v1 · padrão canônico

Lesson 110 — Trim 11 Consolidation: Statistical Inference

Trim 11 synthesis workshop: confidence intervals for the mean, z and t tests, ANOVA, chi-square, simple and multiple regression, and Bayesian inference — all the pillars of inferential statistics in one integrated map.

Used in: 3.º ano do EM / Stochastik LK alemão · Math B japonês (Estatística) · H2 Mathematics (Singapura) — Statistics

\hat\theta \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}};\quad T = \frac{\bar X - \mu_0}{s/\sqrt{n}};\quad P(\theta \mid D) \propto P(D \mid \theta)\,P(\theta)

Trim 11 in three equations: confidence interval to estimate, $T$ statistic to test, and Bayes' rule to update beliefs. All statistical inference is a variation of these three forms.

Choose your door

Rigorous notation, full derivation, hypotheses

Formal synthesis of trim 11

The three pillars of statistical inference

Definition· Pillar 1 — Estimation by confidence interval

Let $X_1, \ldots, X_n \overset{iid}{\sim} N(\mu, \sigma^2)$ .

Two-sided CI of level $1-\alpha$ for $\mu$ ( $\sigma$ known):

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

what this means · Confidence interval for the mean with known population standard deviation.

CI with $\sigma$ unknown — uses Student's t distribution with $n-1$ degrees of freedom:

\bar X \pm t_{\alpha/2;\,n-1}\frac{s}{\sqrt{n}}

what this means · Confidence interval for the mean with sample standard deviation (most common practical application).

Sample size for maximum margin of error $E$ :

n \geq \left(\frac{z_{\alpha/2}\,\sigma}{E}\right)^2

what this means · Sample size formula: z squared times sigma, divided by E squared.

"A confidence interval provides a range of plausible values for the population parameter. The correct interpretation: if we repeat the procedure many times, $(1-\alpha)100\%$ of the constructed intervals will contain the true parameter." — OpenIntro Statistics §5.2

Definition· Pillar 2 — Hypothesis testing (unified view)

Every hypothesis test follows the same scheme:

State $H_0$ and $H_1$ .
Compute the test statistic under $H_0$ .
Compute the p-value = $P(\text{result as or more extreme than observed} \mid H_0)$ .
If $p \leq \alpha$ , reject $H_0$ .

Map of tests:

Scenario	Statistic	Distribution
1 mean, $\sigma$ known	$Z = (\bar X - \mu_0)/(\sigma/\sqrt{n})$	$N(0,1)$
1 mean, $\sigma$ unknown	$T = (\bar X - \mu_0)/(s/\sqrt{n})$	$t_{n-1}$
2 independent means	Welch $T$	$t_{\nu}$ (Welch degrees of freedom)
$k$ means	$F = MS_B/MS_W$	$F_{k-1,\,N-k}$
Contingency table	$\chi^2 = \sum (O-E)^2/E$	$\chi^2_{(r-1)(c-1)}$
Proportion vs reference value	$Z = (\hat p - p_0)/\sqrt{p_0(1-p_0)/n}$	$N(0,1)$

Definition· Pillar 3 — Regression and predictive modeling

Simple linear regression $\hat Y = \hat\beta_0 + \hat\beta_1 X$ :

\hat\beta_1 = \frac{S_{xy}}{S_{xx}},\quad \hat\beta_0 = \bar Y - \hat\beta_1 \bar X

what this means · Ordinary least squares estimators for simple regression.

Multiple regression $\hat{\mathbf{y}} = X\hat{\boldsymbol\beta}$ :

\hat{\boldsymbol\beta} = (X^\top X)^{-1} X^\top \mathbf{y}

what this means · Vector least squares estimator for multiple regression.

Coefficient of determination and its adjusted version:

R^2_{\text{adj}} = 1 - \frac{(1-R^2)(n-1)}{n-p-1}

what this means · Adjusted R-squared penalizes extra predictors that don't contribute to fit.

Theorem· Bayes' rule for updating beliefs

Let $\theta$ be the parameter of interest, $D$ the observed data:

P(\theta \mid D) = \frac{P(D \mid \theta)\,P(\theta)}{P(D)} \propto P(D \mid \theta)\,P(\theta)

what this means · The posterior distribution is proportional to the likelihood times the prior.

Beta-Binomial conjugate: if $\theta \sim \text{Beta}(\alpha, \beta)$ and $X \mid \theta \sim \text{Bin}(n, \theta)$ , then:

\theta \mid X \sim \text{Beta}(\alpha + X,\; \beta + n - X)

what this means · Posterior distribution when the prior is Beta and the likelihood is binomial — the prior is updated by adding successes and failures.

Trim 11 decision flow — each statistical question has its method.

Worked examples

Example— 1· 95% CI for average ENEM score (unknown standard deviation)

Problem. A sample of $n = 36$ ENEM 2023 candidates in Mathematics had mean $\bar X = 580$ points and sample standard deviation $s = 90$ points. Construct a 95% CI for the population mean $\mu$ .

Strategy. Since $\sigma$ is unknown, use the $t$ distribution with $n - 1 = 35$ degrees of freedom. Look up $t_{0.025;\,35}$ in a table.

Solution.

$t_{0.025;\,35} = 2.030$ .

Margin of error: $E = 2.030 \times \frac{90}{\sqrt{36}} = 2.030 \times 15 = 30.45$ .

95% CI: $(580 - 30.45\;;\; 580 + 30.45) = (549.55\;;\; 610.45)$ .

Check. The margin grows with $s$ and shrinks with $\sqrt{n}$ . For $n = 144$ , the margin would be $2.030 \times 90/12 = 15.2$ — half, because we quadrupled $n$ .

Source. OpenIntro Statistics §5.2 — Exercise 5.5, p. 222 · CC-BY-SA.

Example— 2· One-sample t test vs government target

Problem. The government claims the average wait time at health clinics is $\mu_0 = 20$ minutes. An audit samples $n = 25$ visits: $\bar X = 24.8$ min, $s = 7.2$ min. At the $\alpha = 5\%$ level, is there evidence the true mean exceeds 20 minutes?

Strategy. Right-tailed $t$ test ( $H_1: \mu > 20$ ), with $df = 24$ .

Solution.

$T = \frac{24.8 - 20}{7.2/\sqrt{25}} = \frac{4.8}{1.44} = 3.33$ .

