AP Statistics Quick Review

Every stats question starts by correctly classifying variables and identifying how the data were produced, since that classification determines every graph, statistic, and inference procedure that follows.

Categorical (qualitative) variables place individuals into groups (e.g., color, yes/no); quantitative variables take numerical values that can be measured or averaged, and split into discrete (countable) vs. continuous (measurable). Data can come from a census (entire population), a sample survey, an observational study (variables recorded without imposing treatments), or an experiment (researchers impose treatments). Population $=$ ALL individuals of interest; sample $=$ the subset actually measured.

A response variable (the outcome measured, ``$y$'') vs. an explanatory variable (used to explain or predict changes in the response, ``$x$'') --- in an experiment, the explanatory variable is the treatment researchers manipulate.

Treating a numerically-coded categorical variable (e.g., zip code, jersey number) as quantitative just because it ``looks like a number'' --- ask whether averaging the values would be meaningful; if not, it's categorical.

Categorical $arrow$ counts/proportions; quantitative $arrow$ measures/averages; census $=$ whole population.

Categorical data is summarized with counts or percentages in frequency tables and bar charts --- never with a histogram.

Frequency table $=$ raw counts; relative frequency table $=$ counts $$ total (as a %). Bar chart: a bar for each category, gaps between bars, height $=$ frequency or relative frequency; bar order is arbitrary. Pie chart: shows parts of a whole (percentages must sum to 100%).

A bar chart (categorical data, gaps between bars, bar ORDER is arbitrary) vs. a histogram (quantitative data, bars touch, order is fixed by numerical value).

Ranking or comparing bar-chart categories as though the horizontal axis order carries numerical meaning --- categorical axis order is arbitrary, so never describe a bar chart's shape as ``skewed'' or ``increasing.''

Categorical $arrow$ bar chart (gaps, order doesn't matter); never a histogram.

A quantitative distribution is fully described in context using shape, outliers, center, and spread (SOCS) --- never just one number in isolation.

Displays: dotplot, stemplot, histogram. Shape: symmetric, skewed left (tail points left, mean $<$ median), skewed right (tail points right, mean $>$ median), unimodal/bimodal. Always describe distributions IN CONTEXT with units.

Skewed left (long tail toward LOW values, mean pulled DOWN below the median) vs. skewed right (long tail toward HIGH values, mean pulled UP above the median).

Describing a distribution's shape, center, and spread with no context or units (``center is 20'') --- full credit requires stating what ``20'' means in the problem's context (e.g., ``a typical commute is about 20 minutes'').

Skew name $=$ direction of the LONG TAIL; always describe shape + outliers + center + spread, in context.

Which measure of center and spread to report depends entirely on shape: symmetric data uses mean/standard deviation, while skewed data or outliers call for median/IQR.

$$x=/ x_in s=sqrt/ (x_i-x)^2n-1$$ Median $=$ middle value (average of the two middle values if $n$ is even). Range $=$ max $-$ min. IQR $=Q_3-Q_1$ (spread of the middle 50%). Variance $=s^2$.

Standard deviation $s$ (typical distance of DATA POINTS from the mean, uses $n-1$) vs. standard error (spread of a SAMPLE STATISTIC's distribution, e.g. $s/sqrtn$) --- these get mixed up constantly once inference begins.

Dividing by $n$ instead of $n-1$ when computing the sample standard deviation by hand --- the $n-1$ (``degrees of freedom'') correction is required for the sample formula.

Mean/SD $arrow$ symmetric, no outliers; median/IQR $arrow$ skewed or outliers present.

The Normal distribution is a mathematical model for symmetric, mound-shaped data, and z-scores standardize any value into ``number of standard deviations from the mean'' so probabilities can be found on any Normal curve.

$$z=/x-$$ 68--95--99.7 (Empirical) Rule: about 68% of values fall within 1 SD of $$, 95% within 2 SD, 99.7% within 3 SD. Use normalcdf for area (probability) and invNorm for a boundary value given an area.

A z-score (how many SDs a value is from the mean --- can be negative) vs. a percentile (the percent of the distribution BELOW that value --- always between 0 and 100).

Using invNorm when the problem gives a value and asks for a probability (should use normalcdf), or vice versa --- read carefully whether the unknown is an area (probability) or a boundary (a value of $x$).

$z=(x-)/$; 68--95--99.7 for $±1,2,3$ SDs; area $arrow$ normalcdf, boundary $arrow$ invNorm.

The five-number summary and its boxplot let you compare center, spread, and outliers across several groups at a glance using resistant (outlier-proof) measures.

