportion with x = 58 successes out of n = 100, p0 = 0.5, two-sided.")

→ z = 1.6, p_value = 0.109599, proportion = 0.58

One-sample t-testt_test_one_sample

The one-sample t-test of H0: mu = mu0 from sample data: t statistic, degrees of freedom and p-value.

t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}},\ \nu = n - 1

Inputs

x the data values (at least two) vector
mu0 hypothesised mean
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

t (mean - mu0)/(s/sqrt(n))
df n - 1
p_value p-value
mean sample mean
standard_deviation sample standard deviation

Applies when

random sample from a roughly normal population (or large n)
the alternative must be stated for a one-sided claim

Example

m.ask("Perform a one-sample t-test of [5.1, 4.9, 5.3, 5.5, 5.0, 5.2] against mu0 = 5, two-sided.")

→ t = 1.88982, df = 5, p_value = 0.117387, mean = 5.16667, standard_deviation = 0.216025

One-sample z-test for a mean (sigma known)z_test_mean

The z statistic and p-value for H0: mu = mu0 when the population standard deviation is known.

z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}

Inputs

xbar sample mean
mu0 hypothesised mean
sigma population standard deviation (known)
n sample size integer
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

z (xbar - mu0) / (sigma / sqrt(n))
p_value p-value for the chosen alternative

Applies when

sigma known
normal population or large n
the alternative must be stated for a one-sided claim

Example

m.ask("Perform a z-test of H0: mu = 50 with sample mean = 52, sigma = 5 and n = 36, two-sided.")

→ z = 2.4, p_value = 0.0163951

Paired t-testpaired_t_test

The paired (dependent samples) t-test on the differences x - y: t statistic, df and p-value.

t = \frac{\bar{d}}{s_d/\sqrt{n}},\ \nu = n - 1

Inputs

x first measurement of each pair vector
y second measurement of each pair vector
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

t mean difference / (s_d / sqrt(n))
df n - 1
p_value p-value
mean_difference mean of x - y

Applies when

pairs measured on the same units
differences roughly normal (or many pairs)

Example

m.ask("Perform a paired t-test on before [200, 190, 210, 220, 205] and after [195, 185, 200, 214, 200], two-sided.")

→ t = 6.39481, df = 4, p_value = 0.00307015, mean_difference = 6.2

Two-proportion z-testtwo_proportion_z_test

The pooled z-test of H0: p1 = p2 from x1 of n1 and x2 of n2 successes.

z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}

Inputs

x1 successes in sample 1 integer
n1 size of sample 1 integer
x2 successes in sample 2 integer
n2 size of sample 2 integer
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

z z statistic (pooled standard error)
p_value p-value
pooled_proportion (x1 + x2)/(n1 + n2)
difference x1/n1 - x2/n2

Applies when

independent samples
at least about 10 successes and failures in each sample

Example

m.ask("Two-proportion z-test with x1 = 45, n1 = 100, x2 = 30, n2 = 100, two-sided.")

→ z = 2.19089, p_value = 0.0284597, pooled_proportion = 0.375, difference = 0.15

Two-sample t-test (pooled variances)t_test_two_sample_pooled

Two-sample t-test (pooled variances) comparing the means of two independent samples: t statistic, degrees of freedom, p-value.

t = \frac{\bar{x}_1 - \bar{x}_2}{s_p\sqrt{1/n_1 + 1/n_2}},\ \nu = n_1 + n_2 - 2

Inputs

x first sample vector
y second sample vector
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

t t statistic
df degrees of freedom
p_value p-value
difference mean(x) - mean(y)

Applies when

independent random samples
roughly normal populations or large samples
equal population variances assumed

Example

m.ask("Perform a two-sample t-test with pooled variances on [12, 15, 11, 18, 14] and [10, 9, 13, 11, 8], two-sided.")

→ t = 2.53898, df = 8, p_value = 0.0347658, difference = 3.8

Two-sample t-test from summary statisticst_test_summary_statistics

The two-sample t-test from the means, sample standard deviations and sizes of two groups (Welch or pooled).

t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}}

Inputs

m1 mean of group 1
s1 sample standard deviation of group 1
n1 size of group 1 integer
m2 mean of group 2
s2 sample standard deviation of group 2
n2 size of group 2 integer
variance welch (unequal variances) or pooled (equal variances) choice
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

t t statistic
df degrees of freedom
p_value p-value
difference m1 - m2

Applies when

independent random samples
roughly normal populations or large samples

Example

m.ask("Two-sample t-test from summary statistics with pooled variances: mean1 = 20, sd1 = 4, n1 = 15, mean2 = 17, sd2 = 5, n2 = 12, two-sided.")

→ t = 1.73379, df = 25, p_value = 0.0952738, difference = 3

Welch's two-sample t-test (unequal variances)welch_t_test

Welch's two-sample t-test (unequal variances) comparing the means of two independent samples: t statistic, degrees of freedom, p-value.

t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}}

Inputs

x first sample vector
y second sample vector
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

t t statistic
df degrees of freedom
p_value p-value
difference mean(x) - mean(y)

Applies when

independent random samples
roughly normal populations or large samples
Welch-Satterthwaite degrees of freedom

Example

m.ask("Perform a two-sample t-test with unequal variances on [12, 15, 11, 18, 14] and [10, 9, 13, 11, 8], two-sided.")

→ t = 2.53898, df = 7.17415, p_value = 0.0379455, difference = 3.8

Nonparametric tests 7

Friedman testfriedman_test

The Friedman rank test for k >= 3 related treatments measured on the same blocks (subjects): rows are blocks, columns are treatments.

