Lesson 22: One Proportion Z-Test

What We Did: Lessons 17–21

The Central Limit Theorem (CLT): If \(X_1, X_2, \ldots, X_n\) are iid with mean \(\mu\) and standard deviation \(\sigma\), then for large \(n\):

\[\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)\]

Standard Error of the Mean: \(\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\)

Rule of thumb: \(n \geq 30\) unless the population is already normal.

Confidence Interval for a Mean:

Formula When to Use
Large sample (\(n \geq 30\)) \(\bar{X} \pm z_{\alpha/2} \cdot \dfrac{\sigma}{\sqrt{n}}\) Random sample, independence, \(s \approx \sigma\)
Small sample (\(n < 30\)) \(\bar{X} \pm t_{\alpha/2, n-1} \cdot \dfrac{s}{\sqrt{n}}\) Random sample, independence, population ~ Normal

Key ideas: Higher confidence → wider interval. Larger \(n\) → narrower interval.

Confidence Interval for a Proportion:

\[\hat{p} \pm z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]

Conditions: \(n\hat{p} \geq 10\) and \(n(1-\hat{p}) \geq 10\)

Interpretation: “We are C% confident that [interval] captures the true [parameter in context].” The confidence level describes the method’s long-run success rate, not the probability any single interval is correct.

Every hypothesis test follows four steps:

  1. State hypotheses: \(H_0\) (null — status quo) vs. \(H_a\) (alternative — what we want to show)
  2. Compute a test statistic: How far is our sample result from what \(H_0\) predicts?
  3. Find the \(p\)-value: If \(H_0\) were true, how likely is a result this extreme or more?
  4. Make a decision: If \(p \leq \alpha\), reject \(H_0\). If \(p > \alpha\), fail to reject \(H_0\).

One-sample \(t\)-test for a mean (\(\sigma\) unknown):

\[t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}, \qquad df = n - 1\]

Conditions: Random sample, independence (\(n < 10\%\) of population), population approximately normal or \(n \geq 30\).

\(p\)-value: Use pt() with \(df = n-1\).

What We’re Doing: Lesson 22

Objectives

  • Review the one-proportion \(z\)-test
  • Test \(\mu_1 - \mu_2\) with independent samples
  • Choose pooled vs. Welch procedures appropriately
  • Construct and interpret CIs for \(\mu_1 - \mu_2\)

Required Reading

Devore, Sections 8.4, 9.2

Break!

Reese

Cal

DMath Basketball!!

Math vs DELPHI

12-5

DMath Basketball!!

Math vs DELPHI

13-5

Review: The Inference Toolkit

NoteSummary of Confidence Intervals
One-Sample Mean (Large Sample) One-Sample Mean (Small Sample) One Proportion
Parameter \(\mu\) \(\mu\) \(p\)
Formula \(\bar{x} \pm z_{\alpha/2} \cdot \dfrac{\sigma}{\sqrt{n}}\) \(\bar{x} \pm t_{\alpha/2,\, n-1} \cdot \dfrac{s}{\sqrt{n}}\) \(\hat{p} \pm z_{\alpha/2} \cdot \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\)
Conditions \(n \geq 30\) \(n \leq 30\) \(n\hat{p} \geq 10\) & \(n(1-\hat{p}) \geq 10\)
NoteSummary of Hypothesis Tests
One-Sample Mean (Large Sample) One-Sample Mean (Small Sample) One Proportion Two-Sample Mean Paired Mean Two Proportions
Parameter \(\mu\) \(\mu\) \(p\)
\(H_0\) \(\mu = \mu_0\) \(\mu = \mu_0\) \(p = p_0\)
\(H_a\) \(\mu \neq, <, > \mu_0\) \(\mu \neq, <, > \mu_0\) \(p \neq, <, > p_0\)
Test Statistic \(z = \dfrac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}\) \(t = \dfrac{\bar{x} - \mu_0}{s / \sqrt{n}}\) \(z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}}\)
Distribution \(N(0,1)\) \(t_{n-1}\) \(N(0,1)\)
Left-tailed \(p\)-value pnorm(z) pt(t, df=n-1) pnorm(z)
Right-tailed \(p\)-value 1 - pnorm(z) 1 - pt(t, df=n-1) 1 - pnorm(z)
Two-tailed \(p\)-value 2*(1 - pnorm(abs(z))) 2*(1 - pt(abs(t), df=n-1)) 2*(1 - pnorm(abs(z)))
Conditions \(n \geq 30\) Normal pop or \(n \geq 30\) \(np_0 \geq 10\) & \(n(1-p_0) \geq 10\)

