Power & Sample Size

Estimating Sample Size and Power from Parameters

Author

03-Power and Sample Size

Conventions: Two‑sided tests at α = 0.05 unless stated. Use R where helpful. Symbols: δ = effect size in the outcome scale; σ = SD; n = total sample unless noted. For two‑sample tests, assume equal allocation and equal variances unless stated.

Using agentic AI: You may use an agent to derive a formula, draft R code, or suggest a simulation. Treat it as an unverified collaborator: state your assumptions, run the code, and reconcile any disagreement with an independent calculation. For the design challenge, include the prompt(s) used and a concise verification log.

1) One‑sample t‑test power & n (10 minutes)

Setup. You expect observations $Y\sim \mathcal{N}(\mu,\sigma^2)$ with $\sigma=10$. You will test $H_0: \mu=0$ vs $H_1: \mu\neq 0$. The scientifically relevant effect is $\delta = 3$ (i.e., $\mu=3$).

1A. Compute power for $n=50$.

Answer

Analytic normal approx: $Z = \bar Y/(\sigma/\sqrt n)$, noncentrality $\lambda = \delta\sqrt n/\sigma$.

\[ \lambda = 3\cdot\sqrt{50}/10 = 3\cdot7.071/10 \approx 2.121. \]

Two‑sided power (z‑approx): \[ 1 - \Phi(z_{0.975}-\lambda) + \Phi(-z_{0.975}-\lambda) \approx 1-\Phi(1.96-2.121)+\Phi(-1.96-2.121) \approx 1-\Phi(-0.161)+\Phi(-4.081). \] This is approximately $1-(0.436)+\text{tiny} \approx 0.564$. Using exact t power:

Code

power.t.test(n = 50, delta = 3, sd = 10, sig.level = 0.05,
             type = "one.sample", alternative = "two.sided")$power

1B. Required (n) for 80% power.

Answer

Closed‑form (z‑approx): $n \approx (z_{0.975}+z_{0.8})^2 \frac{\sigma^2}{\delta^2}$.

\[ n \approx (1.96+0.84)^2\frac{10^2}{3^2} = (2.80)^2 \cdot 100/9 \approx 7.84 \cdot 11.11 \approx 87.1. \]

Exact via R:

Code

power.t.test(power = 0.8, delta = 3, sd = 10, sig.level = 0.05,
             type = "one.sample", alternative = "two.sided")$n

2) Two‑sample t‑test (equal variance) (8 minutes)

Setup. Two arms, equal allocation, common $\sigma=12$, difference of means $\delta = 5$.

2A. Per‑group (n) for 90% power.

Answer

Z‑approx per‑group (n): $n \approx 2 (z_{0.975}+z_{0.90})^2 \frac{\sigma^2}{\delta^2}$.

\[ n \approx 2(1.96+1.282)^2(12^2)/5^2 = 2(3.242)^2\cdot144/25 \approx 2\cdot10.52\cdot5.76 \approx 121.1 \]

So about 122 per group. Exact:

Code

power.t.test(power = 0.9, delta = 5, sd = 12, sig.level = 0.05,
             type = "two.sample", alternative = "two.sided")$n

2B. Power if $n=40$ per group.

Answer

Code

power.t.test(n = 40, delta = 5, sd = 12, sig.level = 0.05,
             type = "two.sample", alternative = "two.sided")$power

3) Paired t‑test (5 minutes)

Setup. Pre/post measurements with $\sigma_{\text{pre}}=\sigma_{\text{post}}=8$, correlation $\rho=0.6$. Mean change $\delta=2$.

Question. Required number of pairs for 80% power.