$t_{0.05;\,24} = 1.711$ (one-tailed). Since $3.33 > 1.711$ , reject $H_0$ .

p-value: $P(t_{24} \geq 3.33) \approx 0.001 < 0.05$ . There is evidence the average wait time exceeds 20 minutes.

Check. 95% one-sided lower CI: $24.8 - 1.711 \times 1.44 = 22.33$ min. The value $\mu_0 = 20$ is below this limit, confirming rejection.

Source. Statistics, OpenStax §9.4 — adapted from example 9.10, p. 476 · CC-BY.

Example— 3· One-way ANOVA — productivity with 3 fertilizers

Problem. Three fertilizers were tested on groups of $n_i = 10$ plants each. Productivity (kg): $\bar X_1 = 12.0$ , $\bar X_2 = 14.5$ , $\bar X_3 = 13.2$ . Overall mean $\bar X = 13.23$ . $SS_W = 162$ (within-group sum of squares). At the 5% level, is there a difference between fertilizers?

Strategy. ANOVA: compute $SS_B$ , obtain $F$ , compare with $F_{0.05;\,2,\,27}$ .

Solution.

$SS_B = 10[(12.0 - 13.23)^2 + (14.5 - 13.23)^2 + (13.2 - 13.23)^2]$ $= 10[1.513 + 1.613 + 0.001] = 31.27$ .

$MS_B = 31.27/2 = 15.64$ ; $MS_W = 162/27 = 6.00$ .

$F = 15.64 / 6.00 = 2.61$ .

$F_{0.05;\,2,\,27} \approx 3.35$ . Since $2.61 < 3.35$ , do not reject $H_0$ — no evidence of difference among fertilizers at the 5% level.

Check. p-value $\approx 0.092 > 0.05$ . Result consistent. With $n_i = 20$ per group, power would be higher and rejection might occur.

Source. OpenIntro Statistics §7.5 — structure based on ANOVA example, p. 319 · CC-BY-SA.

Example— 4· Simple linear regression — salary vs years of experience

Problem. With $n = 20$ professionals: $\bar X = 8$ years, average salary US$ 6,200 ( $\bar Y = 6200$ ), $S_{xx} = 180$ , $S_{xy} = 54000$ . Fit the regression model and interpret the coefficients.

Strategy. Compute $\hat\beta_1$ and $\hat\beta_0$ using least squares formulas; interpret in context.

Solution.

$\hat\beta_1 = S_{xy}/S_{xx} = 54000/180 = 300$ .

$\hat\beta_0 = \bar Y - \hat\beta_1 \bar X = 6200 - 300 \times 8 = 3800$ .

Model: $\hat Y = 3800 + 300\,X$ (in US dollars, $X$ in years).

Interpretation. Each additional year of experience is associated with an average increase of US$ 300 in salary. The intercept US$ 3,800 is the estimated salary for 0 years of experience (extrapolation — use with caution).

Check. For $X = 10$ years: $\hat Y = 3800 + 3000 = 6800$ dollars. Plausible given $\bar Y = 6200$ for $\bar X = 8$ .

Source. OpenIntro Statistics §8.2 — based on Exercise 8.13 (salary data), p. 360 · CC-BY-SA.

Example— 5· Bayesian update — proportion of defects in production line

Problem. A quality engineer believes a priori that the proportion of defects $\theta$ in a line has distribution $\text{Beta}(3, 17)$ (prior mean $= 3/20 = 15\%$ ). He inspects $n = 40$ parts and finds $X = 4$ defects. What is the posterior distribution and what is the new point estimate for $\theta$ ?

Strategy. Use Beta-Binomial conjugate: posterior $= \text{Beta}(\alpha + X, \beta + n - X)$ .

Solution.

Posterior: $\text{Beta}(3 + 4,\; 17 + 40 - 4) = \text{Beta}(7, 53)$ .

Posterior mean: $7/(7+53) = 7/60 \approx 11.7\%$ .

95% credible interval: 2.5th and 97.5th quantiles of $\text{Beta}(7, 53)$ , approximately $[0.049;\; 0.221]$ .

Check. Before: prior mean 15%. After seeing 4 defects in 40 (MLE = 10%), the posterior shifts to 11.7% — a weighted average of prior and data. With larger samples, the posterior would move ever closer to the MLE.

Source. Statistical Thinking for the 21st Century, Ch. 15 — structure of Beta-Binomial example, §15.3 · CC-BY-NC.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 20Understanding 6Modeling 9Challenge 4Proof 1

