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Lesson 103 — Hypothesis testing: structure and logic

Formal structure of hypothesis testing: H0 vs H1, test statistic, p-value, significance level, Type I and Type II errors, and test power.

Used in: 3.º ano do EM (17-18 anos) · Equiv. Stochastik LK alemão · Equiv. Math B japonês · H2 Statistics singapurense

p\text{-value} = P(T \geq t_{\mathrm{obs}} \mid H_0) \leq \alpha \Rightarrow \text{reject } H_0

The p-value measures the probability of observing a result as extreme or more extreme than what was obtained, assuming that $H_0$ is true. When $p \leq \alpha$ , the data are incompatible with $H_0$ at the chosen significance level, and we reject $H_0$ .

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Rigorous notation, full derivation, hypotheses

Rigorous definition

The five elements of a hypothesis test

"The null hypothesis $H_0$ represents a claim of skepticism. It is the status quo that would be maintained unless there is sufficient evidence against it." — OpenIntro Statistics, §5.1

Errors and test power

Definition· Type I Error, Type II Error, and Power

	$H_0$ true	$H_0$ false
Reject $H_0$	Type I Error ( $\alpha$ )	Correct decision (power $= 1-\beta$ )
Fail to reject $H_0$	Correct decision	Type II Error ( $\beta$ )

Type I Error (false positive): reject $H_0$ when it is true. Probability controlled by $\alpha$ .
Type II Error (false negative): fail to reject $H_0$ when it is false. Probability $\beta$ (depends on $H_1$ , $\sigma$ , $n$ ).
Power $= 1 - \beta$ : probability of detecting the true effect.

For fixed sample size, decreasing $\alpha$ increases $\beta$ (trade-off). To increase power without sacrificing $\alpha$ : increase $n$ .

Formal definition of p-value

"The p-value measures how consistent the data are with $H_0$ . A small p-value indicates that the data are incompatible with $H_0$ — not that $H_0$ is false with probability $1-p$ ." — OpenIntro Statistics, §5.1

Types of alternative hypothesis

Worked examples

Example— 103.1· Two-tailed test for mean with z (basic)

Problem. A company claims the average weight of its coffee packages is $\mu_0 = 500$ g. A sample of $n = 36$ packages gives $\bar X = 492$ g with $\sigma = 24$ g (known). At level $\alpha = 0.05$ , do the data contradict the claim?

Strategy. $H_0: \mu = 500$ , $H_1: \mu \neq 500$ (two-tailed). Use the z-statistic since $\sigma$ is known.

Solution.

$Z = \frac{\bar X - \mu_0}{\sigma/\sqrt{n}} = \frac{492 - 500}{24/\sqrt{36}} = \frac{-8}{4} = -2.00$

Two-tailed p-value: $p = 2\,P(Z \leq -2.00) = 2 \times 0.0228 = 0.0456$ .

Since $p = 0.0456 < \alpha = 0.05$ , we reject $H_0$ . The data contradict the company's claim at the 5% level.

Verification. The critical value for two-tailed with $\alpha = 0.05$ is $z_{0.025} = 1.960$ . Since $|-2.00| = 2.00 > 1.960$ , rejection by the statistic agrees with the p-value. Consistent.

Source. OpenStax Statistics, §9.2, Example 9.3 — CC-BY.

Example— 103.2· Identification of Type I and II errors (conceptual)

Problem. A quality test checks whether a batch of medications has average active ingredient concentration of 50 mg ( $H_0$ ). At level 5%, the batch is approved or rejected. (a) What constitutes a Type I Error in this context? (b) What about Type II Error? (c) Which is more serious?

Strategy. Map the formal definitions to the specific context.

Solution.

(a) Type I Error: reject $H_0$ when $\mu = 50$ mg — that is, reject a batch that is actually in specification. Consequence: waste of good product, rework costs.

(b) Type II Error: fail to reject $H_0$ when $\mu \neq 50$ mg — that is, approve a batch out of specification. Consequence: sub-dosed or over-dosed medication reaches the patient.

(c) In most pharmaceutical contexts, Type II Error is more serious: a medication out of specification can harm the patient. For this reason, clinical trials use small $\alpha$ but require high power (low $\beta$ ), increasing $n$ .

Verification. The asymmetry between the errors justifies calibrating $\alpha$ and $\beta$ differently by context — medicine uses $\alpha = 0.01$ and power of 80-90%.

Source. OpenIntro Statistics, §5.2, Example 5.4 — CC-BY-SA.

Example— 103.3· Power calculation and sample size (intermediate)

Problem. A researcher wants to detect that the average response time at an urgent care center changed from $\mu_0 = 30$ min to $\mu_1 = 27$ min ( $\delta = 3$ min), with $\sigma = 10$ min, $\alpha = 0.05$ (two-tailed), and 80% power. What is the minimum $n$ ?

Strategy. Apply the sample size formula for power: $n = (z_{\alpha/2} + z_\beta)^2 \sigma^2/\delta^2$ .

Solution.

$z_{0.025} = 1.960$ , $z_{0.20} = 0.842$ (80% power $\Rightarrow \beta = 0.20$ ).

$n = \frac{(1.960 + 0.842)^2 \times 10^2}{3^2} = \frac{(2.802)^2 \times 100}{9} = \frac{7.851 \times 100}{9} \approx 87.2$

Round: $n = 88$ visits.

Verification. If $\delta = 6$ min (doubled effect): $n = 7.851 \times 100/36 \approx 22$ . Larger effect requires smaller sample — consistent.

Source. OpenIntro Statistics, §5.3, Example 5.7 — CC-BY-SA.

Example— 103.4· One-tailed test — medication (intermediate)

Problem. A new anticoagulant claims to reduce mean clotting time from $\mu_0 = 12$ s to below that. A sample of $n = 20$ patients gives $\bar X = 11.2$ s and $s = 2$ s. At level $\alpha = 0.05$ , is the medication effective?

Strategy. $H_0: \mu \geq 12$ , $H_1: \mu < 12$ (one-tailed to the left). t-statistic with 19 degrees of freedom.

Solution.

$T = \frac{11.2 - 12}{2/\sqrt{20}} = \frac{-0.8}{0.4472} = -1.789$

For $H_1: \mu < 12$ , p-value $= P(t_{19} \leq -1.789)$ . From t-table: $P(t_{19} \leq -1.729) = 0.05$ and $P(t_{19} \leq -2.093) = 0.025$ . So $p \approx 0.045$ .

Since $p = 0.045 < 0.05$ , we reject $H_0$ . Evidence that the medication reduces clotting time.

Verification. One-tailed critical value: $t_{0.05,\,19} = -1.729$ . Since $T = -1.789 < -1.729$ , rejection by the statistic agrees. Consistent.