Five-number summary: Min, $Q_1$, Median, $Q_3$, Max. $$IQR=Q_3-Q_1$$ Outlier rule: a value is a suspected outlier if it is below $Q_1-1.5(IQR)$ or above $Q_3+1.5(IQR)$.

A resistant statistic (median, IQR --- barely affected by outliers/skew) vs. a non-resistant statistic (mean, standard deviation, range --- pulled strongly by outliers).

Forgetting to apply the $1.5×IQR$ rule and instead calling any visually ``far'' point an outlier --- always compute the fences before labeling a point an outlier on a free-response answer.

Outlier if outside $Q_1-1.5 IQR$ or $Q_3+1.5 IQR$; median/IQR resist outliers, mean/SD don't.

How a sample is selected determines whether results can be trusted to represent the population; random selection is the only defense against bias.

Simple random sample (SRS): every group of $n$ individuals is equally likely to be chosen. Stratified random sampling: divide the population into homogeneous strata, then SRS within each. Cluster sampling: divide into heterogeneous clusters, then randomly select WHOLE clusters. Systematic sampling: select every $k$th individual after a random start. Bias types: undercoverage (some group left out of the sampling frame), nonresponse (selected individuals don't respond), response bias (wording/interviewer influences answers), voluntary response (self-selected sample).

Stratified sampling (sample FROM EVERY stratum) vs. cluster sampling (sample only SOME clusters, but entirely within them) --- strata are chosen to be similar internally/different from each other; clusters are chosen to each look like a mini-population.

Calling any large sample ``unbiased'' --- sample SIZE does not fix bias; only a properly randomized selection method reduces bias, no matter how large $n$ is.

Stratify $arrow$ sample every group; cluster $arrow$ sample only some whole groups; size $$ lack of bias.

A well-designed experiment uses control, random assignment, and replication to isolate the treatment's effect and rule out confounding.

Key principles: control (compare treatments, often with a control group/placebo), random assignment (equalizes lurking variables across groups), replication (enough subjects per group). Completely randomized design: subjects randomly assigned directly to treatments. Randomized block design: group subjects into blocks of similar individuals first, then randomize within each block. Matched pairs: a block design with blocks of size 2 (or each subject serves as their own control). Double-blind: neither subjects nor those measuring results know who got which treatment.

Blocking (grouping by a variable expected to affect the response, done BEFORE random assignment, to reduce variability) vs. stratifying a sample (grouping before random SELECTION, to improve representativeness) --- blocking is for experiments, stratifying is for sampling, and neither replaces randomization.

Confusing a confounding variable (its effect on the response can't be separated from the treatment's effect) with a lurking variable that merely correlates with both the explanatory and response variables --- random assignment is specifically what neutralizes confounding.

Control + Random assignment + Replicate $arrow$ valid experiment; block first, then randomize within blocks.

Whether you can generalize results to a population and whether you can claim causation depend on two completely separate design choices: random sampling and random assignment.

Random SAMPLING (from a population) $arrow$ allows generalizing results TO that population. Random ASSIGNMENT (to treatments in an experiment) $arrow$ allows a cause-and-effect conclusion. Four combinations: both $arrow$ causal AND generalizable; random assignment only $arrow$ causal but NOT generalizable beyond the subjects studied; random sampling only (observational study) $arrow$ generalizable but NOT causal (association only); neither $arrow$ neither.

An observational study (can show association/correlation only) vs. a designed experiment with random assignment (can support a causal claim).

Writing ``causes'' or ``leads to'' in a conclusion based on an observational study --- this is one of the most heavily penalized wording errors on AP Statistics FRQs; observational data supports ``is associated with,'' never ``causes.''

Random sample $arrow$ generalize; random assignment $arrow$ causation; observational $=$ association only, never causation.

Probability quantifies long-run relative frequency, and a short list of rules lets you build up the probability of any complex event from simpler pieces.

$0≤ P(A)≤ 1$; $P(sample space)=1$. Complement rule: $P(A^c)=1-P(A)$. Addition rule (general): $$P(A or B)=P(A)+P(B)-P(A and B)$$ If $A,B$ are mutually exclusive (disjoint): $P(A and B)=0$, so $P(A or B)=P(A)+P(B)$.

Mutually exclusive events (CANNOT both happen, $P(A and B)=0$) vs. independent events (one occurring doesn't change the other's probability, $P(A B)=P(A)$) --- two events with nonzero probability that are mutually exclusive can NEVER be independent.

Adding $P(A)+P(B)$ without subtracting $P(A and B)$ when the events overlap --- double-counts the overlap and inflates the probability.

Mutually exclusive $$ independent; $P(A or B)=P(A)+P(B)-P(A and B)$.