Ex. 110.1Application
$n = 64$ , $\bar x = 50$ , $\sigma = 8$ (known). Construct the 95% CI for $\mu$ .
Select the correct option
95% CI: $(48.04\;; 51.96)$ 95% CI: $(46.08\;; 53.92)$ 95% CI: $(49.02\;; 50.98)$ 95% CI: $(47.44\;; 52.56)$
Select an option first
Solve online OpenIntro Statistics 4th ed. · §5.1
Show solution
Standard error: $\sigma/\sqrt{n} = 8/8 = 1$ . Margin: $1.96 \times 1 = 1.96$ . CI: $(50 - 1.96\;; 50 + 1.96) = (48.04\;; 51.96)$ .
Show step-by-step (with the why)
1. Identify: $n=64$ , $\bar x = 50$ , $\sigma = 8$ (known). Why: $\sigma$ known means use $z$ , not $t$ .
2. Critical value: $z_{0.025} = 1.96$ .
3. Standard error: $\sigma/\sqrt{64} = 8/8 = 1$ .
4. Margin: $E = 1.96 \times 1 = 1.96$ .
5. CI: $(48.04\;; 51.96)$ .
Trick: with $n = 64$ , $\sqrt{n} = 8$ — quick mental math.
Ex. 110.2Application
Using the data from the previous exercise ( $\bar x = 50$ , $s = 8$ , $n = 64$ ), test $H_0: \mu = 48$ vs $H_1: \mu \neq 48$ at the 5% level. Compute $T$ and decide.
Select the correct option
$T = 2.00$ ; reject $H_0$ at 5% level $T = 2.00$ ; do not reject $H_0$ at 5% level $T = 1.00$ ; reject $H_0$ at 5% level $T = 2.00$ ; reject $H_0$ only at 1% level
Select an option first
Solve online OpenIntro Statistics 4th ed. · §5.3
Show solution
$T = (50-48)/(8/8) = 2/1 = 2.00$ . With $df = 63$ , two-sided p-value $\approx 0.046 < 0.05$ . Reject $H_0$ at 5% level.
Ex. 110.3ApplicationAnswer key
Two independent samples: $\bar x_1 = 100$ , $s_1 = 10$ , $n_1 = 50$ ; $\bar x_2 = 95$ , $s_2 = 12$ , $n_2 = 60$ . Perform the Welch $t$ test at the 5% level.
Select the correct option
$T \approx 2.38$ ; reject $H_0$ at 5% level $T \approx 2.38$ ; do not reject $H_0$ at 5% level $T \approx 1.67$ ; do not reject $H_0$ at 5% level $T \approx 3.12$ ; reject $H_0$ at 1% level
Select an option first
Solve online Statistics, OpenStax · §10.1
Show solution
Welch: $T = (100-95)/\sqrt{10^2/50 + 12^2/60} = 5/\sqrt{2 + 2.4} = 5/\sqrt{4.4} \approx 2.38$ . Two-sided p-value $\approx 0.019 < 0.05$ . Reject $H_0$ .
Ex. 110.4Application
Three groups with $n_i = 20$ plants each, means $\bar X_1 = 10$ , $\bar X_2 = 12$ , $\bar X_3 = 14$ . Assuming $\sigma^2 = 4$ (within-group variance), compute $F$ from ANOVA and decide at the 5% level.
Select the correct option
$F = 20$ ; reject $H_0$ $F = 20$ ; do not reject $H_0$ $F = 10$ ; reject $H_0$ $F = 4$ ; do not reject $H_0$
Select an option first
Solve online OpenIntro Statistics 4th ed. · §7.5
Show solution
$SS_B = 20[(10-12)^2+(12-12)^2+(14-12)^2] = 20(4+0+4) = 160$ . $MS_B = 80$ . $F = 80/4 = 20$ . Critical value $F_{0.05;\,2,\,57} \approx 3.16$ . Since $20 \gg 3.16$ , reject $H_0$ .
Show step-by-step (with the why)
1. Overall mean: $\bar X = (10+12+14)/3 = 12$ .
2. $SS_B = 20\sum(\bar X_i - \bar X)^2 = 20(4+0+4) = 160$ .
3. $MS_B = SS_B/(k-1) = 160/2 = 80$ .
4. $F = MS_B/MS_W = 80/4 = 20$ . Strongly reject since $20 \gg 3.16$ .
Note: With $F = 20$ and critical value $\approx 3.16$ , the p-value is tiny.
Ex. 110.5ApplicationAnswer key
Fitted model: $\hat Y = 5 + 2X$ . What is the point prediction for $X = 10$ ?
Select the correct option
$\hat Y = 25$ $\hat Y = 20$ $\hat Y = 15$ $\hat Y = 30$
Select an option first
Solve online OpenIntro Statistics 4th ed. · §8.1
Show solution
$\hat Y = 5 + 2 \times 10 = 5 + 20 = 25$ .
Ex. 110.6Application
$R^2 = 0.8$ , $n = 50$ , $p = 4$ predictors. Compute $R^2_{\text{adj}}$ .
Select the correct option
$R^2_{\text{adj}} \approx 0.782$ $R^2_{\text{adj}} \approx 0.800$ $R^2_{\text{adj}} \approx 0.756$ $R^2_{\text{adj}} \approx 0.810$
Select an option first
Solve online Statistics, OpenStax · §12.4
Show solution
$R^2_{\text{adj}} = 1 - \frac{(1 - 0.8)(50-1)}{50-4-1} = 1 - \frac{0.2 \times 49}{45} = 1 - 0.2178 \approx 0.782$ .
Ex. 110.7Application
A $2 \times 3$ contingency table produced $\chi^2 = 8$ . How many degrees of freedom? Is there association at the 5% level?
Select the correct option
$df = 2$ ; there is association at 5% level $df = 5$ ; no association at 5% level $df = 2$ ; no association at 5% level $df = 4$ ; there is association at 5% level
Select an option first
Solve online OpenIntro Statistics 4th ed. · §6.4
Show solution
Table $2 \times 3$ : $df = (2-1)(3-1) = 2$ . Critical value $\chi^2_{0.05;\,2} = 5.991$ . Since $\chi^2 = 8 > 5.991$ , reject independence hypothesis.
Ex. 110.8Application
Prior $\theta \sim \text{Beta}(2, 2)$ . Observe $X = 8$ successes in $n = 10$ trials. What is the posterior distribution of $\theta$ ?
Select the correct option
$\text{Beta}(10, 4)$ $\text{Beta}(8, 2)$ $\text{Beta}(10, 6)$ $\text{Beta}(2, 4)$
Select an option first
Solve online Statistical Thinking for the 21st Century · Ch. 15
Show solution
Prior $\text{Beta}(2, 2)$ , $X = 8$ successes in $n = 10$ . Posterior: $\text{Beta}(2+8,\; 2+10-8) = \text{Beta}(10, 4)$ .
Ex. 110.9ApplicationAnswer key
With the posterior $\text{Beta}(10, 4)$ from the previous exercise, compute the posterior mean and the MAP (mode of the posterior distribution).
Select the correct option
Mean $\approx 0.714$ ; MAP $= 0.75$ Mean $= 0.75$ ; MAP $\approx 0.714$ Mean $= 0.80$ ; MAP $= 0.80$ Mean $\approx 0.667$ ; MAP $= 0.75$
Select an option first
Solve online Statistical Thinking for the 21st Century · Ch. 15
Show solution
Posterior $\text{Beta}(10, 4)$ : mean $= 10/14 \approx 0.714$ . MAP (mode) $= (10-1)/(14-2) = 9/12 = 0.75$ .
Ex. 110.10Application
Look up in the standard normal distribution table: what is $z_{0.025}$ ?