Answer

SD of differences: $\sigma_D=\sqrt{\sigma^2+\sigma^2-2\rho\sigma^2}=\sqrt{2\sigma^2(1-\rho)} = 8\sqrt{2(1-0.6)}=8\sqrt{0.8}\approx 7.155$.) Use one‑sample t-est on differences with (sd=$\sigma_D$, effect $\delta$:

Code

power.t.test(power = 0.8, delta = 2, sd = 7.155, sig.level = 0.05,
             type = "one.sample", alternative = "two.sided")$n

4) Simple linear regression: test a slope (10 minutes)

Setup. Model $Y=\beta_0+\beta_1 X+\varepsilon$, $\varepsilon \sim N(0,\sigma^2)$ with $\sigma=10$ and (X) standardized: $\bar{X} \approx 0$, $\operatorname{Var}(X)=1$.

Suppose $\beta_1=3$. Test $H_0:\beta_1=0$.

4A. Power at $n=60$ using the noncentral t formulation.

Answer

For centered/standardized (X), $S_{xx} = \sum (x_i-\bar x)^2 \approx n-1$. The test statistic for $\beta_1$ has noncentrality paramter \[ \lambda = \frac{\beta_1}{\sigma}\sqrt{S_{xx}} \approx \frac{3}{10}\sqrt{59} = 0.3 \cdot 7.681 \approx 2.304 \] Power (two‑sided) is $1-\beta$ with df=$n-2=58$:

Code

n <- 60;
sigma <- 10;
beta1 <- 3
lambda <- beta1/sigma * sqrt(n-1)
alpha <- 0.05;
df <- n - 2
c <- qt(1 - alpha/2, df)
# power = P(|T_nc| > c)
1 - (pt(c, df, ncp=lambda) - pt(-c, df, ncp=lambda))

4B. Alternative via correlation. Show equivalence.

Answer

For SLR, $r=\operatorname{Cor}(X,Y)= \frac{\beta_1 \sigma_X}{\sigma_Y}$. With $\sigma_X=1$, $\sigma_Y = \sqrt{\beta_1^2+\sigma^2} = \sqrt{9+100} = \sqrt{109} \approx 10.44$, so $r\approx 3/10.44\approx 0.287$.

Test of $r=0$ uses the same df and noncentrality $\sqrt{n-2}, \frac{r}{\sqrt{1-r^2}}$, which equals $\lambda$ above.

Code

r <- 3/sqrt(9+100)
lambda2 <- sqrt(n-2) * r / sqrt(1 - r^2)
all.equal(lambda, lambda2)

5) Multiple regression: overall $R^2$ (7 minutes)

Setup. You plan (p=4) predictors and expect population $R^2=0.20$. Test $H_0: R^2=0$ (overall model).

Find power at $n=80$ and required (n) for 80% power.

Answer

Use Cohen’s $f^2=\frac{R^2}{1-R^2}=0.25$. The overall F test has df1=(p), df2=(n-p-1), and noncentrality $\lambda=f^2(n-p-1)$.

Power at (n=80):

Code

p <- 4; R2 <- 0.20; f2 <- R2/(1-R2)
n <- 80; df1 <- p; df2 <- n - p - 1
lambda <- f2 * df2
alpha <- 0.05; Fc <- qf(1 - alpha, df1, df2)
# P(F_nc > Fc)
power <- 1 - pf(Fc, df1, df2, ncp=lambda)
power

Solve for (n) (80% power):

Code

f_target <- function(n){
  df2 <- n - p - 1
  if (df2 <= 0) return(-1)  # invalid
  Fc <- qf(0.95, df1=p, df2=df2)
  1 - pf(Fc, df1=p, df2=df2, ncp=f2*df2) - 0.80
}
uniroot(f_target, c(p+5, 1000))$root

6) Sample size for prediction interval width in SLR (10 minutes)

Setup. In $Y=\beta_0+\beta_1 X+\varepsilon$ with $\varepsilon\sim N(0,\sigma^2)$, suppose $\sigma=10$ and you plan $X$ standardized (mean 0, var 1). You want a 95% prediction interval for a future observation at $x_0$ to have total width $W$.

Consider two cases: (i) $x_0=0$ (mean of (X)), (ii) $x_0=1$ (1 SD from mean).