Q = \frac{12}{nk(k+1)}\sum_j R_j^2 - 3n(k+1)

Inputs

data table with one row per block (subject) and one column per treatment matrix

Returns

chi_square Friedman statistic (tie-corrected)
df k - 1
p_value chi-square p-value

Applies when

at least 3 treatments and 2 blocks
chi-square approximation

Example

m.ask("Perform a Friedman test on [[7, 9, 8], [6, 5, 7], [9, 7, 6], [8, 5, 6], [7, 8, 9]].")

→ chi_square = 0.4, df = 2, p_value = 0.818731

Kruskal-Wallis H testkruskal_wallis

The Kruskal-Wallis rank test comparing 2 to 5 independent groups (nonparametric one-way ANOVA).

H = \frac{12}{N(N+1)}\sum \frac{R_i^2}{n_i} - 3(N+1)

Inputs

g1 group 1 vector
g2 group 2 vector
g3 group 3 vector optional
g4 group 4 vector optional
g5 group 5 vector optional

Returns

h H statistic (tie-corrected)
df k - 1
p_value chi-square p-value

Applies when

independent groups, ordinal or continuous data
chi-square approximation (about 5 or more per group)

Example

m.ask("Perform a Kruskal-Wallis test on [27, 2, 4, 18, 7, 9], [20, 8, 14, 36, 21, 22] and [34, 31, 3, 23, 30, 6].")

→ h = 2.8538, df = 2, p_value = 0.240052

Mann-Whitney U test (exact)mann_whitney_exact

The Mann-Whitney U test (exact): compares two independent samples by ranks (statistic U of the first sample and the p-value).

U_1 = R_1 - \frac{n_1(n_1+1)}{2}

Inputs

x first sample vector
y second sample vector
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

u U statistic of the first sample (R1 - n1(n1+1)/2)
p_value p-value

Applies when

independent samples, ordinal or continuous data
no ties
samples of at most 30

Example

m.ask("Perform a exact Mann-Whitney U test on [3.1, 5.2, 4.3, 6.4, 7.5] and [1.5, 2.6, 2.2, 3.9], two-sided.")

→ u = 19, p_value = 0.031746

Mann-Whitney U test (large-sample)mann_whitney_u

The Mann-Whitney U test (large-sample): compares two independent samples by ranks (statistic U of the first sample and the p-value).

U_1 = R_1 - \frac{n_1(n_1+1)}{2}

Inputs

x first sample vector
y second sample vector
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

u U statistic of the first sample (R1 - n1(n1+1)/2)
p_value p-value

Applies when

independent samples, ordinal or continuous data
normal approximation with tie correction and continuity correction

Example

m.ask("Perform a Mann-Whitney U test on [3, 5, 4, 6, 7] and [1.5, 2.6, 2.2, 3.9], two-sided.")

→ u = 19, p_value = 0.0373364

Sign testsign_test

The sign test for the median difference of paired data (or of differences given directly): exact binomial p-value.

P = 2\min\left(P(K \le k), P(K \ge k)\right),\ K \sim \mathrm{Bin}(n, \tfrac12)

Inputs

x differences, or the first measurement of each pair vector
y second measurement of each pair (optional) vector optional
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

positives number of positive differences
n non-zero differences
p_value exact binomial p-value

Applies when

paired data
zero differences dropped

Example

m.ask("Perform a sign test on before [125, 115, 130, 140, 141, 115, 140, 125] and after [110, 122, 125, 120, 140, 124, 123, 137], two-sided.")

→ positives = 5, n = 8, p_value = 0.726562

Wilcoxon signed-rank test (exact)wilcoxon_signed_rank_exact

The Wilcoxon signed-rank test (exact) for paired data (x - y) or for differences given directly (x alone); zero differences are dropped.

R^+ = \sum_{d_i > 0} \operatorname{rank}(|d_i|)

Inputs

x differences, or the first measurement of each pair vector
y second measurement of each pair (optional) vector optional
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

statistic min(R+, R-) for a two-sided test, R+ otherwise
p_value p-value
r_plus sum of ranks of positive differences
n non-zero differences

Applies when

paired data, symmetric distribution of differences
zero differences dropped
no zero and no tied absolute differences
at most 60 differences

Example

m.ask("Perform a exact Wilcoxon signed-rank test on before [125, 115, 130, 140, 141, 115, 140, 125] and after [110, 122, 125, 120, 140, 124, 123, 137], two-sided.")

→ statistic = 12, p_value = 0.460938, r_plus = 24, n = 8

Wilcoxon signed-rank test (large-sample)wilcoxon_signed_rank

The Wilcoxon signed-rank test (large-sample) for paired data (x - y) or for differences given directly (x alone); zero differences are dropped.

R^+ = \sum_{d_i > 0} \operatorname{rank}(|d_i|)

Inputs

x differences, or the first measurement of each pair vector
y second measurement of each pair (optional) vector optional
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

statistic min(R+, R-) for a two-sided test, R+ otherwise
p_value p-value
r_plus sum of ranks of positive differences
n non-zero differences

Applies when

paired data, symmetric distribution of differences
zero differences dropped
normal approximation with tie correction, no continuity correction

Example

m.ask("Perform a Wilcoxon signed-rank test on before [125, 115, 130, 140, 141, 115, 140, 125] and after [110, 122, 125, 120, 140, 124, 123, 137], two-sided.")

→ statistic = 12, p_value = 0.400814, r_plus = 24, n = 8

Power and sample size 5

Power of a one-sample z-testpower_one_sample_z

The power of a one-sample z-test for a standardized effect size d (Cohen's d), sample size n and significance level alpha.

1 - \beta = \Phi\left(d\sqrt{n_{\text{eff}}} - z_{1-\alpha/2}\right) + \Phi\left(-d\sqrt{n_{\text{eff}}} - z_{1-\alpha/2}\right)

Inputs

d standardized effect size (difference / sigma)
n sample size
alpha significance level
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

power probability of rejecting H0 when the effect is d

Applies when

normal approximation (sigma known)
two-sided power includes both rejection tails

Example

m.ask("Find the power of a one-sample z-test with effect size d = 0.5, n = 30 and alpha = 0.05, two-sided.")