Source. OpenStax Statistics, §9.4, Example 9.8 — CC-BY.

Example— 103.5· Misinterpretation of p-value and correction (advanced)

Problem. A researcher obtains $p = 0.03$ in a test of $H_0: \mu = 0$ and claims: "There is a 97% probability that the effect is real." Identify the error and formulate the correct interpretation.

Strategy. Apply the formal definition of p-value and distinguish probabilities on data from probabilities on hypotheses.

Solution.

The statement is incorrect for two reasons:

The p-value is a probability about the data (given $H_0$ ), not about the hypotheses. $P(\text{data} \mid H_0) \neq P(H_0 \mid \text{data})$ — confusing the two is the fallacy of transposed conditionals (base rate neglect).
$1 - p\text{-value} = 0.97$ has no interpretation as probability of $H_1$ . To obtain $P(H_1 \mid \text{data})$ , you would need Bayes's Theorem with a prior on the hypotheses.

Correct interpretation: "If $H_0$ were true, there would be only a 3% probability of observing an effect as large (or larger) as what was observed. The data are statistically incompatible with $H_0$ at the 5% level."

Verification. Two independent studies each with $p = 0.04$ do not imply a third study with $p = 0.04$ — combining evidence is done via meta-analysis, not by multiplying p-values.

Source. OpenIntro Statistics, §5.1, Section "Interpreting p-values" — CC-BY-SA.

Exercise list

42 exercises · 10 with worked solution (25%)

