AP Statistics Unit 7: Inference for Means — Free Review Games.

The t-distribution is used when sigma is unknown and estimated by s.

df = n - 1 = 20 - 1 = 19.

The t-distribution has heavier tails, reflecting extra uncertainty from estimating sigma.

With large df, the t-distribution becomes very close to the z-distribution.

SE = sample SD / \(\sqrt{n}\) = \(s/\sqrt{n}\).

Paired tests compare two related measurements (before/after, matched pairs).

The two-sample t-test compares means from two independent samples.

The \(t^*\) critical value depends on both the desired confidence level and the df.

The t-test is robust to moderate departures from normality, especially with larger n.

A one-sample t-test tests H0: mu = mu0.

Compute the difference for each pair, then do a one-sample t-test on those differences.

Conservative df = smaller n - 1 = min(14,19) = 14.

16 is outside the 90% CI, so we reject H0 at alpha = 0.10.

ME is proportional to \(1/\sqrt{n}\). To halve ME, multiply \(n\) by \(4\).

The t-test assumes normality or relies on CLT for large samples.

The three conditions for a one-sample t-test are randomness for valid inference, independence checked via the 10% condition, and normality of the sampling distribution checked via graphs of the sample or the nearly normal condition for small samples. The distractor mentioning "known population SD" is wrong because the entire reason we use a t-procedure instead of a z-procedure is that \(\sigma\) is unknown. Always verify these three conditions before running any t-test, since violations can invalidate the p-value or interval.

In the one-sample t-statistic, \(s\) is the sample standard deviation, which estimates the unknown population standard deviation from the observed data. It is not "the standard error of the mean," since that quantity is \(s/\sqrt{n}\), the entire denominator, not \(s\) alone. Recognizing which symbol in a formula is estimated from data versus a fixed constant is essential for correctly interpreting t-procedures.

With fewer degrees of freedom, the t-distribution has more variability in the sample standard deviation estimate, producing thicker tails and greater spread than a t-distribution with more degrees of freedom. The choice "the exact same shape" is wrong because t-distributions change shape as degrees of freedom change, becoming narrower as df increases. Remember that low df means more uncertainty and a curve that looks less like the standard normal.

Because t-procedures rely on the t-distribution rather than the standard normal, the critical value is denoted \(t^*\) and is found using degrees of freedom and the desired confidence level. The notation \(z^*\) is incorrect here since that symbol is reserved for procedures where \(\sigma\) is known and the normal distribution applies directly. Always match the critical value symbol to the distribution actually being used for the interval or test.

A paired t-test collapses each pair into a single difference, so the parameter being tested is \(\mu_d\), the true mean of these differences. The option "\(\mu_1 - \mu_2\)" describes the two-sample framework, which is different because that procedure does not require the observations to be linked in pairs. Whenever data come from matched pairs or repeated measures, work with the differences rather than treating the two sets of data as independent samples.

Measuring the same subjects before and after a treatment creates naturally linked pairs, so the correct design is a paired t-test on the before-after differences. Comparing "two different randomly assigned classes" involves independent groups with no natural pairing, making a two-sample t-test the appropriate choice instead. The key signal for pairing is that each observation in one group has a specific, meaningful partner in the other group.

The t-distribution is symmetric and bell-shaped like the normal distribution, but it is centered at 0 with heavier tails to account for the extra uncertainty from estimating \(\sigma\) with \(s\). It is not "skewed right" or "skewed left" because skewness would violate the theoretical symmetry built into the t-distribution's derivation. Students should picture the t-distribution as a normal curve stretched slightly at the tails, not tilted to one side.

Each of the two samples must independently satisfy randomness (from random sampling or random assignment) and the 10% condition, since inference validity depends on both groups being properly collected. "Equal population variances" is not required for the standard (unpooled) two-sample t-procedure, which is the version most commonly used in AP Statistics. Always check conditions separately for each group in a two-sample setting rather than assuming what holds for one group holds for the other.

A higher confidence level requires a larger critical value \(t^*\) to capture more of the distribution, which directly increases the margin of error \(t^* \cdot (s/\sqrt{n})\). "Decrease the sample mean" is wrong because the confidence level has no effect on \(\bar{x}\), which is simply the observed data point around which the interval is centered. Remember the tradeoff: higher confidence produces a wider, less precise interval when all else is held constant.