Select the correct option
$z_{0.025} = 1.960$ $z_{0.025} = 1.645$ $z_{0.025} = 2.326$ $z_{0.025} = 1.282$
Select an option first
Solve online Statistics, OpenStax · Appendix — z table
Show solution
$z_{0.025} = 1.960$ . This is the critical value for two-sided 95% CI. Don't confuse with $z_{0.05} = 1.645$ (90% CI).
Ex. 110.11Application
Using the same data $n = 64$ , $\bar x = 50$ , $\sigma = 8$ , now construct the 90% CI for $\mu$ . Is the interval wider or narrower than the 95% CI?
Select the correct option
90% CI: $(48.36\;; 51.64)$ 90% CI: $(46.08\;; 53.92)$ 90% CI: $(48.04\;; 51.96)$ 90% CI: $(45.00\;; 55.00)$
Select an option first
Solve online OpenIntro Statistics 4th ed. · §5.2
Show solution
For 90% CI, $z_{0.05} = 1.645$ . Standard error: $\sigma/\sqrt{n} = 8/8 = 1$ . Margin: $1.645 \times 1 = 1.645$ . CI: $(50 - 1.645\;; 50 + 1.645) = (48.355\;; 51.645) \approx (48.36\;; 51.64)$ . The 90% CI is narrower than the 95% CI (which had margin 1.96).
Ex. 110.12Application
You want a 95% CI with maximum margin of error $E = 2$ units, knowing $\sigma = 10$ . What is the minimum sample size?
Select the correct option
$n \geq 97$ $n \geq 62$ $n \geq 245$ $n \geq 49$
Select an option first
Solve online Statistics, OpenStax · §8.1
Show solution
Formula: $n \geq (z_{0.025}\,\sigma/E)^2 = (1.96 \times 10/2)^2 = (9.8)^2 = 96.04$ . Round up: $n \geq 97$ .
Show step-by-step (with the why)
1. Identify: $\sigma = 10$ , $E = 2$ , 95% confidence implies $z_{0.025} = 1.96$ .
2. Apply: $n \geq (1.96 \times 10/2)^2 = (9.8)^2 = 96.04$ .
3. Round up (conservative): $n = 97$ .
General rule: to cut the margin in half, quadruple $n$ .
Ex. 110.13Application
A sample of $n = 25$ observations has $\bar x = 25$ and $s = 10$ . Test $H_0: \mu = 20$ vs $H_1: \mu \neq 20$ at the 5% level.
Select the correct option
$T = 2.50$ ; reject $H_0$ at 5% level $T = 2.50$ ; do not reject $H_0$ at 5% level $T = 1.25$ ; do not reject $H_0$ at 5% level $T = 5.00$ ; reject $H_0$ at 1% level
Select an option first
Solve online OpenIntro Statistics 4th ed. · §5.3
Show solution
$T = (\bar x - \mu_0)/(s/\sqrt{n}) = (25 - 20)/(10/\sqrt{25}) = 5/2 = 2.50$ . With $df = 24$ , two-sided critical value $t_{0.025;\,24} = 2.064$ . Since $2.50 > 2.064$ , reject $H_0$ .
Ex. 110.14Application
In a sample of $n = 100$ voters, 60 approve a certain measure. At the 5% level, is there evidence that the true proportion $p \neq 0.50$ ? Use the z test for proportion.
Select the correct option
$Z = 2.00$ ; reject $H_0$ $Z = 1.50$ ; do not reject $H_0$ $Z = 2.00$ ; do not reject $H_0$ $Z = 3.00$ ; reject $H_0$
Select an option first
Solve online Statistics, OpenStax · §9.2
Show solution
$\hat p = 60/100 = 0.60$ . $Z = (0.60 - 0.50)/\sqrt{0.50 \times 0.50/100} = 0.10/\sqrt{0.0025} = 0.10/0.05 = 2.00$ . Two-sided p-value $\approx 0.046 < 0.05$ . Reject $H_0$ .
Ex. 110.15Application
$n = 10$ observations with $\bar X = 4$ , $\bar Y = 10$ , $S_{xx} = 20$ , $S_{xy} = 50$ . Compute $\hat\beta_1$ and interpret.
Select the correct option
$\hat\beta_1 = 2.5$ ; each unit of $X$ increases $Y$ by 2.5 on average $\hat\beta_1 = 0.4$ ; each unit of $X$ increases $Y$ by 0.4 on average $\hat\beta_1 = 2.5$ ; $X$ and $Y$ are negatively correlated $\hat\beta_1 = 5.0$ ; each unit of $X$ increases $Y$ by 5.0 on average
Select an option first
Solve online OpenIntro Statistics 4th ed. · §8.2
Show solution
$\hat\beta_1 = S_{xy}/S_{xx} = 50/20 = 2.5$ . $\hat\beta_0 = \bar Y - 2.5 \bar X = 10 - 2.5 \times 4 = 0$ . Model: $\hat Y = 2.5\,X$ . Each unit increase in $X$ increases $Y$ by 2.5 on average.
Ex. 110.16ApplicationAnswer key
A $2 \times 2$ contingency table produced $\chi^2 = 3.84$ . How many degrees of freedom? Is there association at the 5% level?
Select the correct option
$\chi^2 = 3.84$ ; $df = 1$ ; do not reject $H_0$ at 5% level $\chi^2 = 7.68$ ; $df = 1$ ; reject $H_0$ at 5% level $\chi^2 = 3.84$ ; $df = 3$ ; do not reject $H_0$ at 5% level $\chi^2 = 1.92$ ; $df = 1$ ; do not reject $H_0$ at 5% level
Select an option first
Solve online Statistics, OpenStax · §11.2
Show solution
Table $2 \times 2$ : observed frequencies $O = [30, 20, 20, 30]$ , total 100. Expected frequencies under independence: row 1 has 50 cases, column 1 has 50 cases; each expected cell $E_{11} = 50 \times 50/100 = 25$ . $\chi^2 = 4 \times (30-25)^2/25 = 4$ . But with the problem frequencies: $\chi^2 \approx 3.84$ . $df = (2-1)(2-1) = 1$ . Critical value $\chi^2_{0.05;\,1} = 3.841$ . Since $3.84 \approx 3.841$ , do not reject at the exact threshold.
Ex. 110.17Application
ANOVA with $k = 3$ groups, $N = 27$ total. $SS_B = 90$ , $SS_W = 240$ . Compute $F$ and decide at the 5% level.
Select the correct option
$F = 4.50$ ; reject $H_0$ at 5% level $F = 4.50$ ; do not reject $H_0$ at 5% level $F = 2.25$ ; do not reject $H_0$ at 5% level $F = 9.00$ ; reject $H_0$ at 1% level
Select an option first
Solve online Statistics, OpenStax · §13.1
Show solution
$MS_B = SS_B/(k-1) = 90/2 = 45$ . $MS_W = SS_W/(N-k) = 240/24 = 10$ . $F = 45/10 = 4.50$ . Critical value $F_{0.05;\,2,\,24} \approx 3.40$ . Since $4.50 > 3.40$ , reject $H_0$ .
Ex. 110.18Application
$S_{xx} = 25$ , $S_{yy} = 80$ , $S_{xy} = 40$ . Compute the Pearson correlation coefficient $r$ .
Select the correct option
$r \approx 0.894$ $r \approx 0.750$ $r \approx 0.500$ $r \approx 0.950$
Select an option first
Solve online OpenIntro Statistics 4th ed. · §8.3
Show solution
$r = S_{xy}/\sqrt{S_{xx}\,S_{yy}} = 40/\sqrt{25 \times 80} = 40/\sqrt{2000} = 40/44.72 \approx 0.894$ .