Application 26Understanding 5Modeling 6Challenge 3Proof 2

Ex. 103.1ApplicationAnswer key
A consumer protection agency wants to check whether the average weight of a 500 g flour package is as stated. Formulate the hypotheses $H_0$ and $H_1$ for this scenario.
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$H_0: \mu \neq 500$ g and $H_1: \mu = 500$ g $H_0: \mu = 500$ g and $H_1: \mu \neq 500$ g (two-tailed) $H_0: \mu = 500$ g and $H_1: \mu < 500$ g (one-tailed to the left) $H_0: \mu > 500$ g and $H_1: \mu \leq 500$ g
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Solve online OpenStax Statistics · §9.1
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$H_0: \mu = 500$ g (weight as claimed), $H_1: \mu \neq 500$ g (two-tailed, since weight can be above or below). Two-tailed test because both overweight and underweight are concerns.
Ex. 103.2ApplicationAnswer key
Researchers want to check whether Brazilian teenagers sleep less than the recommended 8 hours per night. Formulate $H_0$ and $H_1$ .
Select the correct option
$H_0: \mu = 8$ h and $H_1: \mu \neq 8$ h (two-tailed) $H_0: \mu \geq 8$ h and $H_1: \mu < 8$ h (one-tailed to the left) $H_0: \mu < 8$ h and $H_1: \mu \geq 8$ h $H_0: \mu = 8$ h and $H_1: \mu > 8$ h (one-tailed to the right)
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OpenIntro Statistics · §5.1
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$H_0: \mu \geq 8$ hours (sufficient sleep), $H_1: \mu < 8$ hours (insufficient sleep). One-tailed test to the left because the concern is with sleep below the recommended amount.
Ex. 103.3Application
$H_0: \mu = 50$ , $H_1: \mu \neq 50$ . Data: $n = 25$ , $\bar X = 52$ , $\sigma = 10$ (known). Calculate the z-statistic and p-value. Conclude for $\alpha = 0.05$ .
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$Z = 1.00$ ; p-value $= 0.32$ ; we reject $H_0$ $Z = 1.00$ ; p-value $= 0.32$ ; we fail to reject $H_0$ $Z = 2.00$ ; p-value $= 0.046$ ; we reject $H_0$ $Z = 1.00$ ; p-value $= 0.16$ ; we fail to reject $H_0$
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Solve online OpenStax Statistics · §9.3
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$Z = (\bar X - \mu_0)/(\sigma/\sqrt{n}) = (52 - 50)/(10/\sqrt{25}) = 2/2 = 1.00$ . Two-tailed p-value: $2 \times P(Z > 1.00) = 2 \times 0.1587 = 0.3174$ . Since $p = 0.32 > 0.05$ , we fail to reject $H_0$ . Insufficient evidence that the mean changed.
Ex. 103.4Application
A manufacturer claims its light bulbs last an average of 1000 h. A sample of $n = 64$ bulbs gives $\bar X = 985$ h with $\sigma = 50$ h (known). At the 5% level, is the mean lifetime less than claimed?
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$Z = -2.40$ ; p-value $= 0.0082$ ; we fail to reject $H_0$ $Z = -2.40$ ; p-value $= 0.0082$ ; we reject $H_0$ $Z = -2.40$ ; p-value $= 0.016$ ; we fail to reject $H_0$ (used two-tailed p-value) $Z = -1.875$ ; p-value $= 0.030$ ; we reject $H_0$
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Solve online OpenStax Statistics · §9.4
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$Z = -2.40$ . One-tailed p-value $= 0.0082 < 0.05$ . We reject $H_0$ : evidence that the mean lifetime is less than 1000 h.
Show step-by-step (with the why)
1. Hypotheses: $H_0: \mu = 1000$ h, $H_1: \mu < 1000$ h (one-tailed to the left).
2. Statistic: $Z = (985 - 1000)/(50/\sqrt{64}) = -15/6.25 = -2.40$ .
3. One-tailed p-value: $P(Z < -2.40) = 1 - \Phi(2.40) = 1 - 0.9918 = 0.0082$ .
4. Since $p = 0.0082 < 0.05$ , we reject $H_0$ .
5. Shortcut: For one-tailed tests, p-value is half the two-tailed p-value for the same $|Z|$ . Here two-tailed would be 0.0164, one-tailed is 0.0082.
Ex. 103.5Application
In a criminal trial, $H_0$ is "the defendant is innocent" and $H_1$ is "the defendant is guilty." Describe Type I and Type II Errors in this context. Which is considered more serious in the Brazilian legal system? Why?
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Type I: acquit a guilty person; Type II: convict an innocent; Type II is more seriousType I: convict an innocent; Type II: acquit a guilty; Type I is more seriousType I: convict an innocent; Type II: acquit a guilty; Type II is more seriousType I: acquit a guilty; Type II: convict an innocent; both equally serious
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Solve online OpenIntro Statistics · §5.2
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Type I Error: convict an innocent (legal system says "guilty" when actually innocent). Type II Error: acquit a guilty person (system says "innocent" when actually guilty). In a democratic legal system, Type I is more serious because "it is better to acquit 100 guilty people than to convict 1 innocent" — the requirement level ("beyond reasonable doubt") is very high, equivalent to very small $\alpha$ .
Ex. 103.6Understanding
A test yields $p = 0.03$ . Which statement below is correct?
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There is a 97% probability that H1 is trueIf H0 were true, there would be only a 3% probability of observing data as extreme or more extremeH0 is false with probability 97%The observed effect is real with probability 97%
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Solve online OpenIntro Statistics · §5.1
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The p-value is a probability about the data (given $H_0$ ), not about hypotheses. The correct interpretation is always: "probability of observing a result as extreme or more extreme, assuming $H_0$ is true".
Ex. 103.7Understanding
A test with $n = 10$ yields $p = 0.12$ . The researcher concludes "the effect does not exist." What may be wrong?
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The effect does not exist; the data confirm H0The sample is too small to detect a true effect that may existThe significance level alpha should be increased to 0.10The alternative hypothesis is false
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Solve online OpenIntro Statistics · §5.2
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Failing to reject $H_0$ does not mean $H_0$ is true. The effect may exist but the sample may be too small (insufficient power) to detect it. Absence of evidence is not evidence of absence.
Ex. 103.8Application
A school implemented a new methodology. The historical average score is $\mu_0 = 35$ points. After intervention, $n = 40$ students achieved $\bar X = 37$ and $\sigma = 8$ (known). At the 5% level, did the score improve?
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$Z = 1.581$ ; p-value $= 0.057$ ; we reject $H_0$ — the score improved $Z = 1.581$ ; p-value $= 0.057$ ; we fail to reject $H_0$ — insufficient evidence $Z = 1.581$ ; p-value $= 0.114$ (two-tailed); we fail to reject $H_0$ $Z = 1.000$ ; p-value $= 0.159$ ; we fail to reject $H_0$
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Solve online OpenStax Statistics · §9.4
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$H_0: \mu = 35$ points, $H_1: \mu > 35$ (one-tailed to the right). $Z = (37 - 35)/(8/\sqrt{40}) = 2/(8/6.324) = 2/1.265 = 1.581$ . p-value: $P(Z > 1.581) = 1 - 0.943 = 0.057$ . Since $p = 0.057 > 0.05$ , we fail to reject $H_0$ at the 5% level. Marginally non-significant.
