Statistics & Probability Statistical Inference Syllabus

Define the statistical power of a test.

Power is the probability of correctly rejecting a false null hypothesis: $$\text{Power} = 1 - \beta = P(\text{reject } H_0 \mid H_0 \text{ false}).$$ It measures the test's ability to detect a true effect.

List four factors that increase the power of a statistical test.

(1) Larger sample size $n$; (2) larger true effect size (bigger departure from $H_0$); (3) larger significance level $\alpha$ (less strict); (4) smaller variability $\sigma$. Using a one-tailed test (when appropriate) also raises power.

Compare one-tailed and two-tailed tests, including how the rejection region is placed.

A two-tailed test ($H_a: \theta \neq \theta_0$) splits $\alpha$ into both tails ($\alpha/2$ each) and detects deviations in either direction. A one-tailed test ($H_a: \theta > \theta_0$ or $<\theta_0$) puts all $\alpha$ in one tail, giving more power to detect an effect in that specified direction only.

Write the test statistic for a one-sample z-test for a mean.

$$z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$$ used when $\sigma$ is known (or $n$ is large) and the sampling distribution of $\bar{x}$ is approximately normal.

Write the test statistic and degrees of freedom for a one-sample t-test for a mean.

$$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}, \qquad df = n-1$$ used when $\sigma$ is unknown and estimated by the sample standard deviation $s$.

Give the pooled two-sample t-test statistic used when equal population variances are assumed.

$$t = \frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}, \quad s_p^{2}=\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}}{n_1+n_2-2}$$ with $df = n_1+n_2-2$.

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.