Use normal sampling distributions and standard errors

Identify the random object

The population distribution describes individual observations X. One sample contains n observations; its realised mean x̄ is one number. The sampling distribution describes the random mean X̄ across repeated samples under one fixed selection rule. A population parameter μ stays fixed even when it is unknown

Use the existing finite-sampling foundation

The companion revisits exact and empirical sampling distributions, independent-draw mean/variance rules and the distinction between n observations per sample and R repeated samples. Here we add normal distributions for means, standard-error calculations and a reproducible experiment across different populations

Normal inputs give exact normal means

If X₁,...,Xₙ are independent with the same N(μ,σ²) distribution, their mean is exactly N(μ,σ²/n) for every positive n. Its SD, called the standard error, is σ/√n. This follows from the independent normal combination rule. Dividing the sum by n divides its variance by n²

Work a normal mean model

Let independent measurements have distribution N(18,36). The population SD remains 6. These are properties across samples; one observed sample mean need not be 18.

Independent observations have distribution N(18,36). Give the standard error of their mean when n=16.

One observation and one sample mean have different spreads

For that population and a sample size of 16, compare the score of the boundary 21 for an individual observation and for the sample mean. The mean distribution is narrower; the same physical deviation is more standard errors away. An observed sample SD is a third object and estimates population spread

A sample mean has mean 18 and standard error 3/2. Give the standard score of 21.

Compute a tail using the standard error

For X̄~N(18,9/4), Pr(X̄>21)=Pr(Z>2)=1-Φ(2)≈0.0228. Using the individual SD6 would answer the different question Pr(X>21). State whether normality is exact from normal inputs or approximate under a justified large-sample argument

Standardise both interval endpoints

For the same sample-mean model, Pr(16.5<X̄<21)=Pr(-1<Z<2)=Φ(2)-Φ(-1)≈0.8186. The probability outside is≈0.1814. Retain guard digits through subtraction and complement before rounding to four decimal places. Equality at a point does not change these continuous probabilities

A sample mean has mean 18 and standard error 3/2. Give the standard score of the lower boundary 16.5

Choose n to meet a theoretical spread target

For iid observations with σ=7, requiring SE≤3/2 means7/√n≤3/2, or n≥196/9. The smallest integer is22. Check n=21 fails and n=22 succeeds. Normality is not needed for the finite-variance identity, and controlling SE does not guarantee a particular sample mean’s error

For iid observations with population SD 7, give the smallest integer n for which the standard error is at most 3/2.

Estimate an unknown standard error

When σ is unknown, use the observed sample SD s to estimate it. The estimated SE is s/√n, not s/n. For n=49 and s=7/2, estimated SE=1/2 and estimated variance of the mean=1/4. These estimates vary from sample to sample. Replacing σ by s does not turn an arbitrary small-sample population into a known normal model

Give the estimated standard error for n=49 and s=7/2.

Large-sample normality is an approximation

For iid observations with finite positive variance, the standardised sample mean approaches a standard normal distribution as n increases. The usefulness of a finite-n approximation depends on population shape and tails. There is no universal n=30 guarantee. Increasing the repeat count R improves the empirical display of a fixed-n distribution; it does not change that distribution

A highly skewed population has a rare large value. What follows merely from n=30?

A finite rare-value example exposes the boundary

Suppose each independent observation is 0 with probability 0.99 or 10 with probability 0.01. Then μ=0.1 and σ²=0.99. For a sample size of 64, more than half the exact mass is at zero. The mean remains unbiased and has standard error given below, yet a smooth normal approximation to this particular finite distribution needs caution

Technology: define the experiment before running it

The runnable Python program uses four finite populations: symmetric values[-2,0,0,2], uniform[-3,-1,1,3], skew[0,0,0,0,5], and a rare-value model with99zeros and one10. Choosing a list entry uniformly implements its stated probabilities. Every sample uses fresh draws with replacement. Each run stores one mean per repetition, for n=1,4,16,64 and R=800.

Technology: runnable Python source

import random, math, json MODELS = { 'symmetric': [-2, 0, 0, 2], 'uniform': [-3, -1, 1, 3], 'skew': [0, 0, 0, 0, 5], 'rare': [0] * 99 + [10], } def simulate(model, n, repetitions, seed): if model not in MODELS or type(n) is not int or n < 1: raise ValueError('Use a listed model and a positive integer sample size.') if type(repetitions) is not int or repetitions < 2: raise ValueError('Use at least two repetitions.') population = MODELS[model] mu = sum(population) / len(population) variance = sum((x - mu)**2 for x in population) / len(population) rng = random.Random(seed) # Set once, outside both sampling loops means = [] for repetition in range(repetitions): sample = [rng.choice(population) for _ in range(n)] means.append(sum(sample) / n) centre = sum(means) / repetitions # This describes the realised list of means, using denominator R. empirical_sd = math.sqrt(sum((x-centre)**2 for x in means) / repetitions) return dict(model=model, n=n, R=repetitions, seed=seed, population_mean=mu, population_variance=variance, theoretical_se=math.sqrt(variance/n), means=means, empirical_mean=centre, empirical_sd=empirical_sd) if __name__ == '__main__': runs = [simulate(model, n, 800, 7300 + 100*i + j) for i, model in enumerate(MODELS) for j, n in enumerate([1, 4, 16, 64])] print(json.dumps(runs))