Ex. 103.9Application
An urgent care center wants to detect a 5 min reduction in response time ( $\delta = 5$ , $\sigma = 10$ ). With $\alpha = 0.05$ and 90% power, what is the minimum $n$ ?
Select the correct option
$n = 34$ visits $n = 43$ visits $n = 88$ visits $n = 61$ visits
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Solve online OpenIntro Statistics · §5.3
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$n = (1.960 + 1.282)^2 \times 100/25 \approx 43$ visits.
Show step-by-step (with the why)
1. $n = (z_{\alpha/2} + z_{eta})^2 \sigma^2/\delta^2$ .
2. $\delta = 5$ min, $\sigma = 10$ min, $z_{0.025} = 1.960$ , 90% power: $z_{0.10} = 1.282$ .
3. $n = (1.960 + 1.282)^2 \times 100/25 = (3.242)^2 \times 4 = 10.51 \times 4 = 42.04$ .
4. Round: $n = 43$ visits.
5. Shortcut: Increasing power from 80% to 90% adds approximately 30% to sample size ( $z_{0.10} = 1.282$ vs $z_{0.20} = 0.842$ ).
Ex. 103.10ApplicationAnswer key
A coin is flipped 100 times and lands heads 60 times. At the 5% level, is the coin fair?
Select the correct option
$Z = 2.00$ ; p-value $= 0.023$ (used one-tailed); we reject $H_0$ $Z = 2.00$ ; p-value $= 0.046$ ; we reject $H_0$ — evidence of biased coin $Z = 2.00$ ; p-value $= 0.046$ ; we fail to reject $H_0$ $Z = 1.20$ ; p-value $= 0.230$ ; we fail to reject $H_0$
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Solve online OpenIntro Statistics · §5.1
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$H_0: p = 0.50$ (fair coin), $H_1: p \neq 0.50$ . Statistic: $Z = (0.60 - 0.50)/\sqrt{0.50 \times 0.50/100} = 0.10/0.05 = 2.00$ . Two-tailed p-value: $2 \times 0.0228 = 0.0456 < 0.05$ . We reject $H_0$ : evidence that the coin is biased.
Ex. 103.11Application
A researcher changes the significance level from $\alpha = 0.05$ to $\alpha = 0.01$ while keeping $n$ fixed. Explain the effect on Type II Error and test power.
Select the correct option
Type II Error decreases and power increasesType II Error increases and power decreasesType II Error does not change; only Type I Error decreasesType II Error and power remain unchanged because n is fixed
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Solve online OpenIntro Statistics · §5.2
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For fixed $n$ and $\sigma$ , decreasing $\alpha$ from 0.05 to 0.01 increases the critical value, shrinking the rejection region. This increases $\beta$ (harder to reject $H_0$ when false) and reduces power. The fundamental trade-off: fewer false positives means more false negatives for fixed $n$ .
Ex. 103.12Application
Normal fasting glucose level: $\mu_0 = 120$ mg/dL. A sample of $n = 50$ diabetic patients gives $\bar X = 128$ mg/dL with $\sigma = 20$ mg/dL. At the 1% level, is average glucose elevated?
Select the correct option
$Z = 2.83$ ; p-value $= 0.0023$ ; we fail to reject $H_0$ at 1% level $Z = 2.83$ ; p-value $= 0.0023$ ; we reject $H_0$ at 1% level $Z = 2.83$ ; p-value $= 0.0046$ (two-tailed); we reject $H_0$ at 1% level $Z = 2.00$ ; p-value $= 0.023$ ; we fail to reject $H_0$ at 1% level
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Solve online OpenStax Statistics · §9.4
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$H_0: \mu = 120$ mg/dL, $H_1: \mu > 120$ mg/dL. $Z = (128 - 120)/(20/\sqrt{50}) = 8/2.828 = 2.83$ . p-value: $P(Z > 2.83) = 1 - 0.9977 = 0.0023 < 0.01$ . We reject $H_0$ : strong evidence that average blood glucose is above normal.
Ex. 103.13Understanding
A result is "statistically significant at 5%." What does this correctly mean?
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The effect is large enough to be clinically relevantThe difference observed is unlikely to occur by chance under H0There is a 95% probability that the treatment worksThe study was well-conducted and free of bias
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Solve online Statistical Thinking for the 21st Century · Cap. 10
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"Statistically significant" only means the p-value fell below $\alpha$ . It does not imply practical relevance, causality, or absence of bias. With very large $n$ , trivial differences become "significant".
Ex. 103.14Application
A company wants to detect whether the average weight of its products fell from $\mu_0 = 250$ g to $\mu_1 = 245$ g, with $\sigma = 20$ g, $\alpha = 0.05$ , and 80% power. What is the minimum $n$ ?
Select the correct option
$n = 71$ products $n = 99$ products $n = 43$ products $n = 125$ products
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Solve online OpenIntro Statistics · §5.3
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$H_0: \mu = 250$ g, $H_1: \mu < 250$ g (one-tailed). With $\sigma = 20$ g and $\alpha = 0.05$ , critical value: $\bar X_c = 250 - 1.645 \times 20/\sqrt{n}$ . To detect $\mu_1 = 245$ g with 80% power: $n = (1.645 + 0.842)^2 \times 400/25 = (2.487)^2 \times 16 = 6.185 \times 16 \approx 99$ .
Ex. 103.15ApplicationAnswer key
A genomics study performs 1000 simultaneous tests with $\alpha = 0.05$ . All tested genes are null (no true effect). How many false positives are expected? If 60 genes are "significant," what is the estimated false discovery rate?
Select the correct option
50 false positives expected; estimated FDR = 5%50 false positives expected; estimated FDR $\approx 83\%$ 5 false positives expected; estimated FDR $\approx 8\%$ 50 false positives expected; estimated FDR = 50%
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Solve online Statistical Thinking for the 21st Century · Cap. 11
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Significance level 5%: reject if $p \leq 0.05$ . In 1000 tests, expect $1000 \times 0.05 = 50$ false positives. If only 60 are significant and all are true positives, estimated FDR = 50/60 = 83%. Most "discovered" effects may be false.
Ex. 103.16Application
A coin is flipped 800 times and lands heads 384 times. At the 5% level, is the coin fair?
Select the correct option
$Z = -1.131$ ; p-value $= 0.258$ ; we reject $H_0$ — coin is biased $Z = -1.131$ ; p-value $= 0.258$ ; we fail to reject $H_0$ — data consistent with fair coin $Z = -1.131$ ; p-value $= 0.129$ (one-tailed); we fail to reject $H_0$ $Z = -2.00$ ; p-value $= 0.046$ ; we reject $H_0$
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Solve online OpenStax Statistics · §9.3
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$Z = (0.48 - 0.50)/\sqrt{0.50 \times 0.50/800} = -0.02/0.01768 = -1.131$ . Two-tailed p-value: $2 \times P(Z < -1.131) = 2 \times 0.129 = 0.258$ . Since $p = 0.258 > 0.05$ , we fail to reject $H_0$ . The data are consistent with a fair coin.
Ex. 103.17Application
A survey of $n = 30$ teenagers recorded average sleep of $\bar X = 7.5$ h with $\sigma = 1.5$ h (from prior studies). At the 5% level, do they sleep less than 8 hours?
Select the correct option
$Z = -1.825$ ; p-value $\approx 0.034$ ; we fail to reject $H_0$ at 5% level $Z = -1.825$ ; p-value $\approx 0.034$ ; we reject $H_0$ — they sleep less than 8 h $Z = -1.825$ ; p-value $\approx 0.068$ (two-tailed); we reject $H_0$ $Z = -1.095$ ; p-value $\approx 0.137$ ; we fail to reject $H_0$
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Solve online OpenIntro Statistics · §5.1