→ power = 0.781908

Power of a two-sample z-testpower_two_sample_z

The power of a two-sample z-test for a standardized effect size d (Cohen's d), sample size n per group and significance level alpha.

1 - \beta = \Phi\left(d\sqrt{n_{\text{eff}}} - z_{1-\alpha/2}\right) + \Phi\left(-d\sqrt{n_{\text{eff}}} - z_{1-\alpha/2}\right)

Inputs

d standardized effect size (difference / sigma)
n sample size per group
alpha significance level
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

power probability of rejecting H0 when the effect is d

Applies when

normal approximation (sigma known)
two-sided power includes both rejection tails

Example

m.ask("Find the power of a two-sample z-test with effect size d = 0.5, n = 30 and alpha = 0.05, two-sided.")

→ power = 0.490686

Sample size per group to compare two means (z-test power)sample_size_two_means

The sample size per group for a two-sample z-test to reach a target power for a standardized effect size d.

n \approx 2\left(\frac{z_{1-\alpha/2} + z_{1-\beta}}{d}\right)^2

Inputs

d standardized effect size (difference / sigma)
alpha significance level
power target power (e.g. 0.8)
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

n_per_group required size per group (rounded up)
n_exact solution before rounding

Applies when

equal group sizes
normal approximation
two-sided power includes both tails

Example

m.ask("What sample size per group is needed for two groups with effect size d = 0.5, alpha = 0.05 and power = 0.8, two-sided?")

→ n_per_group = 63, n_exact = 62.7909

Sample size to estimate a meansample_size_mean

The smallest n giving a z confidence interval for a mean with the required margin of error, sigma known.

n = \left\lceil \left(\frac{z_{\alpha/2}\,\sigma}{E}\right)^2 \right\rceil

Inputs

sigma population standard deviation (planning value)
margin margin of error
confidence confidence level as a fraction (0.95 for 95%)

Returns

n required sample size (rounded up)
n_exact (z sigma / E)^2

Applies when

sigma known or a planning estimate

Example

m.ask("What sample size is needed to estimate the mean within margin of error = 2 with sigma = 15 at 95% confidence?")

→ n = 217, n_exact = 216.082

Sample size to estimate a proportionsample_size_proportion

The smallest n giving a confidence interval for a proportion with the required margin of error (p = 0.5 if no estimate).

n = \left\lceil \frac{z_{\alpha/2}^2\,p(1-p)}{E^2} \right\rceil

Inputs

margin margin of error (as a proportion, 0.03 for 3 points)
confidence confidence level as a fraction (0.95 for 95%)
p planning estimate of the proportion (default 0.5) optional

Returns

n required sample size (rounded up)
n_exact z^2 p(1-p)/E^2 before rounding

Applies when

large-sample (Wald) interval
simple random sample

Example

m.ask("What sample size is needed to estimate a proportion within margin of error = 0.03 at 95% confidence?")

→ n = 1068, n_exact = 1067.07

Probability 1

Probability of at least one occurrenceprobability_at_least_one

The probability that an event of probability p occurs at least once in n independent trials.

P(\text{at least one}) = 1 - (1 - p)^n

Inputs

p probability of the event on one trial
n number of independent trials integer

Returns

probability 1 - (1 - p)^n
none probability that it never occurs, (1 - p)^n

Applies when

independent trials with the same probability p

Example

m.ask("What is the probability of at least one success in n = 5 independent trials with probability = 0.2?")

→ probability = 0.67232, none = 0.32768

Probability distributions 17

Binomial probability: at least k successesbinomial_at_least

The probability of at least k successes in n independent trials with success probability p.

P(X \ge k) = \sum_{j=k}^{n} \binom{n}{j} p^j (1-p)^{n-j}

Inputs

n number of trials integer
k number of successes integer
p probability of success on each trial

Returns

probability P(X >= k)

Applies when

n independent trials, constant success probability 0 < p < 1
0 <= k <= n

Example

m.ask("Find the binomial probability of at least k = 3 successes in n = 10 trials with p = 0.4.")

→ probability = 0.83271

Binomial probability: at most k successesbinomial_at_most

The probability of at most k successes in n independent trials with success probability p.

P(X \le k) = \sum_{j=0}^{k} \binom{n}{j} p^j (1-p)^{n-j}

Inputs

n number of trials integer
k number of successes integer
p probability of success on each trial

Returns

probability P(X <= k)

Applies when

n independent trials, constant success probability 0 < p < 1
0 <= k <= n

Example

m.ask("Find the binomial probability of at most k = 3 successes in n = 10 trials with p = 0.4.")

→ probability = 0.382281

Binomial probability: exactly k successesbinomial_exactly

The probability of exactly k successes in n independent trials with success probability p.

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

Inputs

n number of trials integer
k number of successes integer
p probability of success on each trial

Returns

probability P(X = k)

Applies when

n independent trials, constant success probability 0 < p < 1
0 <= k <= n

Example

m.ask("Find the binomial probability of exactly k = 3 successes in n = 10 trials with p = 0.4.")

→ probability = 0.214991

Expected value and variance of a discrete random variablediscrete_random_variable

The expected value, variance and standard deviation of a discrete random variable given its values and probabilities.

E[X] = \sum x_i p_i,\quad \operatorname{Var}(X) = \sum (x_i - \mu)^2 p_i

Inputs

x the values of the random variable vector
p their probabilities (summing to 1) vector

Returns

expected_value E[X] = sum x p
variance Var[X] = sum (x - E[X])^2 p
standard_deviation sqrt(Var[X])

Applies when

probabilities non-negative, summing to 1 (within 1e-9)

Example

m.ask("Find the expected value of a random variable with values [0, 1, 2, 3] and probabilities [0.1, 0.3, 0.4, 0.2].")