The p-value is the probability, computed under the assumption that \(H_0\) is true, of getting a test statistic at least as extreme as the one observed, which quantifies how surprising the sample result is. It is not "the probability that the null hypothesis is true," a common misconception, since p-values are conditional probabilities about data, not direct probabilities about hypotheses. Keep this conditional definition in mind whenever interpreting p-values on the AP exam, as loose language will cost credit.

Since the researcher specifically wants to know if the fertilizer "increases" height, this is a one-sided test with \(H_a: \mu > 25\), testing against the null value of the standard fertilizer's mean. The option using \(\bar{x}\) instead of \(\mu\) is incorrect because hypotheses are always statements about population parameters, not sample statistics. Always translate directional language like "increases" or "greater than" into a one-sided alternative hypothesis about the parameter of interest.

The standard error is computed as \(s/\sqrt{n} = 20/\sqrt{16} = 20/4 = 5\), giving the typical spread of sample means around the true mean. The choice "\(20\)" is simply the sample standard deviation itself and ignores the division by \(\sqrt{n}\) that converts individual variability into variability of the sample mean. Always divide by the square root of the sample size when moving from \(s\) to the standard error used in t-procedures.

Because the interval \((-2.1, 1.4)\) contains 0, a plausible value for the true mean difference is no difference at all, so we cannot conclude there is a statistically significant effect. The claim that "the treatment definitely decreases reaction time" is wrong because the interval includes both negative and positive plausible values, meaning an increase is also plausible. When a confidence interval for a difference contains 0, that is the standard signal that the difference is not statistically significant.

Since the students in School A and School B are independently sampled with no natural link between individuals in the two groups, a two-sample t-test is the correct procedure. "Comparing left-hand versus right-hand grip strength in the same individuals" is incorrect for a two-sample test because each person contributes a matched pair of measurements, which calls for a paired t-test instead. The determining factor is whether observations across the two groups are linked by design, not simply whether there are two groups.

With \(df = 14\), the two-sided critical value at \(\alpha = 0.05\) is approximately \(t^* = 2.145\), and since the observed \(t = 2.85\) exceeds this threshold, we reject \(H_0\) in favor of \(H_a\). The first option states the correct comparison direction backward, incorrectly concluding to fail to reject even though \(2.85 > 2.145\). When the absolute value of the test statistic exceeds the critical value, that is the standard rejection rule for a t-test.

The standard two-sample t-procedure (unpooled) uses \(\sqrt{s_1^2/n_1 + s_2^2/n_2}\), which correctly combines the individual sampling variability of each group's mean under independence. The option adding the standard errors directly, \(s_1/\sqrt{n_1} + s_2/\sqrt{n_2}\), is wrong because variances of independent quantities add, not standard deviations, so you must add the squared terms before taking the square root. Memorize this variance-addition rule since it applies broadly whenever combining independent sources of variability.

The t-statistic is \(t = \bar{x}_d / (s_d/\sqrt{n}) = 3.2 / (2.1/\sqrt{8}) = 3.2/0.742 \approx 4.31\), reflecting how many standard errors the sample mean difference is from zero. The value "\(1.52\)" incorrectly treats \(s_d\) itself as the standard error rather than dividing it by \(\sqrt{n}\) first. Always compute the standard error of the differences before forming the t-statistic in a paired test.

Overlap of two separate confidence intervals is not equivalent to testing the difference directly, because the correct standard error for a difference combines variability from both samples in a specific way that visual overlap does not replicate. The claim that this approach "is actually a valid and preferred method" is false since statisticians specifically recommend a two-sample procedure or interval for the difference instead. Always construct a confidence interval or test directly on the difference of means when comparing two groups rather than relying on separately built intervals.

Since \(p = 0.03\) is less than \(\alpha = 0.05\), we reject \(H_0\) and conclude there is convincing evidence that the true mean caloric intake differs from 2000 calories. The option stating we should "reject \(H_0\)" but conclude the mean "equals 2000" is self-contradictory, since rejecting the null means we no longer believe the mean equals 2000. Always pair a decision (reject or fail to reject) with a conclusion phrased in the context of the alternative hypothesis, never the null.