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$Z = -1.825$ , p-value $\approx 0.034 < 0.05$ . We reject $H_0$ : evidence that teenagers sleep less than 8 hours.
Show step-by-step (with the why)
1. Hypotheses: $H_0: \mu = 8$ h, $H_1: \mu < 8$ h.
2. Statistic: $Z = (7.5 - 8)/(1.5/\sqrt{30}) = -0.5/0.274 = -1.825$ .
3. One-tailed p-value: $P(Z < -1.825) = 1 - \Phi(1.825) \approx 0.034$ .
4. Since $p = 0.034 < 0.05$ , we reject $H_0$ . Teenagers sleep significantly less than 8 hours.
5. Note: p = 0.034 is close to 0.05. A responsible researcher would report the exact value, not just "significant." With $\alpha = 0.01$ we would not reject.
Ex. 103.18Understanding
Which statement about statistical significance is correct?
Select the correct option
A result with p = 0.001 is always more important than a result with p = 0.04A significant result with n = 1000000 may have negligible practical effectp-value is a direct measure of effect sizeNon-significant results should not be published
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Solve online Statistical Thinking for the 21st Century · Cap. 10
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With very large $n$ , tiny differences become "statistically significant." The p-value depends on both the effect size AND the sample size. With $n = 1{,}000{,}000$ , an effect of 0.001 units may have $p < 0.001$ but be clinically irrelevant.
Ex. 103.19Modeling
A clinical trial tests 20 endpoints simultaneously with $\alpha = 0.05$ . What is the probability of at least one false positive without correction? Describe how Bonferroni correction solves the problem and discuss its limitation.
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FWER $\approx 64\%$ ; Bonferroni uses $\alpha = 0.05/20 = 0.0025$ per test; limitation: more conservative than FDRFWER $= 5\%$ ; Bonferroni uses $\alpha = 0.05$ per test; no limitationFWER $\approx 64\%$ ; Bonferroni uses $\alpha = 0.05 \times 20 = 1$ per test; always controlsFWER $\approx 26\%$ ; Bonferroni uses $\alpha = 0.05/20$ per test; limitation: inflates power
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Solve online Statistical Thinking for the 21st Century · Cap. 11
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FWER with Bonferroni correction: uses $\alpha_{\text{corrected}} = 0.05/20 = 0.0025$ per test. Controls the FWER at 5%. Alternative: FDR (Benjamini-Hochberg) is more powerful when many effects are expected. In the clinical trial with 20 endpoints, Bonferroni may be excessively conservative if endpoints are correlated — the Holm-Bonferroni method is considered a compromise.
Ex. 103.20Application
The historical ENEM pass rate at a school is 30%. After a new methodology, 38 of 100 students passed. At the 5% level, did the rate improve?
Select the correct option
$Z = 1.748$ ; p-value $= 0.040$ ; we fail to reject $H_0$ $Z = 1.748$ ; p-value $= 0.040$ ; we reject $H_0$ — the rate improved $Z = 1.748$ ; p-value $= 0.080$ (two-tailed); we reject $H_0$ $Z = 2.00$ ; p-value $= 0.023$ ; we reject $H_0$
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Solve online OpenStax Statistics · §9.3
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$H_0: p = 0.30$ (historical approval rate), $H_1: p > 0.30$ . $\hat p = 38/100 = 0.38$ . $Z = (0.38 - 0.30)/\sqrt{0.30 \times 0.70/100} = 0.08/0.0458 = 1.748$ . p-value: $P(Z > 1.748) = 0.040$ . Since $p = 0.040 < 0.05$ , we reject $H_0$ : the new methodology appears to improve approval rate.
Ex. 103.21Application
Test $H_0: \mu = 50$ vs $H_1: \mu \neq 50$ with $\sigma = 10$ and $\bar X = 51$ . Calculate the p-value for $n = 10$ and $n = 10000$ . What does this reveal about p-value and effect size?
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For $n=10$ : $p=0.752$ (do not reject); for $n=10000$ : $p\approx 0$ (reject). The p-value measures incompatibility with $H_0$ , not effect sizeFor $n=10$ : $p=0.376$ (reject); for $n=10000$ : $p\approx 0$ (do not reject)Both p-values are equal since $\bar X$ and $\sigma$ do not change with $n$ For $n=10$ : $p=0.076$ ; for $n=10000$ : $p=0.076$ . The p-value is independent of $n$
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Solve online Statistical Thinking for the 21st Century · Cap. 10
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With $n = 10$ : $Z = (51 - 50)/(10/\sqrt{10}) = 1/3.162 = 0.316$ . Two-tailed p-value: $2 \times P(Z > 0.316) = 2 \times 0.376 = 0.752$ . Do not reject. With $n = 10000$ : $Z = (51 - 50)/(10/\sqrt{10000}) = 1/0.1 = 10.0$ . $p \approx 0$ . Reject with certainty. A difference of 1 unit: trivial with $n = 10$ , "highly significant" with $n = 10000$ . Illustrates that p-value measures incompatibility with $H_0$ , not effect size.
Ex. 103.22ApplicationAnswer key
Normal systolic pressure: $\mu_0 = 120$ mmHg. Sample of $n = 60$ sedentary adults: $\bar X = 125$ mmHg, $\sigma = 15$ mmHg. At the 1% level, is average pressure elevated?
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$Z = 2.583$ ; p-value $= 0.0049$ ; we fail to reject $H_0$ at 1% level $Z = 2.583$ ; p-value $= 0.0049$ ; we reject $H_0$ at 1% level — elevated pressure $Z = 2.583$ ; p-value $= 0.0098$ (two-tailed); we fail to reject $H_0$ at 1% level $Z = 1.936$ ; p-value $= 0.027$ ; we fail to reject $H_0$ at 1% level
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Solve online OpenStax Statistics · §9.4
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$H_0: \mu = 120$ mmHg, $H_1: \mu > 120$ . $Z = (125 - 120)/(15/\sqrt{60}) = 5/1.936 = 2.583$ . p-value: $P(Z > 2.583) = 0.0049 < 0.01$ . We reject $H_0$ : strong evidence that average blood pressure is elevated in the sample.
Ex. 103.23ApplicationAnswer key
A veterinary study wants to detect that the average weight of pigs of a breed changed from 125 kg to 120 kg ( $\delta = 5$ , $\sigma = 15$ ). With $\alpha = 0.05$ two-tailed and 80% power, how many animals are needed?
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$n = 43$ animals $n = 71$ animals $n = 99$ animals $n = 138$ animals
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Solve online OpenIntro Statistics · §5.3
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$n = (1.960 + 0.842)^2 \times 225/25 \approx 71$ animals.
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1. To detect $\mu_1 = 120$ kg vs $\mu_0 = 125$ kg with $\sigma = 15$ : $\delta = 5$ .
2. 80% power: $z_{eta} = 0.842$ . Two-tailed 5% level: $z_{\alpha/2} = 1.960$ .
3. $n = (1.960 + 0.842)^2 \times 225/25 = 7.851 \times 9 = 70.7$ . Round: $n = 71$ .
4. Shortcut: The formula $n = (z_{\alpha/2} + z_{eta})^2 \sigma^2/\delta^2$ shows that sample size scales quadratically with $\sigma/\delta$ (inverse signal-to-noise ratio).
Ex. 103.24Modeling
ENEM at a school has $\bar X = 52$ points against $\mu_0 = 50$ from state average, with $s = 10$ and $n = 10000$ students. The result is "highly significant" ( $p < 0.001$ ). Calculate Cohen's effect size $d$ . Is a 2-point difference educationally relevant? Discuss.
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$d = 2.0$ (large effect); the difference is educationally relevant $d = 0.2$ (small effect); statistically significant but likely educationally irrelevant $d = 0.02$ (negligible effect); the test is invalid with such large $n$ $d = 0.5$ (medium effect); the difference justifies policy change