Technology: retain identity and distinguish replay

The displayed reference results were generated by this exact program using Python3.14.7 and its random.Random implementation, with seeds7300+100i+j in the displayed model/size order. Retain the program, runtime version, model, n,R,seed and full means list. Re-executing the same seeded stream is replay of the same run; it is not additional independent evidence

Read the recorded summaries

Each row below is a completed run of800 sample means. The empirical SD uses denominator R for the realised list of means; it is not the original observations’ sample SD s. Theoretical SE is calculated from the population variance and n. The table reports rounded summaries; the program retains all means and full internal precision

What does one row of this summary table describe?

Read the reference runs without forcing perfect agreement

The symmetric runs have theoretical SE√(2/n); the uniform runs√(5/n); the skew runs2/√n; and the rare runs√(0.99/n). Their empirical centres and spreads vary around theoretical values. For the skew model at n=64 the recorded centre is approximately0.9951 and empirical SD0.2527, close to μ=1 and SE=0.25. Agreement in a finite simulation illustrates the model; it does not prove a sampling theorem or real-world independence

In this n=4 plot, what does each dot represent?

How should this n=64 plot be compared with the n=4 plot?

Keep n and R separate when analysing a run

If R=200 samples each contain n=25 observations, there are5000 draws and200 plotted means. If37 means meet an event, its observed relative frequency is37/200. If σ=10, SE=2; changing n to100 halves it to1. Merely taking more repetitions at n=25 does not change SE

A run contains 200 repeated samples, and 37 sample means meet an event. What denominator belongs to the event’s observed relative frequency?

Validate a normal sampling claim, ungraded

A report uses a convenience sample of40 repeated readings from one instrument and claims exact normality with standard error s/√40. Identify the target parameter, which sampling assumptions need support, and which conclusions are exact, estimated or only approximate

Model: separate the claims

The target must be a stated population mean, not automatically the mean of the40 selected readings. Convenience selection can introduce bias; repeated readings can share a common error and fail independence. If observations were iid normal with known σ, X̄ would be exactly normal with SE=σ/√40. With unknown σ, s/√40 is an estimated SE. For a non-normal iid population, a large-sample normal approximation requires assessment of shape and tails;40 does not establish exact normality. The calculation alone validates none of the sampling assumptions

Run and interpret the sampling experiment, ungraded

Execute the supplied program or independently implement the same four models and sample sizes. Validate model probabilities, record every predeclared run and retain all means. Compare empirical means/SDs with μ and σ/√n; plot comparable scales, distinguish R from n and explain limitations of claims about shape. Then use distinct documented seeds for additional runs

Model: test the mechanism before interpreting the plots

The list entries give exact population moments (μ,σ²)=(0,2),(0,5),(1,4),(0.1,0.99). For n=1, sampled means must be population values; for any n each mean is the sum of exactly n fresh draws divided by n. There are R means per run. Check small-n probabilities by exact enumeration before comparing empirical histograms. Retain the complete16-run output rather than choosing only close summaries. Increasing n reduces theoretical SE; increasing R improves a finite display. The rare model can remain strongly discrete and skewed at64 despite its correct mean and variance. A simulation demonstrates the assumed model and implementation, not the validity of that model for a real population

Analyse rare outcomes and dependence, ungraded

For a population with Pr(X=0)=0.99 and Pr(X=10)=0.01, derive μ, σ² and Pr(X̄=0) for n independent observations. Contrast the variance of X̄ with a design that repeats one draw n times. Explain what this shows about normal approximation and the independence requirement

Model: retain the joint sampling law

Calculate the population moments and variance from the two outcomes. Under iid sampling, the sample mean has the population mean, its variance is divided by the sample size, and all observations can be zero. At a sample size of 64, the zero outcome has probability about 0.5256, making exact discreteness visible. If one draw is copied across the sample, the mean equals that draw and retains variance 0.99. Matching marginals and a large entry count cannot substitute for fresh independent observations

Assessment boundary

Marked questions check specified mean/SE calculations, normal probabilities, integer size bounds and supplied-run accounting. Independent simulation implementation, model justification, proof and interpretation remain ungraded. Confidence intervals next use the distinction between known and estimated standard errors