Facts. The half‑width (HW) at $x_0$ is \[ \text{HW}(x_0) = t_{0.975, n-2} \sigma \sqrt{1+\frac{1}{n}+\frac{(x_0-\bar x)^2}{S_{xx}}},\quad S_{xx}=\sum (x_i-\bar x)^2\approx n-1. \]

The minimum achievable half‑width as $n\to\infty$ is $t_{\infty,0.975}\sigma$, where $t_{\infty,0.975}=1.96$.

6A. Feasibility check. With $\sigma=10$, what is the smallest possible total width $W_\text{min}$?

Answer

$W_\text{min}=2\times 1.96\times 10 \approx 39.2$ Any target width $W<39.2$ is impossible regardless of $n$.

6B. Find $n$ to achieve $W=48$ (half‑width 24) at $x_0=0$. You can brute force it, but also try using root-finding/optimization procedures.

Answer

Solve $t_{0.975,n-2} 10 \sqrt{1+1/n} = 24$. Use numeric root‑finding:

Code

sigma <- 10; HW <- 24
f <- function(n){
  df <- n - 2
  if (df <= 0) return(1e6)
  t <- qt(0.975, df)
  t * sigma * sqrt(1 + 1/n) - HW
}
ceiling(uniroot(f, c(5, 1e5))$root)

6C. Repeat for $x_0=1$. Use $S_{xx}\approx n-1$, $\bar x\approx 0$.

Answer

Half‑width equation: ( $t \sigma\sqrt{1+1/n+1/(n-1)} = 24$ ).

Code

sigma <- 10; HW <- 24
f <- function(n){
  df <- n - 2
  if (df <= 2) return(1e6)
  t <- qt(0.975, df)
  Sxx <- n - 1
  t * sigma * sqrt(1 + 1/n + 1/Sxx) - HW
}
ceiling(uniroot(f, c(6, 1e6))$root)

Note the larger (n) due to being 1 SD away from the design mean.

6D. (Optional) Design‑sensitive planning. If you can spread (X) to increase $S_{xx}$ (e.g., choose equally spaced design over a range), how does that change (n) for fixed (W)?

Answer

Increasing $S_{xx}$ shrinks the term $\frac{(x_0- \bar{x})^2}{S_{xx}}$, reducing the half‑width for off‑mean predictions. For a fixed $n$, wider X range lowers HW at $x_0\neq\bar x$. Conversely, for fixed HW, larger $S_{xx}$ allows smaller (n).

7) (Optional) Monte Carlo verification (5 minutes)

Simulate to verify analytic power for a two‑sample t‑test or slope test.

Answer

Code

set.seed(1)
B <- 5000
n <- 40; sd <- 12; delta <- 5; alpha <- 0.05
p <- replicate(B, {
  x1 <- rnorm(n, 0, sd); x2 <- rnorm(n, delta, sd)
  t.test(x2, x1, var.equal = TRUE)$p.value
})
mean(p < alpha)  # ~ power

8) Design-review challenge: plan under uncertainty (20 minutes)

You are planning a two-arm study with a continuous primary outcome. A pilot suggests an SD between 10 and 14; the smallest effect worth detecting is 4 units; you can recruit at most 130 participants total; and anticipate 15% dropout. The primary analysis is a two-sided two-sample t-test at α = 0.05 with equal allocation.

Deliverable. Write a one-page design recommendation that:

calculates the analyzable and enrolled sample sizes needed for 80% power at SD = 10, 12, and 14;
reports achieved power at the maximum feasible enrollment after dropout;
gives a transparent recommendation and one contingency if the pilot SD is too optimistic; and
verifies one analytic result with a simulation (state the number of replicates and Monte Carlo uncertainty).

You may ask an AI agent for a first-pass calculation or code, but include your prompt, independently check one output, and explain any correction you made. The grade is based on the reasoning and audit trail, not on whether an agent produced code quickly.

8) Design-review challenge: plan under uncertainty (20 minutes)

Deliverable. Write a one-page design recommendation that:

calculates the analyzable and enrolled sample sizes needed for 80% power at SD = 10, 12, and 14;
reports achieved power at the maximum feasible enrollment after dropout;
gives a transparent recommendation and one contingency if the pilot SD is too optimistic; and
verifies one analytic result with a simulation (state the number of replicates and Monte Carlo uncertainty).