Increasing the sample size decreases the standard error \(s/\sqrt{n}\) and also tends to reduce \(t^*\) slightly as degrees of freedom rise, both of which shrink the margin of error and narrow the interval. "Increasing the confidence level" is incorrect because a higher confidence level requires a larger critical value, which widens rather than narrows the interval. The general lesson is that larger samples produce more precise estimates, while higher confidence levels trade precision for certainty.

A small test statistic magnitude combined with a large p-value indicates the observed difference in sample means is easily explained by chance variation, so we fail to reject \(H_0\) and lack convincing evidence of a true difference. The claim that "the null hypothesis is proven true" overstates what a large p-value shows, since failing to reject \(H_0\) never proves it, it only indicates insufficient evidence against it. Keep in mind that statistical tests never prove a null hypothesis true, they only fail to provide evidence against it.

Taking the difference within each pair collapses two dependent measurements into one number per subject, reducing the paired design to a familiar one-sample t-procedure applied to those differences. The option about eliminating "the need for random sampling" is wrong because random sampling or random assignment is still required for valid inference, pairing does not remove that condition. This transformation is the core reason paired designs are analyzed differently from two independent samples.

A small p-value here means that, assuming the null hypothesis \(\mu = 50\) is true, obtaining a sample mean of 47 or something even more extreme would be unlikely due to random sampling variation alone. The statement "the probability that \(\mu = 50\) is very small" misinterprets the p-value as a probability about the parameter rather than a conditional probability about the data given the null. Correct p-value interpretations must always be phrased in terms of the probability of the observed (or more extreme) sample result, assuming the null hypothesis is true.

When sample sizes and variances differ substantially, using the Welch approximation for degrees of freedom (the standard unpooled two-sample t-procedure) properly reflects the added uncertainty from the smaller, less reliable sample, giving conservative and valid results. Using "the pooled t-procedure regardless of the discrepancy" is risky because pooling assumes equal population variances, an assumption clearly violated when \(s_1 = 15\) and \(s_2 = 2\) differ so much. In practice, most software and the AP exam default to the unpooled two-sample t-test specifically to handle situations like unequal variances and sample sizes.

With a very small sample, the Central Limit Theorem does not have enough observations to guarantee an approximately normal sampling distribution, so a strongly skewed distribution with an outlier seriously threatens the normality condition and the validity of the t-procedure. The claim that "\(n \geq 5\) is always sufficient" is false, since small samples require the underlying population to be close to normal, which strong skewness and outliers contradict. When conditions are questionable with small samples, students should note the concern explicitly and consider that inference may not be reliable rather than proceeding blindly.

AP Statistics generally uses the unpooled (Welch) two-sample t-procedure, whose degrees of freedom formula is more complex and depends on both sample variances and sizes, unlike the simpler pooled formula \(n_1+n_2-2\) that assumes equal population variances. The claim that "\(n_1+n_2-2\) is always larger than the true df needed" is not a fixed rule, since the relationship between pooled and Welch df depends on the specific variances and sizes involved. Students should know that calculators typically compute the Welch df automatically, so memorizing the exact formula is less important than understanding why it differs from the pooled version.

Because the entire interval \((1.2, 5.8)\) lies above 0, all plausible values for the true mean difference favor Method A, giving convincing evidence that Method A produces higher mean scores. The claim that "we can conclude Method A causes exactly a 3.5 point improvement" is wrong because a confidence interval gives a plausible range, not a single exact causal value, and 3.5 is simply the point estimate at the interval's center, not a guaranteed effect size. When an entire confidence interval for a difference falls on one side of 0, that indicates a statistically significant difference in that direction.

A Type I error occurs when we reject a null hypothesis that is actually true, so here it means concluding \(\mu \neq 100\) when in fact \(\mu = 100\). The option "failing to reject \(H_0\) when \(\mu\) actually equals 100" describes a correct decision, not an error, since \(H_0\) is true and we correctly fail to reject it. Keep the two error types distinct: Type I is a false rejection of a true null, while Type II is a failure to reject a false null.