Conditional probability updates the probability of an event given that another event is known to have occurred, and checking independence formally means checking whether that update changes anything.

$$P(A B)=/P(A and B)P(B)$$ General multiplication rule: $P(A and B)=P(A)· P(B A)$. $A$ and $B$ are independent iff $P(A B)=P(A)$ (equivalently $P(A and B)=P(A)P(B)$).

$P(A B)$ (probability of $A$ given $B$ already happened) vs. $P(B A)$ (probability of $B$ given $A$) --- these are generally NOT equal; reversing them is a classic error (e.g., confusing $P(disease test)$ with $P(positive test)$).

Assuming independence to multiply $P(A)× P(B)$ when sampling WITHOUT replacement from a small population --- removing an individual changes the probabilities for the next draw, so events are not truly independent unless the population is large relative to the sample (10% condition).

$P(A B)=P(A and B)/P(B)$; independent $⇒ P(A B)=P(A)$.

A two-way table displays how two categorical variables relate, and computing marginal, joint, and conditional distributions from it is the foundation for later testing that relationship formally.

Marginal distribution: row or column totals $$ grand total (distribution of ONE variable alone). Joint distribution: each cell $$ grand total (both variables together). Conditional distribution: a row (or column) $$ that row's (column's) total (one variable, GIVEN a specific value of the other).

A conditional distribution (restricted to one row/column, denominator is that row/column's total) vs. a marginal distribution (uses the grand total as the denominator) --- picking the wrong denominator is the single most common two-way-table error.

Comparing raw counts across groups of very different sizes instead of converting to conditional percentages --- association must be assessed by comparing conditional distributions (percentages), not raw counts.

Marginal $arrow$ totals$/$grand total; conditional $arrow$ row or column$/$its own total.

A discrete random variable's probability distribution assigns a probability to every possible outcome, and its mean (expected value) is the long-run average outcome, weighted by probability.

$$_X=E(X)= x_i p(x_i) _X^2= (x_i-_X)^2 p(x_i)$$ Requirements for a valid probability distribution: every $p(x_i)≥ 0$ and $ p(x_i)=1$.

Expected value (a weighted AVERAGE across all possible outcomes, may not equal any single possible value) vs. the most likely single outcome (the mode of the distribution) --- $E(X)=2.5$ is a perfectly valid expected value even if $X$ can only take whole-number values.

Computing a simple (unweighted) average of the possible $x$-values instead of weighting each by its probability --- expected value requires multiplying each outcome by its own probability before summing.

$E(X)= x_i p(x_i)$; a weighted average, not the most likely outcome.

The binomial distribution models the count of successes in a fixed number of independent, identical yes/no trials, and it comes with its own mean and standard deviation formulas.

BINS conditions: Binary outcomes, Independent trials, fixed Number $n$, constant Success probability $p$. $$P(X=k)=nkp^k(1-p)^n-k _X=np _X=sqrtnp(1-p)$$

Binomial setting (fixed number of trials $n$, counting successes) vs. geometric setting (trials continue UNTIL the first success, counting trials needed) --- check whether $n$ is fixed or the number of trials itself is random.

Using the binomial formula when sampling without replacement from a SMALL population (trials aren't independent) --- the binomial model requires independence, only approximately satisfied when $n≤10%$ of the population (the 10% condition).

BINS $arrow$ binomial; $=np$, $=sqrtnp(1-p)$.

The mean of a sum or difference of random variables always adds/subtracts directly, but variances only add --- for BOTH sums and differences --- and only when the variables are independent.

$$_X+Y=_X+_Y _X-Y=_X-_Y$$ If $X$ and $Y$ are independent: $$^2_X+Y=_X^2+_Y^2 ^2_X-Y=_X^2+_Y^2$$

Combining means (subtract for a difference) vs. combining variances (ALWAYS add, even for $X-Y$, because variability accumulates from both variables regardless of sign) --- students very often mistakenly subtract the variances for $X-Y$.

Adding variances when $X$ and $Y$ are NOT independent, or when the same random variable is subtracted from itself (e.g., $X-X$) --- the addition rule requires independence, and $^2_X-X=0$, not $2_X^2$.

Means: add/subtract normally. Variances: ALWAYS add (never subtract) if independent.

When a random variable is approximately Normal, z-scores and Normal-curve area calculations become a tool for finding probabilities of random outcomes, not just describing data.

For $X N(,)$: $$P(X<x)=P≤ft(z</x-)$$ A binomial random variable can be approximated by a Normal model when $np≥ 10$ and $n(1-p)≥ 10$ (large counts condition), using $=np$, $=sqrtnp(1-p)$.