Quick reference

One‑sample n (z‑approx): $n = (z_{1-\alpha/2}+z_{1-\beta})^2 \frac{\sigma^2}{\delta^2}$
Two‑sample per‑group n: $n = 2(z_{1-\alpha/2}+z_{1-\beta})^2 \frac{\sigma^2}{\delta^2}$
SLR slope power ncp: $\lambda = (\beta_1/\sigma)\sqrt{S_{xx}} \approx (\beta_1/\sigma)\sqrt{n-1}$
Overall MR test: $f^2=R^2/(1-R^2)$, $\lambda=f^2(n-p-1)$
PI half‑width at $x_0$: $t\sigma\sqrt{1+1/n+(x_0-\bar x)^2/S_{xx}}$

--- title: "Power & Sample Size" subtitle: "Estimating Sample Size and Power from Parameters" author: "03-Power and Sample Size" format: html: toc: true code-fold: true code-tools: true editor: source engine: knitr --- > **Conventions:** Two‑sided tests at α = 0.05 unless stated. Use R where helpful. Symbols: δ = effect size in the outcome scale; σ = SD; n = total sample unless noted. For two‑sample tests, assume equal allocation and equal variances unless stated. > **Using agentic AI:** You may use an agent to derive a formula, draft R code, or suggest a simulation. Treat it as an unverified collaborator: state your assumptions, run the code, and reconcile any disagreement with an independent calculation. For the design challenge, include the prompt(s) used and a concise verification log. ------------------------------------------------------------------------ ## 1) One‑sample t‑test power & n (10 minutes) **Setup.** You expect observations $Y\sim \mathcal{N}(\mu,\sigma^2)$ with $\sigma=10$. You will test $H_0: \mu=0$ vs $H_1: \mu\neq 0$. The scientifically relevant effect is $\delta = 3$ (i.e., $\mu=3$). **1A.** Compute power for $n=50$. <details> <summary>Answer</summary> Analytic normal approx: $Z = \bar Y/(\sigma/\sqrt n)$, noncentrality $\lambda = \delta\sqrt n/\sigma$. $$ \lambda = 3\cdot\sqrt{50}/10 = 3\cdot7.071/10 \approx 2.121. $$ Two‑sided power (z‑approx): $$ 1 - \Phi(z_{0.975}-\lambda) + \Phi(-z_{0.975}-\lambda) \approx 1-\Phi(1.96-2.121)+\Phi(-1.96-2.121) \approx 1-\Phi(-0.161)+\Phi(-4.081). $$ This is approximately $1-(0.436)+\text{tiny} \approx 0.564$. Using exact **t** power: ```{r, eval = FALSE} power.t.test(n = 50, delta = 3, sd = 10, sig.level = 0.05, type = "one.sample", alternative = "two.sided")$power ``` </details> **1B.** Required (n) for 80% power. <details> <summary>Answer</summary> Closed‑form (z‑approx): $n \approx (z_{0.975}+z_{0.8})^2 \frac{\sigma^2}{\delta^2}$. $$ n \approx (1.96+0.84)^2\frac{10^2}{3^2} = (2.80)^2 \cdot 100/9 \approx 7.84 \cdot 11.11 \approx 87.1. $$ Exact via R: ```{r, eval = FALSE} power.t.test(power = 0.8, delta = 3, sd = 10, sig.level = 0.05, type = "one.sample", alternative = "two.sided")$n ``` </details> ------------------------------------------------------------------------ ## 2) Two‑sample t‑test (equal variance) (8 minutes) **Setup.** Two arms, equal allocation, common $\sigma=12$, difference of means $\delta = 5$. **2A.** Per‑group (n) for 90% power. <details> <summary>Answer</summary> Z‑approx per‑group (n): $n \approx 2 (z_{0.975}+z_{0.90})^2 \frac{\sigma^2}{\delta^2}$. $$ n \approx 2(1.96+1.282)^2(12^2)/5^2 = 2(3.242)^2\cdot144/25 \approx 2\cdot10.52\cdot5.76 \approx 121.1 $$ So about **122 per group**. Exact: ```{r, eval = FALSE} power.t.test(power = 0.9, delta = 5, sd = 12, sig.level = 0.05, type = "two.sample", alternative = "two.sided")$n ``` </details> **2B.