→ expected_value = 1.7, variance = 0.81, standard_deviation = 0.9

Exponential probability above a valueexponential_probability_above

P(X > x) for an exponential distribution given its rate lambda or its mean.

P(X > x) = e^{-\lambda x}

Inputs

x the value (>= 0)
rate rate lambda (events per unit) optional
mean mean 1/lambda (instead of the rate) optional

Returns

probability P(X > x)

Applies when

exponential (memoryless) waiting time
x >= 0
exactly one of rate or mean

Example

m.ask("For an exponential distribution with rate = 0.5, find P(X > 3).")

→ probability = 0.22313

Exponential probability below a valueexponential_probability_below

P(X < x) for an exponential distribution given its rate lambda or its mean.

P(X < x) = 1 - e^{-\lambda x}

Inputs

x the value (>= 0)
rate rate lambda (events per unit) optional
mean mean 1/lambda (instead of the rate) optional

Returns

probability P(X < x)

Applies when

exponential (memoryless) waiting time
x >= 0
exactly one of rate or mean

Example

m.ask("For an exponential distribution with rate = 0.5, find P(X < 3).")

→ probability = 0.77687

Geometric distribution: first success on trial kgeometric_probability

The probability that the first success occurs on trial k, and that it occurs on or before trial k.

P(X = k) = (1-p)^{k-1}p,\quad P(X \le k) = 1 - (1-p)^k

Inputs

p probability of success on each trial
k the trial number (1, 2, ...) integer

Returns

probability P(X = k) = (1 - p)^(k-1) p
cumulative P(X <= k) = 1 - (1 - p)^k

Applies when

independent trials, constant p
X counts trials up to and including the first success

Example

m.ask("For a geometric distribution with p = 0.2, find the probability that the first success is on trial k = 4.")

→ probability = 0.1024, cumulative = 0.5904

Hypergeometric probability (sampling without replacement)hypergeometric_probability

The probability of exactly k successes, and of at most k, in n draws without replacement from a population of N items of which K are successes.

P(X = k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}

Inputs

N population size integer
K number of successes in the population integer
n number of draws integer
k number of successes drawn integer

Returns

probability P(X = k)
cumulative P(X <= k)

Applies when

draws without replacement
K <= N, n <= N, k <= n

Example

m.ask("Find the hypergeometric probability with population size = 50, population successes = 5, draws = 10 and k = 2.")

→ probability = 0.20984, cumulative = 0.95174

Inverse normal (normal percentile)inverse_normal

The value x with P(X < x) = p for a normal distribution (percentiles, cut-offs, top and bottom percentages).

x = \mu + \sigma\,\Phi^{-1}(p)

Inputs

p cumulative probability to the left of x (0 < p < 1)
mu the mean
sigma the standard deviation (> 0)

Returns

x the value with P(X < x) = p
z the standard normal quantile

Applies when

0 < p < 1
sigma > 0

Example

m.ask("Find the 90th percentile of a normal distribution with mean = 100 and standard deviation = 15.")

→ x = 119.223, z = 1.28155

Mean and standard deviation of a binomial distributionbinomial_mean_sd

The mean np, variance np(1-p) and standard deviation of a binomial random variable.

\mu = np,\quad \sigma = \sqrt{np(1-p)}

Inputs

n number of trials integer
p probability of success

Returns

mean n p
variance n p (1 - p)
standard_deviation sqrt(n p (1 - p))

Applies when

0 < p < 1

Example

m.ask("Find the mean and standard deviation of a binomial distribution with n = 50 and p = 0.3.")

→ mean = 15, variance = 10.5, standard_deviation = 3.24037

Negative binomial: r-th success on trial nnegative_binomial_probability

The probability that the r-th success occurs on trial n (independent trials with success probability p).

P(N = n) = \binom{n-1}{r-1} p^r (1-p)^{n-r}

Inputs

r number of successes required integer
p probability of success
n the trial number integer

Returns

probability C(n-1, r-1) p^r (1-p)^(n-r)

Applies when

independent trials, constant p
n >= r

Example

m.ask("Negative binomial: find the probability that the r = 3 success occurs on trial n = 7 with p = 0.4.")

→ probability = 0.124416

Normal probability above a valuenormal_probability_above

P(X > x) for a normal distribution with mean mu and standard deviation sigma (the area to the right).

P(X > x) = 1 - \Phi\left(\frac{x - \mu}{\sigma}\right)

Inputs

x the value
mu the mean
sigma the standard deviation (> 0)

Returns

probability P(X > x)
z the z-score of x

Applies when

X normally distributed
sigma > 0

Example

m.ask("For a normal distribution with mean = 100 and standard deviation = 15, find P(X > 130).")

→ probability = 0.0227501, z = 2

Normal probability below a valuenormal_probability_below

P(X < x) for a normal distribution with mean mu and standard deviation sigma (the area to the left).

P(X < x) = \Phi\left(\frac{x - \mu}{\sigma}\right)

Inputs

x the value
mu the mean
sigma the standard deviation (> 0)

Returns

probability P(X < x)
z the z-score of x

Applies when

X normally distributed
sigma > 0

Example

m.ask("For a normal distribution with mean = 100 and standard deviation = 15, find P(X < 120).")

→ probability = 0.908789, z = 1.33333

Normal probability between two valuesnormal_probability_between

P(a < X < b) for a normal distribution with mean mu and standard deviation sigma.

P(a < X < b) = \Phi\left(\frac{b-\mu}{\sigma}\right) - \Phi\left(\frac{a-\mu}{\sigma}\right)

Inputs

a lower bound
b upper bound
mu the mean
sigma the standard deviation (> 0)

Returns

probability P(a < X < b)
z_a z-score of a
z_b z-score of b

Applies when

X normally distributed
sigma > 0
a < b

Example

m.ask("For a normal distribution with mean = 100 and standard deviation = 15, find P(85 < X < 115).")