Using the EXACT binomial formula (small $n$, discrete outcomes) vs. the Normal APPROXIMATION to the binomial (large $n$, satisfies the large counts condition) --- the approximation is only appropriate once $np≥10$ and $n(1-p)≥10$.

Applying a Normal probability calculation to a random variable without first verifying it's actually approximately Normal (e.g., a strongly skewed count with small $np$) --- always state and check the condition before using normalcdf.

Normal model valid only if shape is (approximately) Normal; for binomial, check $np≥10$, $n(1-p)≥10$.

A sampling distribution describes how a statistic (like $x$ or $p$) would vary across all possible samples of a given size, and the Central Limit Theorem guarantees it becomes approximately Normal even when the population isn't.

Sampling distribution of $ p$: mean $=p$, SD $=sqrt/p(1-p)n$; approximately Normal if $np≥10$ and $n(1-p)≥10$ (Large Counts). Sampling distribution of $ x$: mean $=$, SD $=/sqrt n$; approximately Normal if the population is Normal, or (by the CLT) if $n≥30$.

The standard deviation of the POPULATION/sample ($$ or $s$) vs. the standard deviation of a sampling distribution ($/sqrt n$, often called the standard error) --- the sampling distribution's spread shrinks as $n$ grows; the population's spread does not.

Claiming a sampling distribution is ``Normal'' just because $n$ is ``large'' without checking the actual numeric condition ($np≥10$ & $n(1-p)≥10$, or $n≥30$) --- AP graders require the condition to be checked and stated, not assumed.

$ p$: mean $p$, SD $sqrtp(1-p)/n$. $ x$: mean $$, SD $/sqrt n$. CLT $arrow$ Normal shape as $n$ grows.

Inference uses a sample statistic, plus knowledge of its sampling distribution, to draw a conclusion about an unknown population parameter --- either by estimating it (confidence interval) or testing a claim about it (significance test).

Parameter (fixed, usually unknown number describing the POPULATION, e.g. $p$, $$) vs. statistic (computed from a SAMPLE, e.g. $ p$, $ x$, used to estimate the parameter). General confidence interval form: $$statistic ± (critical value)×(standard deviation of the statistic)$$

A parameter (Greek letters: $p$, $$, $$ --- describes the population, fixed but usually unknown) vs. a statistic (Roman letters: $ p$, $ x$, $s$ --- describes a sample, varies from sample to sample).

Using a Greek letter ($p$, $$) to describe a sample result, or a Roman letter ($ p$, $ x$) to describe the population truth --- rubrics specifically check that hypotheses and conclusions use the correct symbol for parameter vs. statistic.

Parameter (Greek, population, fixed) vs. statistic (Roman, sample, varies); CI $=$ statistic $±$ margin of error.

Every confidence interval and significance test requires the same three types of conditions to be checked and justified in context before the mechanics can be trusted.

Random: data from a random sample or randomized experiment. 10%: when sampling without replacement, $n≤0.10 N$ (ensures near-independence). Large Counts (categorical) / Normal-Large Sample (quantitative): for proportions, $n p≥10$ and $n(1- p)≥10$ (or $np_0, n(1-p_0)≥10$ for a test); for means, population Normal or $n≥30$.

Checking conditions using the null value $p_0$ (required for a SIGNIFICANCE TEST's Large Counts condition) vs. using the sample proportion $ p$ (required for a CONFIDENCE INTERVAL's Large Counts condition) --- tests assume $H_0$ is true, so they check counts using $p_0$, not $ p$.

Listing condition NAMES (``Random, 10%, Large Counts'') without plugging in the problem's numbers to verify each one --- full credit requires the numeric check (e.g., ``$n p=30(0.4)=12≥10$''), not just naming the condition.

Random + 10% + Large Counts/Normal --- always check with the actual numbers, in context.

A one-sample z-interval for $p$ uses the sample proportion plus a margin of error built from the standard error and a critical z-value to capture the true population proportion with a stated confidence level.

$$ p ± z^*sqrt/ p(1- p)n$$ Common critical values: $z^*=1.645$ (90%), $1.96$ (95%), $2.576$ (99%). Correct interpretation: ``We are C% confident that the interval from \_\_\_ to \_\_\_ captures the true population proportion of \_\_\_ (in context).''

The correct interpretation of confidence (``if we repeated this sampling process many times, about C% of the resulting intervals would capture the true parameter'') vs. the common WRONG interpretation (``there is a C% probability the true parameter is in this ONE interval'') --- a specific computed interval either does or does not contain the fixed parameter; the probability statement applies to the METHOD, not this one result.

