DSST Principles of Statistics common mistakes
DSST Principles of Statistics · Understand the traps and how to avoid them.
When a question asks for the probability of 'at least one' success, the candidate computes np or looks up P(X = 1) instead of using the complement 1 − P(X = 0).
The phrase 'at least one' is read as 'one', and the candidate reaches for the binomial formula with k = 1 because that is the formula most recently practised. The complement shortcut is a habit that only forms after repeated exposure to 'at least one' phrasing.
What to do instead
Before computing anything, translate the English into an inequality: 'at least one' means X ≥ 1, so compute 1 − P(X = 0). Drill a short set of problems where the same scenario is asked four ways (exactly one, at least one, at most one, none) and write the inequality above each before solving.
The candidate treats a small p-value as evidence that the effect is large or important, writing conclusions such as 'the treatment strongly improves scores' from p = 0.001 with a trivial mean difference.
p-values and effect sizes are both printed in the same output block, and the smaller number attracts attention. The candidate has memorised 'small p means significant' without attaching the word 'significant' to the null hypothesis rather than to the size of the effect.
What to do instead
State in one sentence what the p-value measures (probability of data this extreme or more, assuming H0 true) and what it does not measure (magnitude). On every practice problem, report the point estimate and a confidence interval alongside the p-value, and phrase the conclusion in units of the response variable.
The candidate checks normality of the raw response variable before running a one-sample t-test, or checks normality of both groups before a two-sample t-test, instead of checking the sampling distribution assumption.
Software tutorials often plot histograms of the outcome first, and the candidate conflates 'the data should look normal' with 'the sampling distribution of the mean is approximately normal'. The central limit theorem's role is underweighted.
What to do instead
Memorise the assumption as being about the sampling distribution of the statistic, which the central limit theorem makes approximately normal for moderate n even when the population is skewed. Practise deciding, for n = 40 with mild skew, that the t-test is still appropriate, and reserve normality checks for small n or strong skew and outliers.
When asked for the standard deviation of a binomial count, the candidate reports np(1 − p) rather than the square root of that quantity, or reports p as the standard deviation of a proportion.
Variance and standard deviation formulas are learned as a pair and stored together; under time pressure the candidate recalls the expression containing p but not the square root symbol that distinguishes the two.
What to do instead
Anchor the pair with units: a count has standard deviation in 'successes', a proportion in probability units, and both are square roots of the corresponding variance. Write Var(X) = np(1 − p) and SD(X) = √(np(1 − p)) side by side, then compute one numeric example such as n = 100, p = 0.3, where the variance is 21 and the SD is about 4.58.
The candidate applies a Poisson model to a count with a fixed upper bound, such as the number of correct answers out of 20 items, or applies a binomial model to events occurring over a continuum of time.
Both distributions are discrete and both appear in word problems, so the candidate pattern-matches on 'count' rather than on the structure of the random process. The Poisson's lack of an upper bound and its rate-per-unit-exposure framing are not actively checked.
What to do instead
Use a two-question test before choosing: is there a fixed number of trials with a success/failure outcome (binomial), or is it events per unit of time, length, or area with no fixed trial count (Poisson)? Also check the mean-variance relationship, since Poisson requires mean ≈ variance, and flag overdispersion when the sample variance far exceeds the mean.
The candidate interprets a 95% confidence interval for a mean as '95% probability that the true mean lies in this interval' after the interval has been computed.
The number 95% is visually attached to the interval, and everyday language treats probability as a property of a specific statement. The frequentist framing, in which the parameter is fixed and the interval is random, is counterintuitive and needs deliberate rehearsal.
What to do instead
Practise the correct sentence: the procedure captures the true parameter in 95% of repeated samples. Reinforce it by simulating many samples, computing an interval for each, and observing that about 95% of the resulting intervals cover the fixed true value while any single computed interval either does or does not.
The candidate computes a prediction interval for a new individual observation but reports it as a confidence interval for the mean response, or uses the mean-response interval when the question asks about one future unit.
Both intervals share the same fitted value and both are printed by regression software; the extra variability of an individual observation is easy to drop because the formula differs only by an added term inside the square root.
What to do instead
Tie the choice to the wording: 'average', 'mean response', or 'expected value' calls for the confidence interval; 'a new', 'an individual', or 'one future observation' calls for the prediction interval. Remember the prediction interval is always the wider of the two and never collapses to zero width as n grows.
The candidate reads a regression slope as a correlation, claiming that a slope of 2.5 means a strong relationship, or compares slopes across models with different units as if they were comparable.
Both quantities describe the association between x and y and both appear in regression output. Slope carries units of y per unit of x, while correlation is unitless and bounded, and this distinction is rarely made explicit in practice problems.
What to do instead
State the units aloud when interpreting a slope: 'each additional year of experience is associated with 2.5 more thousand dollars of salary'. When a question asks about strength of linear association, compute or use r, and remember r is unchanged by rescaling while the slope is not.
The candidate concludes that because a correlation or regression slope is statistically significant, changes in the predictor cause changes in the response.
Statistical significance is the only thing the output formally tests, and causal language is the default in ordinary speech. Confounding and study design are treated as background rather than as the factor that limits the conclusion.
What to do instead
Attach the study design to every regression conclusion: observational data support association only, randomised assignment supports causal language. Practise rewriting causal-sounding conclusions into association language and identifying at least one plausible confounder for each observational example.
The candidate computes a confidence interval for a proportion using p̂ ± z√(p̂(1 − p̂)/n) with a very small sample and few successes, or uses the normal approximation when np̂ or n(1 − p̂) is below about 10.
The Wald formula is the first one learned and is easy to apply mechanically; the small-sample failure and the awkward behaviour near 0 or 1 are covered later and are easy to forget under exam pressure.
What to do instead
Check np̂ and n(1 − p̂) before using the normal approximation, and switch to an exact or score-based interval when either is small. Also notice that the Wald interval can extend below 0 or above 1 and can have zero width when p̂ = 0, which signals that the approximation is inappropriate.