Ex. 110.19Application
Two independent groups: $\bar x_1 = 30$ , $s_1 = 4$ , $n_1 = 16$ ; $\bar x_2 = 20$ , $s_2 = 5$ , $n_2 = 25$ . Construct the 95% CI for $\mu_1 - \mu_2$ using Welch's statistic.
Select the correct option
95% CI: $(\bar x_1 - \bar x_2) \pm t \cdot SE \approx (7.1\;; 12.9)$ 95% CI: $(\bar x_1 - \bar x_2) \pm z \cdot SE \approx (4.0\;; 16.0)$ 95% CI: $(3.6\;; 16.4)$ 95% CI: $(2.1\;; 17.9)$
Select an option first
Solve online Statistics, OpenStax · §8.3
Show solution
Observed difference: $\bar x_1 - \bar x_2 = 30 - 20 = 10$ . Standard error of difference: $SE = \sqrt{s_1^2/n_1 + s_2^2/n_2} = \sqrt{16/16 + 25/25} = \sqrt{1 + 1} = \sqrt{2} \approx 1.414$ . Welch degrees of freedom: $df \approx 37$ , $t_{0.025;\,37} \approx 2.026$ . Margin: $2.026 \times 1.414 \approx 2.865$ . CI: $(10 - 2.865\;; 10 + 2.865) \approx (7.1\;; 12.9)$ .
Ex. 110.20Application
Define Type I and Type II error in a hypothesis test. What is the relationship between them and the parameters $\alpha$ and $\beta$ ?
Select the correct option
Type I error: reject true $H_0$ ; Type II error: fail to reject false $H_0$ Type I error: fail to reject false $H_0$ ; Type II error: reject true $H_0$ Type I and Type II errors are the same concept with different namesType I error: $\alpha$ ; Type II error: $1 - \beta$
Select an option first
Solve online Statistics, OpenStax · §9.5
Show solution
Type I error occurs when $H_0$ is true but we reject it (false positive); probability $= \alpha$ . Type II error occurs when $H_0$ is false but we fail to reject it (false negative); probability $= \beta$ . Power of test $= 1 - \beta$ .
Ex. 110.21Understanding
What is the correct interpretation of a 95% CI in the frequentist perspective?
Select the correct option
If we repeat the procedure many times, 95% of the CIs constructed will contain $\mu$ There is a 95% chance $\mu$ is in this specific interval $\mu$ is certainly in the interval with probability 95%The 95% CI is always wider than the 99% CI
Select an option first
Solve online OpenIntro Statistics 4th ed. · §5.2
Show solution
Correct frequentist interpretation: the procedure produces intervals that cover the true $\mu$ in 95% of repetitions. The interval constructed from one specific sample either contains or does not contain $\mu$ — there is no probability associated with a fixed interval.
Ex. 110.22Understanding
What does the p-value of a hypothesis test measure? Choose the correct interpretation.
Select the correct option
The p-value is the probability of obtaining a result as or more extreme given that $H_0$ is trueThe p-value is the probability that $H_0$ is trueThe p-value is the probability that $H_1$ is trueThe p-value is the probability of making a Type I error
Select an option first
Solve online OpenIntro Statistics 4th ed. · §5.3
Show solution
The p-value is NOT the probability that $H_0$ is true. It is $P(\text{data as or more extreme} \mid H_0)$ . Common errors: confusing p-value with the probability of $H_0$ or $H_1$ — both are serious misinterpretations.
Ex. 110.23UnderstandingAnswer key
Explain what $R^2$ measures in linear regression and why high correlation does not imply causation.
Select the correct option
$R^2$ measures the proportion of variability in $Y$ explained by the regression; correlation does not imply causation $R^2 = 1$ implies the model is causal and perfectStrong correlation implies $X$ causes $Y$ $R^2$ high guarantees the model generalizes well to new data
Select an option first
Solve online OpenIntro Statistics 4th ed. · §8.4
Show solution
$R^2$ measures fit on the training set — not causation or generalization. Correlation $\neq$ causation: classic example, ice cream and drowning are correlated (both rise in summer), but one doesn't cause the other. High $R^2$ with small samples can indicate overfitting.
Ex. 110.24Understanding
What is the fundamental difference between the Bayesian and frequentist interpretation of the parameter $\theta$ ?
Select the correct option
Bayesian: $\theta$ has a probability distribution reflecting uncertainty; frequentist: $\theta$ is an unknown constantBayesian and frequentist always produce the same results with large samplesFrequentist: allows direct statements like ' $P(\theta > 0.5) = 0.8$ '; Bayesian: does notBayesian uses the classical CI; frequentist uses the credible interval
Select an option first
Solve online Statistical Thinking for the 21st Century · Ch. 15
Show solution
Fundamental difference: frequentists treat $\theta$ as an unknown constant (without distribution); Bayesians assign a probability distribution to $\theta$ . The Bayesian credible interval allows one to say "with probability 95%, $\theta \in [a,b]$ "; the frequentist CI does not.
Ex. 110.25Understanding
Why use ANOVA instead of multiple $t$ tests to compare $k > 2$ groups?
Select the correct option
ANOVA tests whether all means are equal simultaneously, avoiding Type I error inflationANOVA is equivalent to performing $\binom{k}{2}$ separate $t$ testsANOVA can be used with categorical response variablesANOVA rejecting $H_0$ identifies which pairs of means differ
Select an option first
Solve online OpenIntro Statistics 4th ed. · §7.5
Show solution
With $k$ groups, performing $\binom{k}{2}$ separate $t$ tests at level $\alpha$ inflates familywise Type I error. ANOVA tests $H_0: \mu_1 = \cdots = \mu_k$ in one test at level $\alpha$ . To identify which pairs differ, use post-hoc (Tukey, Bonferroni) after rejecting $H_0$ .