→ probability = 0.682689, z_a = -1, z_b = 1

Poisson probability: at least k eventspoisson_at_least

The probability of at least k events when events occur independently at mean rate lambda per interval.

P(X \ge k) = 1 - \sum_{j=0}^{k-1} \frac{e^{-\lambda}\lambda^j}{j!}

Inputs

lam the mean number of events (lambda > 0)
k number of events integer

Returns

probability the Poisson probability

Applies when

events independent, at a constant mean rate
lambda > 0, k a whole number

Example

m.ask("For a Poisson distribution with lambda = 4, find the probability of at least k = 2 events.")

→ probability = 0.908422

Poisson probability: at most k eventspoisson_at_most

The probability of at most k events when events occur independently at mean rate lambda per interval.

P(X \le k) = \sum_{j=0}^{k} \frac{e^{-\lambda}\lambda^j}{j!}

Inputs

lam the mean number of events (lambda > 0)
k number of events integer

Returns

probability the Poisson probability

Applies when

events independent, at a constant mean rate
lambda > 0, k a whole number

Example

m.ask("For a Poisson distribution with lambda = 4, find the probability of at most k = 2 events.")

→ probability = 0.238103

Poisson probability: exactly k eventspoisson_exactly

The probability of exactly k events when events occur independently at mean rate lambda per interval.

P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}

Inputs

lam the mean number of events (lambda > 0)
k number of events integer

Returns

probability the Poisson probability

Applies when

events independent, at a constant mean rate
lambda > 0, k a whole number

Example

m.ask("For a Poisson distribution with lambda = 4, find the probability of exactly k = 2 events.")

→ probability = 0.146525

Quality control 6

c chart limits (defects per unit)c_chart

Centre line and 3-sigma limits of a c chart from counts of defects per inspection unit.

\bar{c} \pm 3\sqrt{\bar{c}}

Inputs

counts defects counted in each unit vector

Returns

center c-bar
ucl c-bar + 3 sqrt(c-bar)
lcl max(0, c-bar - 3 sqrt(c-bar))

Applies when

constant inspection unit
Poisson-distributed counts

Example

m.ask("Find the c chart limits for defect counts [3, 5, 2, 6, 4, 7, 3, 4].")

→ center = 4.25, ucl = 10.4347, lcl = 0

Individuals and moving range (I-MR) chart limitsindividuals_moving_range_chart

Control limits of the individuals (X) and moving range charts from individual measurements in time order.

UCL_x = \bar{x} + 2.66\,\overline{MR}

Inputs

x individual measurements in time order vector

Returns

center mean
mean_moving_range average moving range of span 2
x_ucl mean + 2.66 MR-bar
x_lcl mean - 2.66 MR-bar
mr_ucl 3.267 MR-bar
mr_lcl 0

Applies when

measurements in time order
constants E2 = 2.66 (3/d2) and D4 = 3.267 for span 2

Example

m.ask("Find the I-MR chart limits for [10.2, 9.8, 10.5, 10.1, 9.7, 10.3, 10.0].")

→ center = 10.0857, mean_moving_range = 0.466667, x_ucl = 11.327, x_lcl = 8.84438, mr_ucl = 1.5246, mr_lcl = 0

p chart limits (fraction defective)p_chart

Centre line and 3-sigma limits of a p chart from counts of defective items in samples of constant size n.

\bar{p} \pm 3\sqrt{\frac{\bar{p}(1-\bar{p})}{n}}

Inputs

defectives number of defective items in each sample vector
n sample size (constant) integer

Returns

center p-bar
ucl upper control limit (at most 1)
lcl lower control limit (at least 0)
sigma sqrt(p-bar (1 - p-bar)/n)

Applies when

constant sample size
LCL truncated at 0

Example

m.ask("Find the p chart limits for defectives [4, 6, 3, 5, 7, 2, 5] with sample size n = 100.")

→ center = 0.0457143, ucl = 0.108374, lcl = 0, sigma = 0.0208865

Process capability (Cp, Cpk) and expected nonconformingprocess_capability

Process capability indices Cp, Cpk (with Cpu, Cpl) and the expected parts per million outside the specification for a normal process with mean mu and standard deviation sigma.

C_p = \frac{USL - LSL}{6\sigma},\quad C_{pk} = \min\left(\frac{USL - \mu}{3\sigma}, \frac{\mu - LSL}{3\sigma}\right)

Inputs

mu process mean
sigma process standard deviation
lsl lower specification limit
usl upper specification limit

Returns

cp (USL - LSL)/(6 sigma)
cpk min(Cpu, Cpl)
cpu (USL - mu)/(3 sigma)
cpl (mu - LSL)/(3 sigma)
ppm_out expected nonconforming parts per million

Applies when

normal, stable (in-control) process
USL > LSL

Example

m.ask("Find Cp and Cpk for a process with mean = 10.2, standard deviation = 0.1, LSL = 9.8 and USL = 10.5.")

→ cp = 1.16667, cpk = 1, cpu = 1, cpl = 1.33333, ppm_out = 1381.57

X-bar and R control chart limitsxbar_r_chart

Centre lines and 3-sigma control limits of X-bar and R charts from subgroups (rows) of size 2 to 10.

UCL_{\bar{x}} = \bar{\bar{x}} + A_2\bar{R},\quad UCL_R = D_4\bar{R}

Inputs

data one row per subgroup, one column per observation (subgroup size 2-10) matrix

Returns

center_xbar grand mean
mean_range average range
xbar_ucl X-bar UCL
xbar_lcl X-bar LCL
r_ucl R chart UCL
r_lcl R chart LCL
A2 A2
D3 D3
D4 D4

Applies when

equal subgroup sizes 2 to 10
standard 3-decimal control chart constants (A2, D3, D4)

Example

m.ask("Find the X-bar and R chart limits for subgroups [[10.2, 9.9, 10.1, 10.0], [10.3, 10.1, 9.8, 10.2], [9.9, 10.0, 10.1, 9.7]].")