Since the observed test statistic magnitude \(1.9\) exceeds the critical value \(t^* \approx 1.860\) at \(\alpha = 0.10\) with \(df = 8\), the engineer should reject \(H_0\) in favor of the alternative hypothesis. The option citing "the sample size is too small" is a distraction, since \(n=9\) is sufficient as long as normality conditions are reasonably met and does not by itself prevent a valid decision. Comparing the absolute value of the observed t-statistic to the critical value is the direct way to make a rejection decision without needing to look up an exact p-value.

The correct standard error for the difference of two independent sample means is \(\sqrt{s_1^2/n_1 + s_2^2/n_2} = \sqrt{36/20 + 81/25}\), which correctly combines each group's variance divided by its own sample size. The option multiplying variance by sample size, \(\sqrt{6^2\cdot20+9^2\cdot25}\), inverts the correct operation and produces a nonsensical result unrelated to standard error. Always divide each squared standard deviation by its respective sample size before adding, never multiply.

Because the interval \((-1.5, 8.2)\) spans a wide range of plausible differences besides 0, including differences up to about 8.2 mg/dL, we cannot conclude the diets have equal effects, only that we lack convincing evidence they differ. The claim that "the interval containing 0 shows the diets are proven equal" confuses failing to detect a significant difference with proving no difference exists, a critical distinction in inference. Remember that statistical non-significance never proves equality, it only reflects insufficient evidence to reject the null of no difference.

Since \(s\) is a random variable estimated from the sample rather than a known constant like \(\sigma\), the t-distribution incorporates extra variability through its heavier tails, resulting in a critical value \(t^*\) that is always at least as large as the corresponding \(z^*\) for the same confidence level. The claim about "a larger mean than normal distributions" is false because both distributions are centered at 0; the difference lies in spread, not location. This extra width in t-intervals reflects the genuine additional uncertainty of not knowing the true population standard deviation.

A t-statistic near 0 means the sample mean \(\bar{x}\) is very close to the hypothesized value \(\mu_0\) relative to the standard error, giving essentially no evidence against \(H_0\), so we fail to reject it. The claim that such a small \(t\) value is "statistically impossible" is false, since t-statistics can take any real value, including values very near zero when sample results closely match the null hypothesis. A t-statistic close to zero is a completely normal and expected outcome whenever the null hypothesis is approximately or exactly true.

The correct interpretation of a confidence interval is that we are 95% confident the method used produces an interval, in this case \((2.48, 2.52)\), that captures the true unknown population mean bolt diameter. The statement "95% of all bolts have diameters between 2.48 and 2.52 cm" incorrectly applies the interval to individual values rather than to the population mean, confusing a confidence interval with a prediction or tolerance interval. Confidence interval interpretations must always reference capturing the true parameter, never individual data values or the observed sample mean itself.

Since the 95% confidence interval \((-1.8, 0.8)\) contains 0, a mean difference of zero is a plausible value, so we fail to reject \(H_0: \mu_d = 0\) at the corresponding significance level. The option based on "\(\bar{x}_d\) is negative" ignores that the sign of the point estimate alone does not determine statistical significance, what matters is whether the entire interval excludes the hypothesized value. The direct link between confidence intervals and two-sided hypothesis tests is a key connection to exploit on the AP exam to save calculation time.

A 99% confidence level requires capturing a larger proportion of the sampling distribution, which demands a larger critical value \(t^*\) and therefore produces a wider interval than a 90% confidence level, holding sample size and standard deviation fixed. The claim that width "depends entirely on the sample size, not the confidence level" is false because both factors affect the margin of error independently: sample size affects the standard error while confidence level affects \(t^*\). Recognizing this direct relationship between confidence level and interval width is a common conceptual point tested on the AP exam.

When measurements are actually linked (the same 30 patients under two conditions), applying a two-sample t-test ignores the positive correlation between paired observations, resulting in an inflated standard error and incorrect degrees of freedom that can distort the true p-value. The claim that "two-sample and paired tests always give identical results" is false, because pairing typically reduces variability by removing subject-to-subject differences, often making the paired test more powerful and giving a different p-value. Correctly identifying whether data are paired or independent is essential, since applying the wrong procedure can lead to invalid conclusions even when sample sizes match.