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Solve online Statistical Thinking for the 21st Century · Cap. 10
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Cohen's effect size: $d = |\bar X - \mu_0|/s = |52 - 50|/10 = 0.2$ (small effect by Cohen's convention). With $n = 10000$ , even $d = 0.2$ generates $p < 0.0001$ . In educational research, a 2-point difference on a 100-point scale may not justify policy change, despite "significance." Reporting confidence intervals and effect size is essential.
Ex. 103.25Challenge
Show that, under $H_0$ true, the p-value has Uniform $(0,1)$ distribution for continuous tests. Use this result to verify that $P(\text{reject } H_0 \mid H_0) = \alpha$ .
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Under $H_0$ , $p \sim \text{Uniform}(0,1)$ , therefore $P(p \leq \alpha \mid H_0) = \alpha$ Under $H_0$ , $p \sim \text{Normal}(0.5; 1/12)$ , therefore $P(p \leq \alpha \mid H_0) \approx \alpha$ for small $\alpha$ Under $H_0$ , $p$ is deterministic and equals $\alpha$ , so $P(p \leq \alpha \mid H_0) = 1$ Under $H_0$ , $p \sim \text{Uniform}(0,1)$ and therefore $P(p \leq \alpha \mid H_0) = \alpha^2$
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Solve online Statistical Thinking for the 21st Century · Cap. 10
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Under $H_0$ true, the p-value has distribution $\text{Uniform}(0,1)$ . Therefore $P(p \leq 0.05 \mid H_0) = 0.05$ exactly — the level $\alpha$ controls Type I error rate exactly. Under $H_1$ , the p-value distribution concentrates near 0, with the fraction below $\alpha$ equal to power $1-\beta$ .
Ex. 103.26Proof
Use the Neyman-Pearson Lemma to show that the one-tailed z-test (reject if $\bar X > c$ ) is the most powerful level $\alpha$ test for $H_0: \mu = \mu_0$ vs $H_1: \mu = \mu_1 > \mu_0$ with normal data and $\sigma$ known.
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By NP Lemma, the optimal region is based on likelihood ratio; for $\mu_1 > \mu_0$ with normals, this ratio is increasing in $\bar X$ , so reject if $\bar X > c$ By NP Lemma, the optimal region is to reject if $\bar X < c$ because $\mu_1 > \mu_0$ reverses the ratioNP Lemma does not apply to composite hypotheses; one-tailed z-test is chosen by conventionBy NP Lemma, the optimal region is two-tailed (reject if $|\bar X - \mu_0| > c$ ) for any $\mu_1$
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Solve online OpenIntro Statistics · §5.3
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The optimal rejection region by Neyman-Pearson Lemma for $H_0: \mu = \mu_0$ vs $H_1: \mu = \mu_1 > \mu_0$ with normal data is: reject if $\bar X > c$ where $c$ is determined by $P(\bar X > c \mid H_0) = \alpha$ . This is equivalent to the one-tailed z-test. Proof: the likelihood ratio $L(\mu_1)/L(\mu_0) = \exp((\mu_1 - \mu_0)(\bar X - (\mu_0 + \mu_1)/2)/\sigma^2/n)$ is monotonically increasing in $\bar X$ for $\mu_1 > \mu_0$ .
Ex. 103.27Application
An organization reported that teenagers spent an average of 4.5 hours weekly on their phones. You believe the current average is higher. Formulate $H_0$ and $H_1$ and justify the test type (one-tailed/two-tailed).
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$H_0: \mu = 4.5$ h; $H_1: \mu \neq 4.5$ h; two-tailed because direction is unknown $H_0: \mu = 4.5$ h; $H_1: \mu > 4.5$ h; one-tailed to the right because the suspicion is of an increase $H_0: \mu > 4.5$ h; $H_1: \mu \leq 4.5$ h; one-tailed to the left $H_0: \mu = 4.5$ h; $H_1: \mu < 4.5$ h; one-tailed to the left because teens use phones less
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Solve online OpenStax Statistics · §9.1
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$H_0: \mu = 4.5$ hours, $H_1: \mu > 4.5$ (one-tailed — you believe usage has increased). One-tailed test to the right because the interest is in whether current average exceeds the historical value.
Ex. 103.28Application
Researchers study app usage among 273 teenagers: 63 use the app. They test whether more than 30% of teenagers use the app. Formulate $H_0$ and $H_1$ , then describe Type I and Type II Errors in this context.
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$H_0: p = 0.30$ ; $H_1: p > 0.30$ ; Type I = fail to detect use above 30%; Type II = falsely conclude use is above 30% $H_0: p = 0.30$ ; $H_1: p > 0.30$ ; Type I = falsely conclude use is above 30%; Type II = fail to detect true use above 30% $H_0: p > 0.30$ ; $H_1: p \leq 0.30$ ; Type I = conclude use is 30%; Type II = fail to reject $H_0: p = 0.30$ ; $H_1: p \neq 0.30$ ; Type I = wrongly reject; Type II = wrongly fail to reject
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Solve online OpenStax Statistics · §9.2
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$H_0: p = 0.30$ , $H_1: p > 0.30$ . Type I Error: conclude that more than 30% use the app when actually they do not. Type II Error: fail to detect that more than 30% use the app when they actually do.
Ex. 103.29Application
Engineers at a start-up work an average of 60 hours per week. A new hire suspects the average is less. Surveys 10 colleagues and obtains: 70, 45, 55, 60, 65, 55, 55, 60, 50, 55 hours. At the 5% level, is the average less than 60?
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$t = -1.342$ ; p-value $\approx 0.106$ ; we reject $H_0$ — average less than 60 h $t = -1.342$ ; p-value $\approx 0.106$ ; we fail to reject $H_0$ — insufficient evidence $t = -1.342$ ; p-value $\approx 0.212$ (two-tailed); we fail to reject $H_0$ $t = -1.500$ ; p-value $\approx 0.084$ ; we fail to reject $H_0$
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Solve online OpenStax Statistics · §9.4
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One-tailed test to the left: $H_0: \mu \geq 60$ , $H_1: \mu < 60$ . Sample of 10 engineers: hours worked = 70, 45, 55, 60, 65, 55, 55, 60, 50, 55. $\bar X = 57$ , $s \approx 7.07$ . $t = (57 - 60)/(7.07/\sqrt{10}) = -3/2.236 = -1.342$ . p-value: $P(t_9 < -1.342) \approx 0.106$ . Do not reject at $\alpha = 0.05$ .
Ex. 103.30Application
A Gallup survey indicates that 40% of Americans fear public speaking. A student believes the percentage at her school is lower. She surveys 361 peers and 141 report fearing public speaking. At the 5% level, is the proportion less than 40%?
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$Z = -0.364$ ; p-value $\approx 0.358$ ; we reject $H_0$ — proportion less than 40% $Z = -0.364$ ; p-value $\approx 0.358$ ; we fail to reject $H_0$ — no evidence of proportion less than 40% $Z = -0.364$ ; p-value $\approx 0.716$ (two-tailed); we fail to reject $H_0$ $Z = -1.746$ ; p-value $\approx 0.040$ ; we reject $H_0$
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Solve online OpenStax Statistics · §9.3
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$H_0: p = 0.40$ , $H_1: p < 0.40$ . $\hat p = 141/361 \approx 0.3906$ . $Z = (0.3906 - 0.40)/\sqrt{0.40 \times 0.60/361} = -0.0094/0.02580 \approx -0.364$ . p-value: $P(Z < -0.364) \approx 0.358$ . We fail to reject $H_0$ : no evidence that the proportion at the school is less than 40%.