→ center_xbar = 10.025, mean_range = 0.4, xbar_ucl = 10.3166, xbar_lcl = 9.7334, r_ucl = 0.9128, r_lcl = 0, A2 = 0.729, D3 = 0, D4 = 2.282

X-bar and S control chart limitsxbar_s_chart

Centre lines and 3-sigma control limits of X-bar and S charts from subgroups (rows) of size 2 to 10.

UCL_{\bar{x}} = \bar{\bar{x}} + A_3\bar{s},\quad UCL_s = B_4\bar{s}

Inputs

data one row per subgroup, one column per observation (subgroup size 2-10) matrix

Returns

center_xbar grand mean
mean_sd average subgroup standard deviation
xbar_ucl X-bar UCL
xbar_lcl X-bar LCL
s_ucl S chart UCL
s_lcl S chart LCL
A3 A3
B3 B3
B4 B4

Applies when

equal subgroup sizes 2 to 10
standard 3-decimal constants (A3, B3, B4)

Example

m.ask("Find the X-bar and S chart limits for subgroups [[10.2, 9.9, 10.1, 10.0], [10.3, 10.1, 9.8, 10.2], [9.9, 10.0, 10.1, 9.7]].")

→ center_xbar = 10.025, mean_sd = 0.171969, xbar_ucl = 10.305, xbar_lcl = 9.74503, s_ucl = 0.389681, s_lcl = 0, A3 = 1.628, B3 = 0, B4 = 2.266

Regression and correlation 10

Confidence interval for a correlation (Fisher z)correlation_confidence_interval

The Fisher z-transformation confidence interval for the population correlation from paired data.

\tanh\left(\operatorname{atanh}(r) \pm \frac{z_{\alpha/2}}{\sqrt{n-3}}\right)

Inputs

x values of the first variable (x) vector
y paired values of the second variable (y) vector
confidence confidence level as a fraction (0.95 for 95%)

Returns

r sample correlation
lower lower limit
upper upper limit

Applies when

bivariate normal data
at least 4 pairs
|r| < 1

Example

m.ask("Find the 95% confidence interval for the correlation of x = [1, 2, 3, 4, 5, 6] and y = [2, 3, 5, 4, 6, 7].")

→ r = 0.942857, lower = 0.559149, upper = 0.9939

Covariancecovariance

The sample (n - 1) and population (n) covariance of paired data.

s_{xy} = \frac{1}{n-1}\sum (x_i - \bar{x})(y_i - \bar{y})

Inputs

x values of the first variable (x) vector
y paired values of the second variable (y) vector

Returns

sample_covariance divisor n - 1
population_covariance divisor n

Example

m.ask("Find the covariance of x = [2, 4, 6, 8] and y = [1, 3, 2, 5].")

→ sample_covariance = 3.66667, population_covariance = 2.75

Durbin-Watson statisticdurbin_watson

The Durbin-Watson statistic of regression residuals (about 2: no first-order autocorrelation; towards 0: positive; towards 4: negative).

d = \frac{\sum_{t=2}^{n} (e_t - e_{t-1})^2}{\sum_{t=1}^{n} e_t^2}

Inputs

residuals regression residuals in time order vector

Returns

durbin_watson sum of squared successive differences / sum of squares

Applies when

residuals in time order

Example

m.ask("Compute the Durbin-Watson statistic for residuals [0.5, -0.2, 0.3, -0.4, 0.1, 0.6, -0.5].")

→ durbin_watson = 2.53448

Kendall's tau-b rank correlationkendall_tau

Kendall's tau-b (tie-adjusted) between paired data with its two-sided large-sample p-value.

\tau_b = \frac{C - D}{\sqrt{(n_0 - n_1)(n_0 - n_2)}}

Inputs

x values of the first variable (x) vector
y paired values of the second variable (y) vector

Returns

tau Kendall's tau-b
p_value two-sided p-value (normal approximation with tie correction)
concordant concordant pairs
discordant discordant pairs

Applies when

paired ordinal or continuous data
large-sample normal approximation for the p-value

Example

m.ask("Find Kendall's tau for x = [1, 2, 3, 4, 5, 6] and y = [2, 1, 4, 3, 6, 5].")

→ tau = 0.6, p_value = 0.0908739, concordant = 12, discordant = 3

Multiple linear regressionmultiple_linear_regression

Ordinary least squares with several predictors: coefficients (intercept first), R squared, adjusted R squared and the overall F-test.

\hat{\beta} = (X^\top X)^{-1}X^\top y,\quad R^2 = 1 - \frac{SSE}{SST}

Inputs

X predictor matrix: one row per observation, one column per predictor matrix
y response values vector

Returns

coefficients [intercept, b1, ..., bk]
r_squared R squared
adj_r_squared adjusted R squared
f overall F statistic
f_p_value p-value of the F test
residual_se residual standard error

Applies when

n > k + 1
predictors not collinear
inference assumes independent normal errors

Example

m.ask("Fit a multiple regression of y = [10, 12, 15, 19, 22, 25] on X = [[1, 2], [2, 1], [3, 4], [4, 3], [5, 6], [6, 5]].")

→ coefficients = [6.229166666666668, 3.0625, 0.06249999999999978], r_squared = 0.993659, adj_r_squared = 0.989431, f = 235.038, f_p_value = 0.000504992, residual_se = 0.600925

Pearson correlation coefficientpearson_correlation

Pearson's correlation coefficient r between paired data, r squared, and the two-sided p-value for H0: rho = 0.

r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}}

Inputs

x values of the first variable (x) vector
y paired values of the second variable (y) vector

Returns

r correlation coefficient
r_squared coefficient of determination
p_value two-sided p-value (t with n - 2 df)
n number of pairs

Applies when

paired observations
linear relationship; the p-value assumes bivariate normality

Example

m.ask("Find the correlation coefficient of x = [1, 2, 3, 4, 5] and y = [2, 4, 5, 4, 5].")