Writing the confidence interval conclusion in terms of individual data values or a sample statistic instead of the population parameter (e.g., ``C% of students are in this range'' instead of ``we are C% confident the true proportion of all students is in this range'').

$ p ± z^*sqrt p(1- p)/n$; confidence describes the METHOD, not one specific interval.

A significance test asks ``how surprising would this sample result be if the null hypothesis were actually true?'' and uses that surprise level (the p-value) to decide whether to reject the null hypothesis.

$H_0$: the ``no effect/no difference'' claim, always contains an equality. $H_a$: the claim being tested for, one- or two-sided. P-value $=$ probability of a result at least as extreme as the one observed, ASSUMING $H_0$ is true. Decision rule: if p-value $<$, reject $H_0$ (statistically significant); otherwise, fail to reject $H_0$. Type I error: reject a TRUE $H_0$ (probability $=$). Type II error: fail to reject a FALSE $H_0$ (probability $=$). Power $=1-=$ probability of correctly rejecting a false $H_0$.

Type I error (rejecting a true $H_0$ --- a ``false alarm,'' probability $$) vs. Type II error (failing to reject a false $H_0$ --- a ``missed detection,'' probability $$) --- lowering $$ to reduce Type I error risk always INCREASES the risk of a Type II error, all else equal.

Concluding a test ``proves $H_0$ is true'' after failing to reject it --- a test can never prove the null; ``fail to reject $H_0$'' only means insufficient evidence against it. Also: interpreting the p-value as ``the probability $H_0$ is true'' (it is not).

P-value $< ⇒$ reject $H_0$. Type I $=$ reject true $H_0$; Type II $=$ keep false $H_0$; Power $=1-$.

The one-sample z-test for a proportion measures how many standard errors the sample proportion falls from the hypothesized value, assuming the null is true.

$H_0: p=p_0$. Test statistic: $$z=/ p - p_0sqrt/p_0(1-p_0)n$$ Find the p-value from the standard Normal distribution based on $H_a$ (one- or two-sided).

The standard error used in a CONFIDENCE INTERVAL for $p$ (uses $ p$: $sqrt p(1- p)/n$) vs. the standard error used in a SIGNIFICANCE TEST for $p$ (uses the null value $p_0$: $sqrtp_0(1-p_0)/n$) --- using $ p$ in a test statistic's denominator is a common but incorrect shortcut.

Reporting only the test statistic and p-value without a full ``State-Plan-Do-Conclude'' write-up, or writing a conclusion that doesn't reference both the p-value/$$ AND the context --- rubrics require the conclusion to link the numeric decision back to the real-world claim.

$z=( p-p_0)/sqrtp_0(1-p_0)/n$ --- denominator uses $p_0$, not $ p$.

Comparing two population proportions uses the same ``statistic $±$ margin of error'' structure, but the standard error now combines variability from both independent samples.

$$( p_1 - p_2) ± z^*sqrt/ p_1(1- p_1)n_1+/ p_2(1- p_2)n_2$$ Conditions: Random (two independent samples/groups), 10% for each sample, Large Counts for BOTH groups ($n_1 p_1, n_1(1- p_1), n_2 p_2, n_2(1- p_2)$ all $≥10$).

An interval that CONTAINS 0 (plausible that $p_1=p_2$; no convincing evidence of a difference) vs. an interval that does NOT contain 0 (0 is not a plausible value for $p_1-p_2$; convincing evidence of a difference, and the sign tells you which group is larger).

Forgetting to check the Large Counts condition separately for BOTH samples (four separate numeric checks total, not two) --- a common shortcut that costs a condition-check point on the rubric.

$( p_1- p_2)± z^*sqrt p_1(1- p_1)/n_1+ p_2(1- p_2)/n_2$; interval containing 0 $arrow$ no evidence of a difference.

Testing whether two population proportions differ requires pooling the two samples into one combined estimate of the common proportion under $H_0: p_1=p_2$.

Pooled proportion: $$ p_c=/X_1+X_2n_1+n_2$$ Test statistic: $$z=/( p_1- p_2)-0sqrt p_c(1- p_c)≤ft(/1n_1+/1n_2)$$

The pooled proportion $ p_c$ (used ONLY in the significance test, because $H_0$ assumes $p_1=p_2$) vs. the separate sample proportions $ p_1, p_2$ (used in the CONFIDENCE INTERVAL, which makes no such assumption) --- pooling in a confidence interval, or failing to pool in a test, is a frequent computational slip.

Forgetting to pool (using $ p_1, p_2$ separately in the test statistic's standard error instead of $ p_c$) --- this produces a numerically different (and incorrect) test statistic and p-value.