Ex. 103.31ApplicationAnswer key
The average US career length before retirement is 34 years. Formulate hypotheses to test whether this mean still holds. Describe Type I and Type II Errors and their practical consequences.
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$H_0: \mu = 34$ years; $H_1: \mu \neq 34$ years; Type I = wrongly conclude mean changed; Type II = fail to detect that mean changed $H_0: \mu \neq 34$ years; $H_1: \mu = 34$ years; Type I = confirm status quo wrongly $H_0: \mu = 34$ years; $H_1: \mu > 34$ years (one-tailed); Type I = overestimate career $H_0: \mu = 34$ years; $H_1: \mu < 34$ years (one-tailed); Type I = claim false reduction
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Solve online OpenStax Statistics · §9.2
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$H_0: \mu = 34$ years of work before retirement, $H_1: \mu \neq 34$ . Type I Error: wrongly claim the mean is not 34 years. Type II Error: fail to detect that the mean has changed when it truly did.
Show step-by-step (with the why)
1. Hypotheses: $H_0: \mu = 34$ years, $H_1: \mu \neq 34$ years (two-tailed).
2. With $\sigma = 8$ years (known): $Z = (\bar X - 34)/(8/\sqrt{n})$ .
3. If $n$ and $\bar X$ are not specified, recognize that for the stated mean of 34 years, the test would be two-tailed with rejection if $|Z| > 1.960$ at 5%.
4. Type I Error is concluding the mean differs from 34 when it actually equals 34 (false alarm). Type II Error is failing to reject when the mean truly differs from 34 (missing a real effect).
Ex. 103.32Application
National data show that 68% of online courses at community colleges are taught by full-time faculty. A researcher tests if this applies to California. Surveys 45 instructors at LBCC: 34 are full-time. At the 5% level, is there evidence of a difference?
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$Z = 1.087$ ; p-value $= 0.277$ ; we reject $H_0$ — proportion differs from 68% $Z = 1.087$ ; p-value $= 0.277$ ; we fail to reject $H_0$ — no evidence of difference $Z = 1.087$ ; p-value $= 0.139$ (one-tailed); we reject $H_0$ $Z = 2.174$ ; p-value $= 0.030$ ; we reject $H_0$
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Solve online OpenStax Statistics · §9.3
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$H_0: p = 0.68$ , $H_1: p \neq 0.68$ . $\hat p = 34/45 \approx 0.7556$ . $Z = (0.7556 - 0.68)/\sqrt{0.68 \times 0.32/45} = 0.0756/0.06954 \approx 1.087$ . Two-tailed p-value: $2 \times P(Z > 1.087) \approx 2 \times 0.1385 = 0.277$ . We fail to reject $H_0$ : no evidence that the proportion in California differs from 68%.
Ex. 103.33Understanding
Which statement about the p-value is correct?
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The p-value is the probability that H0 is trueA small p-value proves the effect is clinically relevantThe p-value depends on both effect size and sample sizeFailing to reject H0 proves H0 is true
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Solve online Statistical Thinking for the 21st Century · Cap. 10
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The p-value is $P(\text{data as extreme} \mid H_0)$ . It depends on the observed effect AND the sample size. With very large $n$ , tiny effects become "significant." The p-value does not measure practical relevance nor the probability of $H_0$ being true.
Ex. 103.34ApplicationAnswer key
A city had 14% of residents who walked for exercise. This year, 9 of 70 surveyed residents walk for exercise. At the 5% level, did the rate decrease?
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$Z = -0.275$ ; p-value $\approx 0.392$ ; we reject $H_0$ — rate decreased $Z = -0.275$ ; p-value $\approx 0.392$ ; we fail to reject $H_0$ — no evidence of decrease $Z = -0.275$ ; p-value $\approx 0.784$ (two-tailed); we fail to reject $H_0$ $Z = -1.746$ ; p-value $\approx 0.040$ ; we reject $H_0$
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Solve online OpenStax Statistics · §9.4
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$H_0: p \leq 0.14$ , $H_1: p < 0.14$ . $\hat p = 9/70 \approx 0.1286$ . $Z = (0.1286 - 0.14)/\sqrt{0.14 \times 0.86/70} = -0.0114/0.04142 = -0.275$ . p-value: $P(Z < -0.275) \approx 0.392$ . We fail to reject: no evidence that the rate fell.
Ex. 103.35Modeling
The average national income of registered nurses is US $69,110 annually. A survey of 41 California nurses obtained$ \bar X = $US$ 71,121 $and$ s = $US$ 7,489$. At the 5% level, do California nurses earn more than the national average?
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$H_0: \mu = 69110$ ; $H_1: \mu > 69110$ ; use one-tailed t-test to the right with $n-1$ df $H_0: \mu = 69110$ ; $H_1: \mu \neq 69110$ ; use two-tailed z-test $H_0: \mu > 69110$ ; $H_1: \mu \leq 69110$ ; use one-tailed t-test to the left $H_0: \mu = 69110$ ; $H_1: \mu < 69110$ ; use one-tailed t-test to the left
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Solve online OpenStax Statistics · §9.4
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See the reference indicated in source for the detailed solution.
Ex. 103.36Modeling
According to the AAA, approximately 54% of fatal accidents are caused by driver error. In a sample of 30 accidents, 14 were caused by driver error. At the 5% level, is there evidence that the proportion in the sample differs from 54%?
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$Z = -0.802$ ; p-value $\approx 0.422$ ; we fail to reject $H_0$ — no evidence of difference from national proportion $Z = -0.802$ ; p-value $\approx 0.422$ ; we reject $H_0$ — proportion differs from 54% $Z = -0.802$ ; p-value $\approx 0.211$ (one-tailed); we reject $H_0$ $Z = -1.604$ ; p-value $\approx 0.054$ ; we fail to reject $H_0$
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Solve online OpenStax Statistics · §9.3
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$Z = -0.802$ , p-value $\approx 0.422$ . We fail to reject $H_0$ : no evidence that the proportion of accidents from driver error differs from 54%.
Show step-by-step (with the why)
1. Hypotheses: $H_0: p = 0.54$ , $H_1: p \neq 0.54$ (two-tailed).
2. With 30 accidents, 14 from driver error: $\hat p = 14/30 \approx 0.467$ .
3. $Z = (0.467 - 0.54)/\sqrt{0.54 \times 0.46/30} = -0.073/0.09105 = -0.802$ .
4. Two-tailed p-value: $2 \times P(Z < -0.802) \approx 2 \times 0.211 = 0.422$ .
5. Since $p = 0.422 > 0.05$ , we fail to reject $H_0$ .
Ex. 103.37Modeling
The New York Times reported an average commute time of 25.4 minutes in major US cities. Austin believes the local commute is shorter. Surveys 25 workers: $\bar X = 24.2$ min, $s = 4.3$ min. At the 5% level, does Austin have a significantly shorter commute?
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$t = -1.395$ ; p-value $\approx 0.088$ ; we reject $H_0$ — Austin has shorter commute $t = -1.395$ ; p-value $\approx 0.088$ ; we fail to reject $H_0$ — no evidence at 5% $t = -1.395$ ; p-value $\approx 0.176$ (two-tailed); we reject $H_0$ $t = -1.730$ ; p-value $\approx 0.049$ ; we reject $H_0$ — Austin has shorter commute
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Solve online OpenStax Statistics · §9.4