** Power if $n=40$ per group. <details> <summary>Answer</summary> ```{r, eval = FALSE} power.t.test(n = 40, delta = 5, sd = 12, sig.level = 0.05, type = "two.sample", alternative = "two.sided")$power ``` </details> ------------------------------------------------------------------------ ## 3) Paired t‑test (5 minutes) **Setup.** Pre/post measurements with $\sigma_{\text{pre}}=\sigma_{\text{post}}=8$, correlation $\rho=0.6$. Mean change $\delta=2$. **Question.** Required number of pairs for 80% power. <details> <summary>Answer</summary> SD of differences: $\sigma_D=\sqrt{\sigma^2+\sigma^2-2\rho\sigma^2}=\sqrt{2\sigma^2(1-\rho)} = 8\sqrt{2(1-0.6)}=8\sqrt{0.8}\approx 7.155$.) Use one‑sample t-est on differences with (sd=$\sigma_D$, effect $\delta$: ```{r, eval = FALSE} power.t.test(power = 0.8, delta = 2, sd = 7.155, sig.level = 0.05, type = "one.sample", alternative = "two.sided")$n ``` </details> ------------------------------------------------------------------------ ## 4) Simple linear regression: test a slope (10 minutes) **Setup.** Model $Y=\beta_0+\beta_1 X+\varepsilon$, $\varepsilon \sim N(0,\sigma^2)$ with $\sigma=10$ and (X) standardized: $\bar{X} \approx 0$, $\operatorname{Var}(X)=1$. Suppose $\beta_1=3$. Test $H_0:\beta_1=0$. **4A.** Power at $n=60$ using the noncentral t formulation. <details> <summary>Answer</summary> For centered/standardized (X), $S_{xx} = \sum (x_i-\bar x)^2 \approx n-1$. The test statistic for $\beta_1$ has noncentrality paramter $$ \lambda = \frac{\beta_1}{\sigma}\sqrt{S_{xx}} \approx \frac{3}{10}\sqrt{59} = 0.3 \cdot 7.681 \approx 2.304 $$ Power (two‑sided) is $1-\beta$ with df=$n-2=58$: ```{r, eval = FALSE} n <- 60; sigma <- 10; beta1 <- 3 lambda <- beta1/sigma * sqrt(n-1) alpha <- 0.05; df <- n - 2 c <- qt(1 - alpha/2, df) # power = P(|T_nc| > c) 1 - (pt(c, df, ncp=lambda) - pt(-c, df, ncp=lambda)) ``` </details> **4B.** Alternative via correlation. Show equivalence. <details> <summary>Answer</summary> For SLR, $r=\operatorname{Cor}(X,Y)= \frac{\beta_1 \sigma_X}{\sigma_Y}$. With $\sigma_X=1$, $\sigma_Y = \sqrt{\beta_1^2+\sigma^2} = \sqrt{9+100} = \sqrt{109} \approx 10.44$, so $r\approx 3/10.44\approx 0.287$. Test of $r=0$ uses the same df and noncentrality $\sqrt{n-2}, \frac{r}{\sqrt{1-r^2}}$, which equals $\lambda$ above. ```{r, eval = FALSE} r <- 3/sqrt(9+100) lambda2 <- sqrt(n-2) * r / sqrt(1 - r^2) all.equal(lambda, lambda2) ``` </details> ------------------------------------------------------------------------ ## 5) Multiple regression: overall $R^2$ (7 minutes) **Setup.** You plan (p=4) predictors and expect population $R^2=0.20$. Test $H_0: R^2=0$ (overall model). Find power at $n=80$ and required (n) for 80% power. <details> <summary>Answer</summary> Use Cohen's $f^2=\frac{R^2}{1-R^2}=0.25$. The overall F test has df1=(p), df2=(n-p-1), and noncentrality $\lambda=f^2(n-p-1)$. Power at (n=80): ```{r, eval = FALSE} p <- 4; R2 <- 0.20; f2 <- R2/(1-R2) n <- 80; df1 <- p; df2 <- n - p - 1 lambda <- f2 * df2 alpha <- 0.05; Fc <- qf(1 - alpha, df1, df2) # P(F_nc > Fc) power <- 1 - pf(Fc, df1, df2, ncp=lambda) power ``` Solve for (n) (80% power): ```{r, eval = FALSE} f_target <- function(n){ df2 <- n - p - 1 if (df2 <= 0) return(-1) # invalid Fc <- qf(0.95, df1=p, df2=df2) 1 - pf(Fc, df1=p, df2=df2, ncp=f2*df2) - 0.80 } uniroot(f_target, c(p+5, 1000))$root ``` </details> ------------------------------------------------------------------------ ## 6) Sample size for **prediction interval width** in SLR (10 minutes) **Setup.