Pool for the TEST ($ p_c$); don't pool for the INTERVAL ($ p_1, p_2$ separately).

The chi-square test for homogeneity compares the distribution of ONE categorical variable across SEVERAL independent populations or treatment groups, extending the two-proportion test to more than two groups.

$H_0$: the distribution of the categorical variable is the SAME across all populations/groups. $$^2= /(Observed-Expected)^2Expected Expected count=/row total×column totalgrand total$$ $df=(rows-1)(columns-1)$.

Homogeneity (SEVERAL separate samples, one from each population/treatment, comparing the SAME variable's distribution across them) vs. independence (ONE single sample, classified by TWO categorical variables, asking whether they're associated) --- the design (how samples were taken) tells you which test name applies, even though the mechanics/formula are identical.

Using $df=n-1$ (as in a goodness-of-fit setting) instead of $(rows-1)(columns-1)$ for a two-way table --- always compute $df$ from the table's dimensions for homogeneity/independence tests.

Several samples, one variable, same shape? $arrow$ homogeneity; $^2=(O-E)^2/E$, $df=(r-1)(c-1)$.

The chi-square test for independence uses ONE sample classified by two categorical variables to test whether those variables are associated in the population.

$H_0$: the two categorical variables are INDEPENDENT (not associated) in the population; $H_a$: they are associated. Same test statistic and $df$ formula as homogeneity: $$^2=/(O-E)^2E, df=(rows-1)(columns-1)$$ Conditions: Random, 10% (if sampling without replacement), Large Counts (every expected count $≥5$).

The Large Counts condition for chi-square tests (every EXPECTED count $≥5$) vs. the Large Counts condition for proportion z-procedures ($n p$ and $n(1- p)≥10$) --- different threshold, different quantity (expected counts, not $n p$).

Writing hypotheses in terms of specific proportions being equal (as in a two-proportion test) instead of the correct chi-square wording (``the two variables are independent/not associated,'' or ``the distributions are the same'') --- rubrics require association/independence language, not equality-of-proportions language.

One sample, two categorical variables, testing association $arrow$ independence; all expected counts $≥5$.

Because the population standard deviation $$ is essentially never known for means, inference for $$ replaces $$ with the sample standard deviation $s$ and uses the wider-tailed t-distribution instead of the Normal to account for that extra uncertainty.

$$t=/ x - s/sqrt n, df=n-1$$ The t-distribution is symmetric, mound-shaped, with heavier tails than Normal; as $df$ increases, $t$ approaches the standard Normal distribution.

A z-procedure (used when $$ is known --- rare on the AP exam, or for proportions whose SD is a function of $p$ itself) vs. a t-procedure (used when $$ is unknown and estimated by $s$ --- essentially always the case for means).

Using a z critical value/table for a mean's confidence interval or test instead of a t critical value with $df=n-1$ --- since $$ is virtually never known for means on the AP exam, means ALWAYS require t-procedures.

Means $arrow$ t-procedures ($$ unknown, use $s$); $df=n-1$; $t$ has heavier tails than $z$.

A one-sample t-interval estimates a population mean using the sample mean plus a margin of error built from the t-distribution.

$$ x ± t^*/ssqrt n, df=n-1$$ Conditions: Random, 10%, and Normal/Large Sample (population approximately Normal, OR $n≥30$, OR a graph of the sample data shows no strong skew/outliers).

The Normal/Large Sample condition for MEANS (checked by graphing the sample data or using $n≥30$) vs. the Large Counts condition for PROPORTIONS (checked with a numeric inequality like $n p≥10$) --- means require looking at a graph or sample size, not a Large Counts calculation.

Using $z^*$ instead of $t^*$ (with the correct $df=n-1$) when finding the critical value from a table --- an easy points-off error when working quickly under time pressure.

$ x ± t^* (s/sqrt n)$, $df=n-1$; check Normal/Large Sample by graphing or $n≥30$.

The one-sample t-test measures how many estimated standard errors the sample mean falls from the hypothesized population mean.

$H_0: =_0$. $$t=/ x-_0s/sqrt n, df=n-1$$ Find the p-value from the t-distribution with $df=n-1$, based on the direction of $H_a$.

A one-sided alternative ($H_a: <_0$ or $>_0$, uses ONE tail's area for the p-value) vs. a two-sided alternative ($H_a:_0$, DOUBLES the one-tail area for the p-value) --- using the wrong number of tails changes the p-value and can flip the decision.

Rejecting $H_0$ purely because the sample mean differs numerically from $_0$, without actually computing the p-value and comparing it to $$ --- ``different'' is not automatically ``statistically significant.''