Because pretest and posttest scores for the same student are highly correlated, treating them as independent samples ignores this correlation and typically inflates the standard error, which reduces the power of the test to detect a true improvement. The claim that the two-sample test "is more powerful because it uses more degrees of freedom" is misleading, since although degrees of freedom are higher, the inflated standard error from ignoring pairing usually outweighs that benefit and lowers overall power. Correctly pairing dependent data is crucial because it isolates the within-subject effect, which typically produces a more precise and powerful test.

For a one-sided test with \(t = 2.0\) and \(df = 15\), the area to the right of \(t=2.0\) under the t-distribution is approximately \(0.032\), which represents the probability of observing a result this extreme or more so if \(H_0\) were true. The value "\(0.064\)" is actually the two-sided p-value (roughly double the one-sided value), which would be incorrect here since the alternative hypothesis \(\mu > 48\) specifies a one-sided test. Always match whether you double the tail probability based on whether the alternative hypothesis is one-sided or two-sided.

With only \(n=10\) observations, the Central Limit Theorem provides little protection against non-normality, so strong skewness combined with an outlier can distort both the sample mean and standard deviation, threatening the accuracy of the stated confidence level. The claim that t-procedures are "always robust to skewness regardless of sample size" is false since robustness improves with larger samples but is weak for very small, heavily skewed data sets. Recognizing the limits of robustness for small, skewed samples with outliers is essential for correctly evaluating whether a confidence interval's stated confidence level can be trusted.

With variances of \(4\) and \(36\), a 9-fold difference, the equal variance assumption underlying the pooled procedure is clearly violated, so the pooled t-test's standard error and df would misrepresent the true variability, making the unpooled Welch procedure more appropriate and trustworthy. The claim that both procedures "will always produce identical p-values" is false because the pooled and unpooled methods use different formulas for standard error and degrees of freedom, which typically yield different results, especially with unequal variances. Recognizing large disparities in sample variances as a red flag for avoiding the pooled procedure is an important synthesis skill for two-sample inference.

Larger sample sizes reduce the standard error \(\sqrt{s_1^2/n_1 + s_2^2/n_2}\), and since the margin of error is directly tied to this standard error, a substantially larger sample would produce a narrower interval around roughly the same point estimate. The claim that "the confidence level would automatically increase to 99%" confuses sample size with confidence level, two entirely separate concepts that a researcher sets independently. Increasing sample size is the primary tool for improving precision without changing the stated confidence level.

Setting \(ME = z^*\cdot s/\sqrt{n} \leq 2\) and solving gives \(n \geq (z^* \cdot s / ME)^2 = (1.96 \cdot 10/2)^2 = 9.8^2 \approx 96.04\), so rounding up gives approximately \(97\). The option "\(49\)" results from a common error of forgetting to square the entire expression or misplacing the division, underestimating the true sample size needed. Always round sample size calculations up to the next whole number, since a smaller sample would not guarantee the desired margin of error.

Modern statistical practice, including the standard AP Statistics approach, recommends defaulting to the unpooled (Welch) two-sample t-procedure in nearly all situations because it performs almost as well as the pooled method when variances are similar, while also protecting against invalid results when variances differ. The claim that "pooling is required whenever \(n_1 = n_2\)" is a misconception, since equal sample sizes do not guarantee equal population variances, and pooling should be based on variance similarity, not sample size alone. Choosing a procedure based on which gives "the smaller p-value" is statistically inappropriate and represents p-hacking rather than sound methodology.

A small p-value only indicates that the observed mean difference is unlikely to be due to chance, but it says nothing directly about the size or real-world importance of that difference, which requires examining \(\bar{x}_d\) and the corresponding confidence interval. The claim that "a t-statistic of 3.1 always corresponds to a large effect size" is false because the t-statistic depends heavily on sample size and standard error, so even a tiny true effect can produce a large t-statistic with enough data. Students should always separate the concepts of statistical significance and practical significance, since large sample sizes can make even trivial effects statistically significant.