Ex. 110.26Understanding
When do you use the chi-square independence test and when do you use ANOVA? Identify the application conditions for each.
Select the correct option
Chi-square tests independence between categorical variables; ANOVA compares means of numeric variablesChi-square and ANOVA test the same type of hypothesisChi-square can be used for continuous numeric variablesANOVA applies when the response variable is categorical
Select an option first
Solve online Statistics, OpenStax · §11.1
Show solution
Chi-square independence: both variables are categorical (nominal or ordinal). ANOVA: one numeric response variable and one categorical factor with $k \geq 2$ levels. Confusing the two is a serious test-selection error.
Ex. 110.27Modeling
An election survey interviewed $n = 500$ voters; 250 support candidate A. Construct a 95% CI for the true proportion $p$ of supporters.
Select the correct option
95% CI: $(\hat p \pm 1.96\,SE) \approx (0.456\;; 0.544)$ ; with 95% confidence, the true proportion is in this range95% CI: $(0.50 \pm 0.010) = (0.490\;; 0.510)$ 95% CI: $(0.400\;; 0.600)$ 95% CI: $(0.443\;; 0.557)$
Select an option first
Solve online OpenIntro Statistics 4th ed. · §5.4
Show solution
$\hat p = 250/500 = 0.500$ . Standard error: $SE = \sqrt{\hat p(1-\hat p)/n} = \sqrt{0.25/500} = \sqrt{0.0005} \approx 0.02236$ . Margin: $1.96 \times 0.02236 \approx 0.0438$ . CI: $(0.500 - 0.044\;; 0.500 + 0.044) \approx (0.456\;; 0.544)$ .
Show step-by-step (with the why)
1. Compute $\hat p = 250/500 = 0.500$ .
2. Compute the standard error: $SE = \sqrt{0.5 \times 0.5 / 500} \approx 0.02236$ .
3. Margin: $1.96 \times 0.02236 \approx 0.044$ .
4. CI: $(0.456\;; 0.544)$ .
Note that the CI for a proportion uses the standard error based on $\hat p$ , not on population $\sigma$ .
Ex. 110.28Modeling
Regression model for energy consumption (kWh/month) as a function of average monthly temperature (°C): $\hat Y = 120 + 4.5X$ , fitted for temperatures between 20 and 60°C. Predict consumption for $X = 40$ °C and for $X = 70$ °C. Which prediction involves extrapolation?
Select the correct option
$\hat Y = 120 + 4.5 \times 40 = 300$ ; there is extrapolation since $X = 70$ is outside the sample domain $\hat Y = 120 + 4.5 \times 40 = 300$ ; the prediction is reliable at any point $\hat Y = 120 + 4.5 \times 70 = 435$ ; the prediction is reliable $\hat Y = 120 + 4.5 \times 70 = 435$ ; there is extrapolation — use with caution
Select an option first
Solve online OpenIntro Statistics 4th ed. · §8.2
Show solution
For $X = 40$ : $\hat Y = 120 + 4.5 \times 40 = 120 + 180 = 300$ (interpolation — safe). For $X = 70$ : $\hat Y = 120 + 4.5 \times 70 = 435$ (extrapolation — outside the sample interval 20 to 60 degrees, use with caution).
Ex. 110.29ModelingAnswer key
A 4-sided die was rolled 100 times. Observed frequencies: face 1 = 20, face 2 = 30, face 3 = 28, face 4 = 22. At the 5% level, is the die fair? Use the chi-square goodness-of-fit test.
Select the correct option
$\chi^2 = 2.72$ ; $df = 3$ ; do not reject $H_0$ at 5% level ( $\chi^2_{crit} = 7.815$ ) $\chi^2 = 2.72$ ; $df = 3$ ; reject $H_0$ at 5% level $\chi^2 \approx 10.0$ ; $df = 3$ ; reject $H_0$ at 5% level $\chi^2 = 2.72$ ; $df = 4$ ; do not reject $H_0$ at 5% level
Select an option first
Solve online Statistics, OpenStax · §11.3
Show solution
Expected frequencies: each category should have $100/4 = 25$ observations. $\chi^2 = (20-25)^2/25 + (30-25)^2/25 + (28-25)^2/25 + (22-25)^2/25 = 1 + 1 + 0.36 + 0.36 = 2.72$ . Degrees of freedom: $df = 4 - 1 = 3$ . Critical value: $\chi^2_{0.05;\,3} = 7.815$ . Since $2.72 < 7.815$ , do not reject $H_0$ — no evidence the die is unfair.
Ex. 110.30ModelingAnswer key
A clinical trial measured healing time (in days) of 25 patients before and after a new ointment formula. Means: before $\bar x = 25$ days, after $\bar x = 20$ days, standard deviation of differences $s_d = 9.14$ days. At the 5% level, did the new formula reduce healing time (one-tailed test)?
Select the correct option
$T \approx 2.74$ ; reject $H_0$ ; the new formulation reduces time on average $T \approx 2.74$ ; do not reject $H_0$ ; no evidence of reduction $T \approx 1.37$ ; do not reject $H_0$ $T \approx 4.11$ ; reject $H_0$ at 1% level
Select an option first
Solve online OpenIntro Statistics 4th ed. · §7.3
Show solution
Paired differences: $\bar d = 25 - 20 = 5$ days, $s_d = 9.14$ days (from problem), $n = 25$ . $T = \bar d/(s_d/\sqrt{n}) = 5/(9.14/5) = 5/1.828 \approx 2.74$ . $t_{0.05;\,24} = 1.711$ (one-tailed). Since $2.74 > 1.711$ , reject $H_0$ — the new formulation is faster on average.
Show step-by-step (with the why)
1. Paired test: each patient has two times (before and after). Compute differences $d_i = \text{before} - \text{after}$ .
2. $\bar d = 5$ days; standard deviation of differences $s_d = 9.14$ days.
3. $T = 5/(9.14/\sqrt{25}) = 5/1.828 \approx 2.74$ .
4. Reject $H_0$ since $T > t_{0.05;\,24} = 1.711$ .
Paired test is more powerful than independent two-sample test when there is correlation within pairs (same patient at both times).
Ex. 110.31Modeling
Fitted regression model: $\hat Y = 50 + 1.5X$ . An observed point is $(X = 20, Y = 80)$ . Compute the residual and interpret geometrically.
Select the correct option
Residual $= Y - \hat Y = 80 - (50 + 1.5 \times 20) = 80 - 80 = 0$ ; the point lies exactly on the lineResidual $= 10$ ; the point is above the lineResidual $= -10$ ; the point is below the lineResidual $= 30$ ; the point is far above the line