** In $Y=\beta_0+\beta_1 X+\varepsilon$ with $\varepsilon\sim N(0,\sigma^2)$, suppose $\sigma=10$ and you plan $X$ standardized (mean 0, var 1). You want a **95% prediction interval** for a *future* observation at $x_0$ to have total width $W$. Consider two cases: (i) $x_0=0$ (mean of (X)), (ii) $x_0=1$ (1 SD from mean). **Facts.** The half‑width (HW) at $x_0$ is $$ \text{HW}(x_0) = t_{0.975, n-2} \sigma \sqrt{1+\frac{1}{n}+\frac{(x_0-\bar x)^2}{S_{xx}}},\quad S_{xx}=\sum (x_i-\bar x)^2\approx n-1. $$ The **minimum achievable** half‑width as $n\to\infty$ is $t_{\infty,0.975}\sigma$, where $t_{\infty,0.975}=1.96$. **6A.** Feasibility check. With $\sigma=10$, what is the smallest possible total width $W_\text{min}$? <details> <summary>Answer</summary> $W_\text{min}=2\times 1.96\times 10 \approx 39.2$ Any target width $W<39.2$ is **impossible** regardless of $n$. </details> **6B.** Find $n$ to achieve $W=48$ (half‑width 24) at $x_0=0$. You can brute force it, but also try using root-finding/optimization procedures. <details> <summary>Answer</summary> Solve $t_{0.975,n-2} 10 \sqrt{1+1/n} = 24$. Use numeric root‑finding: ```{r, eval = FALSE} sigma <- 10; HW <- 24 f <- function(n){ df <- n - 2 if (df <= 0) return(1e6) t <- qt(0.975, df) t * sigma * sqrt(1 + 1/n) - HW } ceiling(uniroot(f, c(5, 1e5))$root) ``` </details> **6C.** Repeat for $x_0=1$. Use $S_{xx}\approx n-1$, $\bar x\approx 0$. <details> <summary>Answer</summary> Half‑width equation: ( $t \sigma\sqrt{1+1/n+1/(n-1)} = 24$ ). ```{r, eval = FALSE} sigma <- 10; HW <- 24 f <- function(n){ df <- n - 2 if (df <= 2) return(1e6) t <- qt(0.975, df) Sxx <- n - 1 t * sigma * sqrt(1 + 1/n + 1/Sxx) - HW } ceiling(uniroot(f, c(6, 1e6))$root) ``` Note the larger (n) due to being 1 SD away from the design mean. </details> **6D. (Optional)** Design‑sensitive planning. If you can spread (X) to increase $S_{xx}$ (e.g., choose equally spaced design over a range), how does that change (n) for fixed (W)? <details> <summary>Answer</summary> Increasing $S_{xx}$ shrinks the term $\frac{(x_0- \bar{x})^2}{S_{xx}}$, reducing the half‑width for off‑mean predictions. For a fixed $n$, wider X range lowers HW at $x_0\neq\bar x$. Conversely, for fixed HW, larger $S_{xx}$ allows smaller (n). </details> ------------------------------------------------------------------------ ## 7) (Optional) Monte Carlo verification (5 minutes) Simulate to verify analytic power for a two‑sample t‑test or slope test. <details> <summary>Answer</summary> ```{r, eval = FALSE} set.seed(1) B <- 5000 n <- 40; sd <- 12; delta <- 5; alpha <- 0.05 p <- replicate(B, { x1 <- rnorm(n, 0, sd); x2 <- rnorm(n, delta, sd) t.test(x2, x1, var.equal = TRUE)$p.value }) mean(p < alpha) # ~ power ``` </details> ------------------------------------------------------------------------ ## 8) Design-review challenge: plan under uncertainty (20 minutes) You are planning a two-arm study with a continuous primary outcome. A pilot suggests an SD between 10 and 14; the smallest effect worth detecting is 4 units; you can recruit at most 130 participants total; and anticipate 15% dropout. The primary analysis is a two-sided two-sample t-test at α = 0.05 with equal allocation. **Deliverable.