A 95% confidence interval that contains the null value corresponds exactly to failing to reject \(H_0\) at the \(\alpha = 0.05\) significance level in a two-sided test, which means the p-value must indeed be greater than \(0.05\). The claim that "confidence intervals and p-values are unrelated procedures" is false, since for two-sided tests they are mathematically linked: whether the interval captures the null value directly determines the reject/fail-to-reject decision at the matching significance level. This duality between confidence intervals and two-sided hypothesis tests is a powerful shortcut for checking consistency between reported results on the AP exam.

Unlike simple df formulas like \(n-1\), the Welch-Satterthwaite approximation for the unpooled two-sample t-test is derived algebraically from the sample variances and sizes, and it commonly produces a non-integer value that software and calculators use directly for accurate p-value computation. The claim that "non-integer df only occurs when sample sizes are unequal" is false, since even with \(n_1 = n_2 = 8\) here, unequal sample variances alone are enough to produce a fractional df under the Welch method. Students should understand that fractional degrees of freedom in two-sample t-tests are mathematically expected and do not indicate an error in the analysis.

When pairs are positively correlated, computing differences removes the shared subject-to-subject variability, which typically shrinks the standard error and increases the magnitude of the t-statistic compared to incorrectly treating the data as independent samples. The claim that both approaches would give "exactly the same result" is false since the two procedures use fundamentally different standard error calculations, even though the raw data values themselves are unchanged. This scenario illustrates why choosing the correct paired or unpaired procedure can materially change statistical conclusions, not just computational details.

Since a 99% confidence level requires a larger critical value \(t^*\) than a 90% level, the resulting interval will be wider on both ends, and because 0 was already just barely outside the lower bound at 90% confidence \((-0.3, 4.1)\), the wider 99% interval is very likely to include 0 as well. The claim that "the interval width will stay exactly the same" is incorrect because interval width is a direct function of the critical value, which changes with the confidence level even when the data remain fixed. This reasoning highlights how increasing confidence level can shift the practical conclusion of a test, sometimes changing a marginally significant result into a non-significant one.

The test statistic is \(t = \bar{x}_d/(s_d/\sqrt{n}) = 4.2/(10.5/\sqrt{25}) = 4.2/2.1 = 2.00\), and since \(2.00\) exceeds the critical value \(t^*_{24} \approx 1.711\) for a one-sided test at \(\alpha = 0.05\), we reject \(H_0\). The last option incorrectly claims we should fail to reject simply because the alternative is one-sided, but a one-sided alternative does not automatically prevent rejection; it actually makes rejection easier since the entire rejection region is on one tail. Always compute the exact test statistic and compare it to the appropriate one-sided or two-sided critical value rather than assuming the test direction alone determines the decision.

As sample size (and thus degrees of freedom) increases, the t-distribution's tails become thinner and it more closely resembles the standard normal distribution, causing \(t^*\) to decrease and converge toward \(z^*\) for the same confidence level. The claim that "\(t^*\) is always exactly double \(z^*\)" is false, since this ratio depends heavily on degrees of freedom and is only roughly true for very small sample sizes, not a fixed universal relationship. Understanding this convergence explains why with large samples, t-procedures and z-procedures yield very similar critical values and interval widths.

The skeptic is correctly invoking the concept of practical significance, since even though the interval \((0.1, 1.5)\) excludes 0 and is statistically significant, the actual size of the effect may still be too small to matter in a real-world context, especially in a large sample of 50 pairs. The claim that "95% confidence always implies practical importance" conflates the confidence level, which reflects the reliability of the interval-generating method, with the substantive size or importance of the effect being estimated. Distinguishing statistical significance from practical significance is a critical high-level skill for interpreting inference results correctly on the AP exam.

With only 15 pairs, the small sample size provides limited protection from the Central Limit Theorem, so a strongly skewed distribution of differences with an outlier could meaningfully distort \(\bar{x}_d\) and threaten the validity of the paired t-procedure, requiring careful assessment before proceeding. The claim that paired t-tests "are always robust to any distribution shape" is false, since robustness improves with larger sample sizes and is weak in small, heavily skewed samples, especially with outliers present. Removing an outlier without statistical or contextual justification is inappropriate practice and can bias results, so any exclusion must be carefully justified rather than done simply to achieve normality.