$t=( x-_0)/(s/sqrt n)$, $df=n-1$; two-sided $H_a⇒$ double the one-tail p-value.

When two measurements are naturally linked (before/after, twins, matched subjects), you must first reduce the two sets of data to ONE set of differences and then run ordinary one-sample t-procedures on those differences.

Compute $d_i = x_i - y_i$ for each pair; then apply the one-sample t-interval/t-test formulas to the $d_i$'s using $ d$ and $s_d$ in place of $ x$ and $s$: $$ d ± t^*/s_dsqrt n t=/ d-_0s_d/sqrt n, \ df=n-1$$ (Typically $_0=0$ to test ``no difference.'')

Paired/matched-pairs data (each subject or pair contributes ONE difference; use one-sample t-procedures on the differences) vs. two independent samples (two SEPARATE, unrelated groups; use two-sample t-procedures) --- treating paired data as two independent samples ignores the correlation between pairs and is a major procedure-selection error.

Running a two-sample t-test on paired data instead of first taking differences --- always ask ``is each observation in group 1 linked to a specific observation in group 2?'' before choosing the procedure.

Linked measurements $arrow$ take differences $d_i$ first, then one-sample t on $ d$.

Comparing two independent population means combines each sample's own variability into a single standard error, without ever pooling the two sample standard deviations.

$$( x_1- x_2)± t^*sqrt/s_1^2n_1+/s_2^2n_2$$ $df$ is computed by technology (a conservative hand estimate: $df=(n_1-1, n_2-1)$). Conditions: Random (independent samples/random assignment), 10% for each group, Normal/Large Sample for each group.

Two-sample t-procedures for means (NEVER pool $s_1,s_2$; each group keeps its own standard deviation in the standard error) vs. the pooled two-proportion z-test (DOES pool $ p_1, p_2$ into $ p_c$) --- pooling rules differ between the two inference settings and are frequently swapped by mistake.

Using a conservative hand-calculated $df=(n_1-1,n_2-1)$ but then looking up a critical value as though it were the (larger) technology-computed $df$ --- this makes the interval falsely narrower/more confident than it should be.

$( x_1- x_2)± t^*sqrts_1^2/n_1+s_2^2/n_2$ --- never pool $s_1,s_2$ for means.

The two-sample t-test asks whether the observed difference between two independent sample means is larger than what sampling variability alone would produce if the population means were equal.

$H_0:_1=_2$ (equivalently $_1-_2=0$). $$t=/( x_1- x_2)-0sqrt/s_1^2n_1+/s_2^2n_2$$

Independent (unpaired) two-sample data (compare using two-sample t, standard errors from BOTH groups combine) vs. paired data (reduce to differences FIRST, then use one-sample t) --- re-check which design applies before selecting a formula; this is the single most common procedure-selection error in Unit 4.

Failing to state/verify that the random assignment or independent random samples condition holds for BOTH groups before running the test --- a two-sample procedure requires the two groups themselves to be independent of each other, not just each individual sample being random.

$t=[( x_1- x_2)-0]/sqrts_1^2/n_1+s_2^2/n_2$; independent groups only (paired data uses differences instead).

A scatterplot reveals the direction, form, and strength of the relationship between two quantitative variables, and the correlation coefficient $r$ numerically summarizes only the strength and direction of a LINEAR relationship.

$$r=/1n-1≤ft(/x_i- xs_x)≤ft(/y_i- ys_y)$$ $-1≤ r≤1$; sign matches the direction of association; $|r|$ closer to 1 means a stronger linear relationship. $r$ has no units and does not depend on which variable is $x$ vs. $y$.

Correlation (association, quantified by $r$) vs. causation (one variable actually CAUSING change in another) --- a strong $r$, even $r=1$, never by itself establishes causation; only a well-designed experiment with random assignment can.

Using $r$ to describe a clearly CURVED (nonlinear) scatterplot --- $r$ only measures the strength of a LINEAR relationship and can be misleadingly close to 0 (or high) even when a strong nonlinear pattern is obviously present; always look at the scatterplot first.

$r$ measures LINEAR strength/direction only; correlation $$ causation, ever.

The least-squares regression line is the unique line that minimizes the sum of squared vertical distances between the data points and the line, giving the ``best'' linear model for predicting $y$ from $x$.

$$ y = b_0+b_1x, b_1=r·/s_ys_x, b_0= y - b_1 x$$ $b_1$ (slope) $=$ predicted change in $y$ for each 1-unit increase in $x$. $b_0$ (y-intercept) $=$ predicted $y$ when $x=0$ (often not meaningful if $x=0$ is outside the data range). The LSRL always passes through $( x, y)$.