** Write a one-page design recommendation that: - calculates the analyzable and enrolled sample sizes needed for 80% power at SD = 10, 12, and 14; - reports achieved power at the maximum feasible enrollment after dropout; - gives a transparent recommendation and one contingency if the pilot SD is too optimistic; and - verifies one analytic result with a simulation (state the number of replicates and Monte Carlo uncertainty). You may ask an AI agent for a first-pass calculation or code, but include your prompt, independently check one output, and explain any correction you made. The grade is based on the reasoning and audit trail, not on whether an agent produced code quickly. ------------------------------------------------------------------------ ## 8) Design-review challenge: plan under uncertainty (20 minutes) You are planning a two-arm study with a continuous primary outcome. A pilot suggests an SD between 10 and 14; the smallest effect worth detecting is 4 units; you can recruit at most 130 participants total; and anticipate 15% dropout. The primary analysis is a two-sided two-sample t-test at α = 0.05 with equal allocation. **Deliverable.** Write a one-page design recommendation that: - calculates the analyzable and enrolled sample sizes needed for 80% power at SD = 10, 12, and 14; - reports achieved power at the maximum feasible enrollment after dropout; - gives a transparent recommendation and one contingency if the pilot SD is too optimistic; and - verifies one analytic result with a simulation (state the number of replicates and Monte Carlo uncertainty). You may ask an AI agent for a first-pass calculation or code, but include your prompt, independently check one output, and explain any correction you made. The grade is based on the reasoning and audit trail, not on whether an agent produced code quickly. ------------------------------------------------------------------------ ### Quick reference - One‑sample n (z‑approx): $n = (z_{1-\alpha/2}+z_{1-\beta})^2 \frac{\sigma^2}{\delta^2}$ - Two‑sample per‑group n: $n = 2(z_{1-\alpha/2}+z_{1-\beta})^2 \frac{\sigma^2}{\delta^2}$ - SLR slope power ncp: $\lambda = (\beta_1/\sigma)\sqrt{S_{xx}} \approx (\beta_1/\sigma)\sqrt{n-1}$ - Overall MR test: $f^2=R^2/(1-R^2)$, $\lambda=f^2(n-p-1)$ - PI half‑width at $x_0$: $t\sigma\sqrt{1+1/n+(x_0-\bar x)^2/S_{xx}}$

1) One‑sample t‑test power & n (10 minutes)

2) Two‑sample t‑test (equal variance) (8 minutes)

3) Paired t‑test (5 minutes)

4) Simple linear regression: test a slope (10 minutes)

5) Multiple regression: overall \(R^2\) (7 minutes)

6) Sample size for prediction interval width in SLR (10 minutes)

7) (Optional) Monte Carlo verification (5 minutes)

8) Design-review challenge: plan under uncertainty (20 minutes)

8) Design-review challenge: plan under uncertainty (20 minutes)

Quick reference