→ r = 0.774597, r_squared = 0.6, p_value = 0.124027, n = 5

Prediction from a least squares lineregression_prediction

The predicted y at a given x0 from the least-squares line fitted to paired data.

\hat{y}_0 = b_0 + b_1 x_0

Inputs

x values of the first variable (x) vector
y paired values of the second variable (y) vector
x0 the x value to predict at

Returns

y_hat b0 + b1 x0
slope slope
intercept intercept

Applies when

x0 within the range of the data (extrapolation is unreliable)

Example

m.ask("Using the regression line for x = [1, 2, 3, 4, 5] and y = [2.2, 4.1, 6.1, 7.9, 10.2], predict y at x0 = 3.5.")

→ y_hat = 7.09, slope = 1.98, intercept = 0.16

Significance test for a correlation coefficientcorrelation_t_test

The t-test of H0: rho = 0 from a sample correlation r and the number of pairs n.

t = r\sqrt{\frac{n-2}{1-r^2}}

Inputs

r sample correlation coefficient
n number of pairs (>= 3) integer
alternative alternative hypothesis: two-sided (default), greater or less choice optional

Returns

t r sqrt((n - 2)/(1 - r^2))
df n - 2
p_value p-value

Applies when

-1 < r < 1
bivariate normal data

Example

m.ask("Test the significance of a correlation r = 0.45 with n = 30 pairs, two-sided.")

→ t = 2.66641, df = 28, p_value = 0.0125911

Simple linear regression (least squares line)simple_linear_regression

The least-squares line y = intercept + slope x: coefficients, r, r squared, standard errors and the slope's p-value.

\hat{y} = b_0 + b_1 x,\quad b_1 = \frac{S_{xy}}{S_{xx}},\quad b_0 = \bar{y} - b_1\bar{x}

Inputs

x values of the first variable (x) vector
y paired values of the second variable (y) vector

Returns

slope b1 = Sxy/Sxx
intercept b0 = ybar - b1 xbar
r correlation
r_squared coefficient of determination
p_value two-sided p-value for H0: slope = 0
slope_se standard error of the slope
intercept_se standard error of the intercept
residual_se residual standard error

Applies when

at least 3 pairs, x not all equal
inference assumes independent normal errors with constant variance

Example

m.ask("Find the least squares regression line for x = [1, 2, 3, 4, 5] and y = [2.2, 4.1, 6.1, 7.9, 10.2].")

→ slope = 1.98, intercept = 0.16, r = 0.999287, r_squared = 0.998574, p_value = 2.28734e-05, slope_se = 0.0432049, intercept_se = 0.143295, residual_se = 0.136626

Spearman rank correlationspearman_correlation

Spearman's rank correlation coefficient (ties given average ranks) with its two-sided p-value.

\rho_s = r\left(\operatorname{rank}(x), \operatorname{rank}(y)\right)

Inputs

x values of the first variable (x) vector
y paired values of the second variable (y) vector

Returns

rho Spearman's rho
p_value two-sided p-value (t approximation, n - 2 df)
n pairs

Applies when

paired ordinal or continuous data
p-value by the t approximation

Example

m.ask("Find the Spearman rank correlation of x = [10, 20, 30, 40, 50] and y = [12, 25, 22, 41, 48].")

→ rho = 0.9, p_value = 0.0373861, n = 5

Reliability and survival 7

Exponential reliability (constant failure rate)exponential_reliability

Reliability R(t) = exp(-lambda t), unreliability and MTTF for a constant failure rate (given the rate or the MTBF).

R(t) = e^{-\lambda t},\quad MTTF = 1/\lambda

Inputs

t mission time
rate failure rate lambda optional
mtbf mean time between failures (instead of the rate) optional

Returns

reliability exp(-lambda t)
unreliability 1 - R(t)
mttf 1/lambda
failure_rate lambda

Applies when

constant failure rate (useful-life period)
exactly one of rate or MTBF

Example

m.ask("Find the reliability at t = 1000 hours for a constant failure rate = 0.0002 per hour.")

→ reliability = 0.818731, unreliability = 0.181269, mttf = 5000, failure_rate = 0.0002

k-out-of-n system reliabilityk_out_of_n_reliability

The reliability of a system that works when at least k of n identical independent components work.

R = \sum_{j=k}^{n}\binom{n}{j} r^j (1-r)^{n-j}

Inputs

n number of components integer
k number that must work integer
r reliability of one component

Returns

reliability P(at least k of n work)

Applies when

identical, independent components

Example

m.ask("Find the reliability of a k-out-of-n system with k = 2, n = 3 and component reliability = 0.9.")

→ reliability = 0.972

Kaplan-Meier survival estimatekaplan_meier

The Kaplan-Meier (product-limit) estimate of the survival function from times with event/censoring indicators, evaluated at time t, and the whole step curve.

\hat{S}(t) = \prod_{t_i \le t}\left(1 - \frac{d_i}{n_i}\right)

Inputs

times follow-up times vector
events 1 = event observed, 0 = censored vector
t time at which to report S(t)

Returns

survival S(t)
event_times distinct event times
survival_curve S just after each event time

Applies when

non-informative censoring
a censoring at an event time counts as still at risk

Example

m.ask("Kaplan-Meier: times [6, 6, 6, 7, 10, 13, 16, 22, 23] with events [1, 1, 1, 0, 1, 1, 1, 1, 1]; find the survival at t = 12.")

