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Sampling and the sample mean · Model a sample mean

Choose the model, then the random quantity

A sample mean is not an individual observation. First justify a normal model; then use the mean and variance of the quantity the question asks about.

  1. Exact normal, approximate normal or neither?

    Assume independent, identically distributed observations with finite positive variance. Is the population normal, or is a CLT approximation justified for this sample and probability?

    The common mean is μ\mu and variance is σ2\sigma^2.

    Normal populationExact normal mean and total

    Independent normal observations give normal sums and means at any positive sample size. You do not need a large-sample argument.

    Non-normal or unspecified population · CLT justifiedUse an approximation and say why

    The sample must be sufficiently large for the population and the event being estimated. A classroom sample-size guideline is not a guarantee, especially for highly skewed populations or far tails.

    No exact model and no adequate CLT justificationDo not invent a normal probability

    Mean and variance alone do not determine a normal distribution. State any additional assumption needed; the individual-observation distribution is not made normal by increasing the sample size.

    Check the sampling assumptions first. Dependent observations need covariance terms; do not silently treat a finite-population draw without replacement as iid observations.

  2. Choose one observation, the mean or the total

    These moment formulas use the iid model above. Apply normal probabilities to the mean or total only after justifying an exact or approximate normal model.

    Xˉ=1n∑iXi\bar X=\frac1n\sum_iX_i, while T=∑iXiT=\sum_iX_i.

    One observationKeep the original population distribution

    Its expectation is μ\mu and variance is σ2\sigma^2. A CLT argument about the sample mean does not change the distribution of an individual observation.

    Sample meanKeep the mean; divide the variance

    E(Xˉ)=μE(\bar X)=\mu and Var⁡(Xˉ)=σ2/n\operatorname{Var}(\bar X)=\sigma^2/n. The standard error is σ/n\sigma/\sqrt n; do not divide the standard deviation by nn.

    Sample totalMultiply both moments by the sample size

    E(T)=nμE(T)=n\mu and Var⁡(T)=nσ2\operatorname{Var}(T)=n\sigma^2. Its standard deviation is σn\sigma\sqrt n. Scale any event boundaries to match the quantity you use.

    The moment formulas do not themselves establish normality. Use ∼ for the exact normal model and make the approximation explicit when using CLT.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

One observation, mean or total?

Illustration: four independent observations come from N(40,100)N(40,100). Predict the distributions of one observation, their mean and their total.

Show reasoning

  1. 1. Keep the individual distribution

    Taking a sample does not change the distribution of each observation.

    Xi∼N(40,100)X_i\sim N(40,100)

  2. 2. Divide the variance for the mean

    Divide the population variance by four: 100/4 = 25. Normality of the population makes this exact at n=4; no CLT approximation is needed.

    Xˉ∼N(40,25)\bar X\sim N(40,25)

  3. 3. Multiply the variance for the total

    Adding four independent observations gives mean 4 × 40 = 160 and variance 4 × 100 = 400. The total is also four times the mean.

    T∼N(160,400)T\sim N(160,400)

Do the moments prove normality?

Illustration: four iid observations now come from an unspecified, strongly right-skewed population with mean 40 and variance 100. Must their mean have the same normal distribution as above?

Show reasoning

  1. 1. Calculate what the assumptions justify

    The common mean and independent finite-variance observations give expectation 40 and variance 100/4 = 25.

    E(Xˉ)=40Var⁡(Xˉ)=25\begin{aligned}E(\bar X)&=40\\\operatorname{Var}(\bar X)&=25\end{aligned}

  2. 2. Keep moments separate from distribution shape

    These two moments do not prove a normal distribution. A sample of four from this population does not justify the normal approximation. A sufficiently large sample may support CLT, with adequacy depending on the population and the tail being estimated.