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Statistics & Probability Statistical Inference Flashcards
50 question-and-answer cards covering Statistical Inference as it is examined in Statistics & Probability. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Statistical Inference deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
What is Welch's two-sample t-test and when is it preferred?
Welch's test compares two means without assuming equal variances: $$t = \frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}$$ with df from the Welch–Satterthwaite approximation. It is preferred (and is the default) whenever the two group variances or sample sizes differ.
When is a paired (dependent-samples) t-test used, and what is its test statistic?
It is used when observations are naturally paired (e.g. before/after on the same subjects). You analyze the differences $d_i$: $$t = \frac{\bar{d} - \mu_{d,0}}{s_d/\sqrt{n}}, \qquad df = n-1,$$ where $n$ is the number of pairs.
State the key assumptions of one-sample and two-sample t-tests.
(1) Random/independent sampling; (2) the data (or, for two samples, each group; for paired, the differences) come from an approximately normal population, with robustness improving as $n$ grows via the CLT; (3) for the pooled two-sample test, equal population variances (homogeneity).
What diagnostics check the normality and variance assumptions before a t-test or ANOVA?
Normality: histograms, normal Q–Q plots (points near the line), and tests like Shapiro–Wilk. Equal variance (homoscedasticity): compare group spreads/boxplots or use Levene's or the F-test. Also check independence and screen for outliers that distort means and variances.
Write the one-proportion z-test statistic.
$$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$$ Note the standard error uses the hypothesized $p_0$ (not $\hat{p}$), since it is computed assuming $H_0$ is true.
Write the two-proportion z-test statistic and its pooled proportion.
$$z = \frac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}, \quad \hat{p}=\frac{x_1+x_2}{n_1+n_2}$$ The pooled $\hat{p}$ is used because $H_0: p_1=p_2$ assumes a common proportion.
Give the chi-square goodness-of-fit test statistic and its degrees of freedom.
$$\chi^{2} = \sum \frac{(O_i - E_i)^{2}}{E_i}, \qquad df = k - 1$$ where $O_i$ and $E_i$ are observed and expected counts across $k$ categories. It tests whether a categorical variable follows a specified distribution.
How are expected counts computed in a chi-square goodness-of-fit test?
For each category, $E_i = n\,p_i$, where $n$ is the total sample size and $p_i$ is the proportion specified by $H_0$ for that category. The test requires all expected counts to be sufficiently large (commonly $E_i \geq 5$).
Give the chi-square test-of-independence statistic and its degrees of freedom.
$$\chi^{2} = \sum \frac{(O - E)^{2}}{E}, \qquad df = (r-1)(c-1)$$ for an $r \times c$ contingency table. It tests whether two categorical variables are associated (dependent).
How is each expected cell count computed in a chi-square test of independence?
$$E_{ij} = \frac{(\text{row } i \text{ total})\times(\text{column } j \text{ total})}{\text{grand total}}.$$ This is the count expected if the row and column variables were independent.
What are the main assumptions/conditions for chi-square tests on counts?
(1) Data are counts (frequencies) from a random sample; (2) observations are independent; (3) categories are mutually exclusive and exhaustive; (4) expected counts are large enough — a common rule is all $E_i \geq 5$ (or at least 80% of cells $\geq 5$).
State the chi-square test statistic and df for a test about a single population variance.
$$\chi^{2} = \frac{(n-1)s^{2}}{\sigma_0^{2}}, \qquad df = n-1$$ used to test $H_0: \sigma^{2} = \sigma_0^{2}$. It requires the population to be (approximately) normal, an assumption to which this test is quite sensitive.
What test compares two population variances, and what is its statistic?
The F-test: $$F = \frac{s_1^{2}}{s_2^{2}}$$ with $df_1 = n_1-1$ and $df_2 = n_2-1$. Under $H_0: \sigma_1^{2}=\sigma_2^{2}$ it follows an F-distribution; it assumes both populations are normal.
State the hypotheses of a one-way ANOVA.
$H_0: \mu_1 = \mu_2 = \cdots = \mu_k$ (all group means equal) versus $H_a:$ at least one mean differs. It compares $k$ group means simultaneously using one categorical factor.
Give the one-way ANOVA F statistic in terms of mean squares.
$$F = \frac{\text{MSB}}{\text{MSW}} = \frac{SSB/(k-1)}{SSW/(N-k)}$$ where MSB is mean square between groups (signal) and MSW is mean square within groups (noise). Large $F$ suggests between-group differences exceed within-group variation.
State the sum-of-squares partition in one-way ANOVA.
Total variation splits into between-group and within-group parts: $$SST = SSB + SSW,$$ i.e. $\sum(x_{ij}-\bar{x})^{2} = \sum n_j(\bar{x}_j-\bar{x})^{2} + \sum(x_{ij}-\bar{x}_j)^{2}.$
What are the degrees of freedom for the F-test in a one-way ANOVA with $k$ groups and $N$ total observations?
Between groups (numerator): $df_1 = k-1$. Within groups / error (denominator): $df_2 = N-k$. Total: $df_{\text{total}} = N-1 = (k-1)+(N-k)$.
List the assumptions of ANOVA.
(1) Independent random samples/observations; (2) each group's population is approximately normal; (3) homogeneity of variance — all groups share a common variance $\sigma^{2}$. ANOVA is fairly robust to mild normality violations, especially with balanced designs.
What is the purpose of a two-way ANOVA and what factors does it examine?
Two-way ANOVA tests the effect of two categorical factors on a continuous response simultaneously. It examines two main effects (one per factor) and their interaction effect, partitioning variance into these components plus error.
What does an interaction effect mean in a two-way ANOVA?
An interaction means the effect of one factor on the response depends on the level of the other factor. Graphically, the lines in an interaction (means) plot are non-parallel. If interaction is significant, main effects must be interpreted cautiously.
What is the standard error of the sample mean, and why does it matter?
$$SE(\bar{x}) = \frac{\sigma}{\sqrt{n}} \quad (\text{estimated by } s/\sqrt{n}).$$ It quantifies the sampling variability of $\bar{x}$ and shrinks as $\sqrt{n}$ grows, forming the denominator of z/t statistics and the basis of margins of error.
Why is the Central Limit Theorem important for inference about a mean?
The CLT states that for large $n$, the sampling distribution of $\bar{x}$ is approximately normal with mean $\mu$ and standard deviation $\sigma/\sqrt{n}$, regardless of the population's shape. This justifies using z/t procedures and normal-based confidence intervals even for non-normal populations.
Why can't a two-tailed hypothesis test 'accept' the null hypothesis?
Failing to reject $H_0$ only means the data provide insufficient evidence against it — absence of evidence is not evidence of absence. The null may still be false with the test simply lacking power, so we say 'fail to reject,' never 'accept.'
After a significant one-way ANOVA, what is needed to identify which means differ, and why?
A post-hoc multiple-comparison procedure (e.g. Tukey's HSD, Bonferroni, Scheffé). ANOVA's F-test only signals that at least one mean differs, not which; post-hoc tests compare pairs while controlling the inflated family-wise Type I error from many comparisons.
What this deck covers
The Statistical Inference deck follows the Statistics & Probability Statistical Inference syllabus — 7 chapters and 27 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 7.1 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 227 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Statistical Inference flashcards FAQ
How many Statistical Inference flashcards are in this Statistics & Probability deck?
50 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these Statistics & Probability flashcards free?
Yes. The preview here is free to read with no signup, and the full 50-card deck is free inside the Examius app.
What do the Statistical Inference cards cover?
They follow the Statistics & Probability Statistical Inference syllabus — 7 chapters and 27 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.