→ survival = 0.533333, event_times = [6.0, 10.0, 13.0, 16.0, 22.0, 23.0], survival_curve = [0.6666666666666667, 0.5333333333333334, 0.4000000000000001, 0.2666666666666668, 0.1333333333333334, 0.0]

Log-rank test comparing two survival curveslog_rank_test

The log-rank (Mantel-Cox) test of equal survival in two groups from times and event indicators.

\chi^2 = \frac{\left(\sum (d_{1i} - e_{1i})\right)^2}{\sum v_i}

Inputs

times1 times, group 1 vector
events1 events (1/0), group 1 vector
times2 times, group 2 vector
events2 events (1/0), group 2 vector

Returns

chi_square (O1 - E1)^2 / V
p_value chi-square p-value (1 df)
observed_1 events in group 1
expected_1 expected events in group 1

Applies when

non-informative censoring
proportional hazards gives the test its best power

Example

m.ask("Log-rank test: group 1 times [6, 6, 6, 7, 10, 13, 16] events [1, 1, 1, 0, 1, 1, 1], group 2 times [1, 1, 2, 2, 3, 4, 4] events [1, 1, 1, 1, 1, 1, 1].")

→ chi_square = 13.8115, p_value = 0.000202097, observed_1 = 6, expected_1 = 10.4222

Reliability of a parallel systemparallel_system_reliability

The reliability of independent components in parallel (at least one must work).

R_p = 1 - \prod (1 - R_i)

Inputs

r component reliabilities (0 to 1) vector

Returns

reliability system reliability

Applies when

independent components

Example

m.ask("Find the reliability of a parallel system with component reliabilities [0.95, 0.9, 0.98].")

→ reliability = 0.9999

Reliability of a series systemseries_system_reliability

The reliability of independent components in series (all must work).

R_s = \prod R_i

Inputs

r component reliabilities (0 to 1) vector

Returns

reliability system reliability

Applies when

independent components

Example

m.ask("Find the reliability of a series system with component reliabilities [0.95, 0.9, 0.98].")

→ reliability = 0.8379

Weibull reliability, hazard rate and MTTFweibull_reliability

Reliability R(t), unreliability, hazard rate and mean time to failure of a two-parameter Weibull distribution.

R(t) = e^{-(t/\eta)^\beta},\quad h(t) = \frac{\beta}{\eta}\left(\frac{t}{\eta}\right)^{\beta-1}

Inputs

beta shape parameter beta
eta scale parameter eta (characteristic life)
t time

Returns

reliability exp(-(t/eta)^beta)
unreliability 1 - R(t)
hazard_rate (beta/eta)(t/eta)^(beta-1)
mttf eta Gamma(1 + 1/beta)

Applies when

two-parameter Weibull (no location shift)
t >= 0

Example

m.ask("For a Weibull distribution with beta = 2 and eta = 1000 hours, find the reliability at t = 500.")

→ reliability = 0.778801, unreliability = 0.221199, hazard_rate = 0.001, mttf = 886.227

Sampling distributions 2

Probability that a sample mean is above a valuesample_mean_probability_above

P(sample mean > x) for samples of size n from a population with mean mu and standard deviation sigma (central limit theorem).

P(\bar{X} < x) = \Phi\left(\frac{x - \mu}{\sigma/\sqrt{n}}\right)

Inputs

x the value for the sample mean
mu population mean
sigma population standard deviation
n sample size integer

Returns

probability the probability
z z = (x - mu)/(sigma/sqrt(n))
standard_error sigma/sqrt(n)

Applies when

population normal, or n large enough for the central limit theorem
independent observations

Example

m.ask("A population has mean = 50 and standard deviation = 10; for a sample of size n = 25, find the probability that the sample mean is greater than 52.")

→ probability = 0.158655, z = 1, standard_error = 2

Probability that a sample mean is below a valuesample_mean_probability_below

P(sample mean < x) for samples of size n from a population with mean mu and standard deviation sigma (central limit theorem).

P(\bar{X} < x) = \Phi\left(\frac{x - \mu}{\sigma/\sqrt{n}}\right)

Inputs

x the value for the sample mean
mu population mean
sigma population standard deviation
n sample size integer

Returns

probability the probability
z z = (x - mu)/(sigma/sqrt(n))
standard_error sigma/sqrt(n)

Applies when

population normal, or n large enough for the central limit theorem
independent observations

Example

m.ask("A population has mean = 50 and standard deviation = 10; for a sample of size n = 25, find the probability that the sample mean is less than 52.")

→ probability = 0.841345, z = 1, standard_error = 2

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docs/statistics_ai999x_capabilities.pdf	3.5 MB
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onnx/search/tokenizer.json	176.4 KB
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Questions

Frequently asked

What does statistics-ai999x do?

Statistics answers questions in statistics: descriptive statistics, probability distributions, confidence intervals, hypothesis tests, regression and correlation, experimental design, quality control and reliability. Each answer gives the value with its unit, the steps and the formula in LaTeX.

How do I install statistics-ai999x?

With pip: pip install statistics-ai999x. The same files are on Hugging Face (hf download AI999X/statistics-ai999x --local-dir statistics-ai999x) and GitHub.

Does it run offline?

Yes. It is a Python package that runs on an ordinary CPU without a network connection (Python >=3.10); no GPU and no API key are needed.

How many capabilities does statistics-ai999x have?

110 capabilities in 16 topics, including Descriptive statistics, Probability distributions, Hypothesis tests, Regression and correlation, Confidence intervals.

Which field and tasks does it cover?

Field: Mathematics & Statistics. Tasks: Question Answering, Token Classification, Text Classification, Tabular Regression, Sentence Similarity, Feature Extraction.

What does an answer look like?

Asked "Perform a one-sample t-test of [5.1, 4.9, 5.3, 5.5, 5.0, 5.2] against mu0 = 5, two-sided.", it answers: t = 1.88982, df = 5, p_value = 0.117387, mean = 5.16667, standard_deviation = 0.216025.

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It says so and names what is missing or out of range, instead of guessing a number.

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