The slope's SIGN (matches the sign of $r$ --- positive $r$ gives positive slope) vs. the slope's SIZE (depends on the units/spread of $x$ and $y$ through $s_y/s_x$, not on $r$ alone) --- a strong correlation does not imply a large slope, and vice versa.

Extrapolating --- using the regression equation to predict $ y$ for an $x$-value far outside the range of the original data --- the linear pattern is not guaranteed to hold outside the observed data range.

$ y=b_0+b_1x$; $b_1=r(s_y/s_x)$; line always passes through $( x, y)$; never extrapolate.

A residual measures the LSRL's prediction error for each point, and a residual plot's overall pattern (or lack of one) is the definitive check for whether a linear model is actually appropriate.

$$residual=y- y \ (observed-predicted)$$ Residuals always sum to (approximately) 0. A residual plot with points scattered randomly around the horizontal line at 0, with no leftover pattern, confirms a linear model is appropriate; a curved pattern in the residual plot means the linear model is NOT appropriate, even if $r$ is high.

A point that is unusual in the $y$-direction only (a large residual, poorly predicted, but doesn't necessarily change the line much) vs. an influential point (usually extreme in the $x$-direction, whose removal would substantially change $b_1$ or $b_0$) --- a point can have a small residual and still be highly influential.

Concluding a linear model ``fits well'' from a high $r$ or $R^2$ alone without checking the residual plot for leftover curvature --- a nonlinear relationship can still produce a high $r$ if it's monotonic enough; the residual plot is the real diagnostic.

Residual $=y- y$; random scatter in residual plot $arrow$ linear model is appropriate; pattern $arrow$ it isn't.

$R^2$ translates model fit into a percentage: the proportion of the variation in $y$ that is explained by the linear relationship with $x$.

For simple linear regression, $$R^2 = r^2$$ Interpretation: ``$R^2×100%$ of the variation in [$y$, in context] is explained by the linear model relating it to [$x$, in context].'' $R^2$ close to 1 $=$ model explains most of the variation; $R^2$ close to 0 $=$ model explains very little.

$r$ (correlation, ranges $-1$ to $1$, has a sign showing direction) vs. $R^2$ (coefficient of determination, ranges $0$ to $1$, always non-negative, describes proportion of variation explained) --- squaring $r$ loses the direction information, so $R^2$ alone can't tell you if the association is positive or negative.

Interpreting $R^2$ without reference to variation (``$R^2=0.81$ means 81% of the points fit the line'') instead of the correct wording (``81% of the variation in $y$ is explained by $x$'') --- rubrics require the ``variation'' wording, in context, for full credit.

$R^2=r^2$ for simple regression; interpret as % of VARIATION in $y$ explained by $x$.

When a scatterplot and residual plot show a clear curve, transforming one or both variables (often with logarithms) can straighten the relationship so linear regression tools can be applied to the transformed data.

Exponential growth pattern ($y=ab^x$): transform by taking $(y)$; if $( y)$ vs. $x$ is linear, the original data follows an exponential model. Power model ($y=ax^p$): transform by taking $(x)$ AND $(y)$; if $( y)$ vs. $(x)$ is linear, the original data follows a power model. After fitting the LSRL to the transformed data, ``undo'' the logarithms algebraically to get a model in terms of the original $x$ and $y$.

An exponential transformation (log the RESPONSE variable only) vs. a power transformation (log BOTH variables) --- using the wrong one leaves curvature in the residual plot of the transformed data.

Fitting a line to the transformed data but then reporting predictions or residuals as though they were still on the ORIGINAL scale --- predictions from a $(y)$-vs-$x$ model must be exponentiated back before being reported as a predicted $y$-value.

Curved data $arrow$ try $ y$ (exponential) or $ x$ & $ y$ (power); check the new residual plot for linearity.

Every inference formula is either ``statistic $±$ margin of error'' or ``statistic $-$ null value, over standard error.''

Ask: how many variables, how many samples, and categorical or quantitative --- then match the procedure.

Formulas/tables are given; conditions, correct notation, and context-based conclusions are entirely on you.

State $arrow$ Plan $arrow$ Do $arrow$ Conclude --- skipping Plan's numeric condition checks is the most common way to lose easy points.

42 MCQ (50%, 90 min) + 4 FRQ, 10 pts each (50%, 90 min) --- FRQ3 is always Inference.

Formulate $arrow$ Collect $arrow$ Analyze $arrow$ Interpret --- the four course practices, in order, on every study.

Count the variables, identify categorical vs. quantitative, then match to the display or procedure.