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Two-tailed mean test with unknown $\sigma$. $H_0: \mu = 25.4$ minutes, $H_1: \mu < 25.4$ . With $n = 25$ , $\bar X = 24.2$ , $s = 4.3$ : $t = (24.2 - 25.4)/(4.3/\sqrt{25}) = -1.2/0.86 = -1.395$ . p-value with $t_{24}$ : $\approx 0.088$ . Do not reject at 5%: no evidence Austin has shorter commute.
Ex. 103.38ModelingAnswer key
A clinical trial tests 20 endpoints with $\alpha = 0.05$ each. (a) What is the probability of at least one false positive, assuming all $H_0$ are true? (b) Apply Bonferroni correction and indicate the new level per test.
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FWER $\approx 64\%$ ; Bonferroni level per test $= 0.0025$ FWER $= 5\%$ (tests are independent, so level does not change); no correction neededFWER $\approx 64\%$ ; Bonferroni level per test $= 0.05 \times 20 = 1.00$ FWER $\approx 26\%$ ; Bonferroni level per test $= 0.0025$
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Solve online Statistical Thinking for the 21st Century · Cap. 11
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Without correction: $P(\geq 1\text{ false positive}) = 1 - (1-0.05)^{20} \approx 1 - 0.358 = 0.642$ . That is, over 60% chance of at least one false positive without correction. With Bonferroni: $\alpha_{\text{corr}} = 0.05/20 = 0.0025$ per test. Application to clinical trial: only results with p-value below 0.0025 would be declared significant, drastically reducing false positives at the cost of less power for each individual test.
Ex. 103.39Challenge
Exercise 103.20 showed approval of 38 of 100 students versus historical rate of 30%, with $p = 0.040$ . Calculate Cohen's effect size $h$ for proportions. Is the result educationally relevant? Compare statistical significance with practical relevance.
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$h \approx 0.172$ (small effect); statistically significant but likely irrelevant in practice $h \approx 0.80$ (large effect); result both statistically and educationally relevant $h \approx 0.172$ (small effect); result statistically non-significant (p-value above 0.05) $h = 0.08$ (absolute difference in proportions); Cohen's medium effect
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Solve online Statistical Thinking for the 21st Century · Cap. 10
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Cohen's effect size for proportions: $h = 2\arcsin(\sqrt{\hat p}) - 2\arcsin(\sqrt{p_0})$ . For $\hat p = 0.38$ and $p_0 = 0.30$ : $h \approx 2(0.666) - 2(0.580) = 0.172$ (small effect). With $n = 100$ : $Z \approx 1.748$ , p-value $\approx 0.040 < 0.05$ . Statistically significant, but effect is small. The pedagogical decision to adopt the methodology should consider cost, scalability, and magnitude of 8 percentage points — not just the p-value.
Ex. 103.40Proof
Derive the analytical expression for the power of a one-tailed z-test ( $H_1: \mu > \mu_0$ ) as a function of $n$ , $\delta = \mu_1 - \mu_0$ , and $\sigma$ . Show that power increases with $n$ and with $\delta/\sigma$ (signal-to-noise ratio).
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Power $= 1 - \Phi(z_{\alpha} - \delta\sqrt{n}/\sigma)$ ; increasing in $n$ and $\delta/\sigma$ Power $= \Phi(z_{\alpha} - \delta\sqrt{n}/\sigma)$ ; decreasing in $n$ Power $= 1 - \Phi(z_{\alpha} + \delta\sqrt{n}/\sigma)$ ; decreasing in $\delta/\sigma$ Power $= \alpha$ regardless of $n$ and $\delta$ — power is fixed by the test level
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Solve online OpenIntro Statistics · §5.3
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Power is $1 - \beta = P(\text{reject } H_0 \mid H_1 \text{ true})$ . For one-tailed test with $\\sigma$ known and $\mu_1 > \mu_0$ : we reject if $Z = (\bar X - \mu_0)/(\sigma/\sqrt{n}) > z_{\alpha}$ . Under $H_1$ : $\bar X \sim \mathcal{N}(\mu_1, \sigma^2/n)$ . Power = $P(Z > z_{alpha} \mid \mu_1) = P\left(\frac{\bar X - \mu_0}{\sigma/\sqrt{n}} > z_{alpha} \mid \mu_1\right) = P\left(Z > z_{alpha} - \frac{(\mu_1 - \mu_0)\sqrt{n}}{\sigma}\right) = 1 - \Phi\left(z_{alpha} - \frac{\delta\sqrt{n}}{\sigma}\right)$ where $\delta = \mu_1 - \mu_0$ . Increases with $n$ , $\delta$ and decreases with $\sigma$ .
Ex. 103.41ChallengeAnswer key
A genome-wide association study (GWAS) tests 5,000 genetic polymorphisms in 10 students, using $\alpha = 0.05$ for each test without correction. (a) How many false positives are expected? (b) What would the Bonferroni-corrected level be? (c) Discuss why GWAS studies with small samples have low reliability.
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250 false positives; $\alpha_{\text{Bonf}} = 10^{-5}$ ; with small $n$ , most significant results are false positives50 false positives; $\alpha_{\text{Bonf}} = 10^{-3}$ ; GWAS studies are reliable because they use multiple testing250 false positives; $\alpha_{\text{Bonf}} = 10^{-5}$ ; Bonferroni is inappropriate for GWAS25 false positives; $\alpha_{\text{Bonf}} = 0.05$ ; the level does not change with multiple tests
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Solve online Statistical Thinking for the 21st Century · Cap. 11
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Type I inflation via multiple testing: with 10 students testing 5,000 genetic polymorphisms each, at 5% level, expect $5000 \times 0.05 = 250$ false positives per student, or 2,500 total. Bonferroni correction: $\alpha = 0.05/5000 = 10^{-5}$ per test. Alternative: FDR (Benjamini-Hochberg) is more powerful in GWAS. In studies with small $n$ relative to the number of tests, most "significant" results are false positives — illustration of the replicability crisis in genomics.
Ex. 103.42Application
The natural birth ratio is approximately 100 girls per 105 boys (proportion of girls ≈ 48.8%). In China it is reported as 100:114 (≈ 46.7% girls). A researcher studies 150 births and finds 60 girls. At the 5% level, is the proportion of girls significantly different from 48.8%?
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$Z = -2.157$ ; p-value $= 0.031$ ; we fail to reject $H_0$ — proportion compatible with 48.8% $Z = -2.157$ ; p-value $= 0.031$ ; we reject $H_0$ — proportion significantly differs from 48.8% $Z = -2.157$ ; p-value $= 0.016$ (one-tailed); we fail to reject $H_0$ $Z = -1.320$ ; p-value $= 0.187$ ; we fail to reject $H_0$
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Solve online OpenStax Statistics · §9.3
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$H_0: p = 0.488$ (natural ratio), $H_1: p \neq 0.488$ . China: $\hat p = 60/150 = 0.40$ . $Z = (0.40 - 0.488)/\sqrt{0.488 \times 0.512/150} = -0.088/0.0408 = -2.157$ . Two-tailed p-value: $2 \times P(Z < -2.157) \approx 2 \times 0.0155 = 0.031$ . We reject at 5%: the proportion of girls in the studied births is significantly different from the natural ratio.

Sources

OpenIntro Statistics (4th ed.) — Diez, Çetinkaya-Rundel, Barr · CC-BY-SA. Sections §5.1–5.3 (test structure, p-value, power, sample size).
Statistics (OpenStax) — Illowsky, Dean · CC-BY. Chapter 9 (null and alternative hypotheses, Type I and II errors, complete examples with z).
Statistical Thinking for the 21st Century — Russell Poldrack · CC-BY-NC. Chapters 10–11 (replicability crisis, responsible p-value use, FDR, effect size).