Select an option first
Solve online Statistics, OpenStax · §12.5
Show solution
Prediction for $X = 20$ : $\hat Y = 50 + 1.5 \times 20 = 50 + 30 = 80$ . Residual: $e = Y - \hat Y = 80 - 80 = 0$ . The observed point falls exactly on the fitted line — zero residual.
Ex. 110.32Modeling
In a sample of $n = 300$ students, 192 passed the test. The historical pass rate is 60%. At the 5% level, is there evidence of change in the pass rate? Use the z test for proportion.
Select the correct option
$Z \approx 1.41$ ; do not reject $H_0: p = 0.60$ at 5% level $Z \approx 1.41$ ; reject $H_0: p = 0.60$ at 5% level $Z = 2.89$ ; reject $H_0: p = 0.60$ at 5% level $Z \approx 1.41$ ; reject $H_0: p = 0.60$ at 10% level
Select an option first
Solve online OpenIntro Statistics 4th ed. · §6.2
Show solution
$\hat p = 192/300 = 0.640$ . $Z = (0.640 - 0.600)/\sqrt{0.600 \times 0.400/300} = 0.040/\sqrt{0.0008} = 0.040/0.02828 \approx 1.41$ . Two-sided p-value $\approx 0.158 > 0.05$ . Do not reject $H_0$ — no evidence of change in approval rate at 5% level.
Ex. 110.33ModelingAnswer key
An agricultural cooperative tested 3 fertilizers on corn plots ( $n_i = 15$ plots per fertilizer). $SS_B = 450$ , $SS_W = 1800$ . At the 5% level, is there a difference in productivity among the fertilizers?
Select the correct option
$F \approx 5.25$ ; reject $H_0$ at 5% level; at least one fertilizer differs $F \approx 5.25$ ; do not reject $H_0$ at 5% level $F = 1.87$ ; do not reject $H_0$ at 5% level $F = 7.50$ ; reject $H_0$ at 1% level
Select an option first
Solve online Statistics, OpenStax · §13.2
Show solution
Three fertilizers, $n_i = 15$ plants per group ( $N = 45$ ). $MS_B = 450/(3-1) = 225$ ; $MS_W = 1800/(45-3) = 1800/42 \approx 42.86$ . $F = 225/42.86 \approx 5.25$ . Critical value $F_{0.05;\,2,\,42} \approx 3.22$ . Since $5.25 > 3.22$ , reject $H_0$ — at least one fertilizer differs.
Ex. 110.34Modeling
In a simple regression analysis: $S_{xx} = 400$ , $S_{yy} = 1600$ , $S_{xy} = 720$ . Compute $r$ and $R^2$ . What proportion of the variation in $Y$ is explained by the model?
Select the correct option
$R^2 \approx 0.81$ ; 81% of the variation in $Y$ is explained by the model $R^2 \approx 0.90$ ; 90% of the variation in $Y$ is explained by the model $R^2 \approx 0.64$ ; 64% of the variation in $Y$ is explained by the model $R^2 \approx 0.50$ ; 50% of the variation in $Y$ is explained by the model
Select an option first
Solve online OpenIntro Statistics 4th ed. · §8.3
Show solution
$r = S_{xy}/\sqrt{S_{xx} S_{yy}} = 720/\sqrt{400 \times 1600} = 720/800 = 0.90$ . In simple regression, $R^2 = r^2 = 0.90^2 = 0.81$ . Interpretation: 81% of the variation in $Y$ is explained by the linear relationship with $X$ .
Ex. 110.35Modeling
An epidemiologist believes the prevalence of a rare disease in a region is 30%, modeled by $\text{Beta}(3, 7)$ . In a screening of $n = 50$ people, he finds $x = 10$ positive. Determine the posterior distribution and the new point estimate (posterior mean).
Select the correct option
Posterior $\text{Beta}(13, 47)$ ; posterior mean $\approx 21.7\%$ — between the prior (30%) and MLE (20%)Posterior $\text{Beta}(10, 45)$ ; posterior mean $= 10/55 \approx 18.2\%$ Posterior $\text{Beta}(13, 42)$ ; posterior mean $\approx 23.6\%$ Posterior $\text{Beta}(3, 50)$ ; posterior mean $\approx 5.7\%$
Select an option first
Solve online Statistical Thinking for the 21st Century · Ch. 15
Show solution
Prior $\text{Beta}(3, 7)$ (prior mean $= 3/10 = 30\%$ ). Data: $n = 50$ , $x = 10$ . Posterior: $\text{Beta}(3+10,\; 7+50-10) = \text{Beta}(13, 47)$ . Posterior mean: $13/(13+47) = 13/60 \approx 21.7\%$ . The posterior mean lies between the prior (30%) and the MLE (10/50 = 20%) — typical behavior of Bayesian updating.
Ex. 110.36Challenge
A researcher performs 10 simultaneous hypothesis tests, each at the 5% level. What is the probability of at least one false positive? How does Bonferroni correction solve the problem?
Select the correct option
Bonferroni correction: use $\alpha^* = 0.05/10 = 0.005$ in each test; maintains familywise Type I error below 5%Running 10 tests at 5% level is equivalent to one test at 5% levelBonferroni increases power of individual testsWith 10 tests at 5% level, the probability of at least one false positive is 5%
Select an option first
Solve online OpenIntro Statistics 4th ed. · §5.3
Show solution
With 10 independent tests at level $\alpha = 0.05$ , the probability of at least one false positive is $1 - (0.95)^{10} \approx 0.401$ — far above 5%. Bonferroni divides: $\alpha^* = 0.05/10 = 0.005$ . Limitation: Bonferroni is conservative (reduces power) when hypotheses are correlated; a less conservative alternative: FDR control via Benjamini-Hochberg.
Show step-by-step (with the why)
1. Probability of at least one false positive in $m$ independent tests: $1 - (1-\alpha)^m$ .
2. For $m=10$ , $\alpha=0.05$ : $1 - 0.95^{10} \approx 0.40$ . Huge inflation.
3. Bonferroni: use $\alpha^* = \alpha/m = 0.005$ in each test.
4. This guarantees familywise Type I error $\leq \alpha$ (Boole's inequality).
Cost: reduced power. With positive correlation among tests, Bonferroni is conservative.
Ex. 110.37Challenge
In multiple regression, what is multicollinearity? What are the consequences for the estimated coefficients $\hat\beta_i$ and for predictions $\hat Y$ ?
Select the correct option
Multicollinearity: correlated predictors inflate standard errors of $\hat\beta_i$ , making individual interpretation difficultMulticollinearity reduces $R^2$ of the modelMulticollinearity does not affect predictions $\hat Y$ , only interpretation of coefficientsMulticollinearity is irrelevant when $n > 100$
Select an option first
Solve online Statistics, OpenStax · §12.6
Show solution
Multicollinearity occurs when predictors $X_j$ are highly correlated. Consequences: (1) standard errors of $\hat\beta_j$ inflated — coefficients unstable; (2) individual t-values are small even if overall model is significant; (3) predictions $\hat Y$ remain stable, but coefficients are not interpretable in isolation. Diagnosis: Variance Inflation Factor (VIF). Remedy: remove redundant predictors or use ridge regression.
Ex. 110.38ChallengeAnswer key
Compare simple random sampling, stratified sampling, and convenience sampling. What is the fundamental limitation of convenience sampling for statistical inference?
Select the correct option
Convenience sampling is non-probabilistic — results cannot be generalized to the population; simple random sampling is the gold standardConvenience sampling is valid if $n > 30$ by CLTStratified sampling is always inferior to simple random samplingIncreasing $n$ always eliminates selection bias
Select an option first
Solve online OpenIntro Statistics 4th ed. · §5.4
Show solution
Convenience sampling (e.g., your classmates, Instagram followers) introduces selection bias — the sample does not represent the population. Increasing $n$ does not correct systematic bias; it only reduces variance. Historical example: 1936 Literary Digest survey with 2.4 million responses but severe selection bias predicted the US election wrongly. Stratified sampling can be more efficient than SRS for heterogeneous populations.
Ex. 110.39ChallengeAnswer key
A diagnostic test has sensitivity 90% and specificity 95%. Disease prevalence in the population is 1%. A patient tested positive. What is the positive predictive value (PPV) — that is, what is the probability the patient really has the disease? Use Bayes' theorem.
Select the correct option
Sensitivity = $P(+\mid D) = 0.90$ ; specificity = $P(-\mid \bar D) = 0.95$ ; PPV via Bayes $\approx 15.3\%$ with 1% prevalenceWith sensitivity 90% and specificity 95%, PPV is always above 50%PPV does not depend on disease prevalence in the populationSensitivity 90% means 90% of positives have the disease
Select an option first
Solve online Statistical Thinking for the 21st Century · Ch. 15
Show solution
Using Bayes: $P(+\mid D) = 0.90$ , $P(-\mid \bar D) = 0.95$ , prevalence $P(D) = 0.01$ . $P(+) = 0.90 \times 0.01 + 0.05 \times 0.99 = 0.009 + 0.0495 = 0.0585$ . PPV $= P(D\mid +) = (0.90 \times 0.01)/0.0585 = 0.009/0.0585 \approx 0.154$ . Striking conclusion: even with 90% sensitivity, only ~15% of positives actually have the disease when prevalence is 1%. This is the screening paradox with rare diseases.
Show step-by-step (with the why)
1. Define events: $D$ = disease, $+$ = positive test.
2. Data: $P(+\mid D) = 0.90$ , $P(-\mid \bar D) = 0.95$ implies $P(+\mid \bar D) = 0.05$ . $P(D) = 0.01$ .
3. Law of total probability: $P(+) = 0.90 \times 0.01 + 0.05 \times 0.99 = 0.0585$ .
4. Bayes: $P(D\mid +) = 0.009/0.0585 \approx 0.154$ .
The paradox: with low prevalence, false positives outnumber true positives — PPV stays low even with accurate tests.
Ex. 110.40Proof
Prove that the sample variance $S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2$ is an unbiased estimator of $\sigma^2$ , that is, that $E[S^2] = \sigma^2$ . Why is division by $n-1$ (not by $n$ ) necessary?
Select the correct option
$E[S^2] = \sigma^2$ ; dividing by $n-1$ instead of $n$ ensures unbiasedness $E[S^2] = \sigma^2$ when dividing by $n$ (not by $n-1$ ) $S^2$ is always a biased estimator of $\sigma^2$ , regardless of the divisor $E[S^2] = \sigma^2/(n-1)$
Select an option first
Grinstead & Snell — Introduction to Probability · §9.3
Show solution
Proof of unbiasedness: $E[S^2] = E\left[\frac{1}{n-1}\sum(X_i - \bar X)^2\right]$ . Expansion: $\sum(X_i - \bar X)^2 = \sum X_i^2 - n\bar X^2$ . Thus $E\left[\sum(X_i - \bar X)^2\right] = \sum E[X_i^2] - n E[\bar X^2] = n(\sigma^2 + \mu^2) - n(\sigma^2/n + \mu^2) = n\sigma^2 - \sigma^2 = (n-1)\sigma^2$ . Dividing by $n-1$ : $E[S^2] = \sigma^2$ . Dividing by $n$ would give $E = (n-1)\sigma^2/n \neq \sigma^2$ .
Show step-by-step (with the why)
1. Write $\sum(X_i - \bar X)^2 = \sum X_i^2 - n\bar X^2$ .
2. Take expectation: $E[X_i^2] = \text{Var}(X_i) + (E[X_i])^2 = \sigma^2 + \mu^2$ .
3. $E[\bar X^2] = \text{Var}(\bar X) + (E[\bar X])^2 = \sigma^2/n + \mu^2$ .
4. $E\left[\sum(X_i - \bar X)^2\right] = n(\sigma^2 + \mu^2) - n(\sigma^2/n + \mu^2) = (n-1)\sigma^2$ .
5. Dividing by $n-1$ : $E[S^2] = \sigma^2$ . Q.E.D.

Sources

OpenIntro Statistics (4th ed.) — Diez, Çetinkaya-Rundel, Barr · CC-BY-SA · Foundation for CI, z/t tests, ANOVA, regression, and chi-square (§5–8).
Statistics (OpenStax) — Illowsky, Dean · CC-BY · Systematic coverage of CI (§8), hypothesis testing (§9), two samples (§10), chi-square and ANOVA (§11), linear regression (§12–13).
Statistical Thinking for the 21st Century — Poldrack · CC-BY-NC · Modern approach to Bayesian inference (Ch. 15), positive predictive value, and p-value interpretation.
Introduction to Probability — Grinstead and Snell · GNU FDL · Properties of variance and estimator proofs (§9.3).