Decision rule is always the same: \(p \leq \alpha\) → Reject \(H_0\). \(p > \alpha\) → Fail to reject \(H_0\).

Last Class: The One-Proportion \(z\)-Test

Last class we tested hypotheses about a population mean \(\mu\). Now we shift to a population proportion \(p\).

ImportantOne-Proportion z-Test

Hypotheses:

\(H_0: p = p_0\)

\(H_a: p < p_0\)
\(H_a: p > p_0\)
\(H_a: p \neq p_0\)

Test statistic:

\[z = \frac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1 - p_0)}{n}}}\]

Conditions: \(np_0 \geq 10\) and \(n(1 - p_0) \geq 10\)

Review Example: DMath Basketball Free Throws

The DMath basketball team claims they shoot better than 50% from the free-throw line. In their last \(n = 80\) free-throw attempts, they made 48.

Check conditions: \(np_0 = 80(0.50) = 40 \geq 10\) and \(n(1-p_0) = 80(0.50) = 40 \geq 10\). Good to go.

Step 1: State the Hypotheses

\[H_0: p = 0.50 \qquad \text{vs.} \qquad H_a: p > 0.50\]

Step 2: Compute the Test Statistic

\[z = \frac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1 - p_0)}{n}}} = \frac{0.60 - 0.50}{\sqrt{\dfrac{0.50(0.50)}{80}}} = \frac{0.10}{0.0559} = 1.79\]

n <- 80; x <- 48
p_hat <- x / n; p0 <- 0.50
se <- sqrt(p0 * (1 - p0) / n)
z <- (p_hat - p0) / se
z
[1] 1.788854

Step 3: Find the \(p\)-Value

This is right-tailed, so we want \(P(Z \geq z)\):

p_value <- 1 - pnorm(z)
p_value
[1] 0.03681914

\[p\text{-value} = P(Z \geq 1.79) = 0.0368\]

Step 4: Decide and Conclude

At \(\alpha = 0.05\): \(p = 0.0368 \leq 0.05\). We reject \(H_0\). At the 5% significance level, there is sufficient evidence that DMath shoots better than 50% from the free-throw line.

New Material: The Two-Sample \(t\)-Test

So far every test we’ve done compares one sample to a fixed value (\(\mu_0\) or \(p_0\)). But what if we want to compare two groups to each other?

  • Do cadets who study with a tutor score higher than those who don’t?
  • Do Soldiers at Fort Bragg run faster than Soldiers at Fort Drum?
  • Does a new rifle cleaning method reduce maintenance time compared to the standard method?

These questions all ask: Is there a difference between two population means?

For example: Do students taught by LTC Turner perform better (or worse) than students taught by MAJ Day?

What are we really asking? \(\mu_T - \mu_D > 0\)?

That sounds like… an alternative hypothesis!

\[H_a: \mu_T - \mu_D > 0\]

And where there’s an alternative, there’s a null:

\[H_0: \mu_T - \mu_D = 0\]

Now what do we need? Data.

LTC Turner MAJ Day
\(n\) 52 54
\(\bar{x}\) 82.3 80.1
\(s\) 10.5 11.2 
# LTC Turner's sections (18 + 18 + 16 = 52 students)
n_turner <- 52; xbar_turner <- 82.3; s_turner <- 10.5

# MAJ Day's sections (18 + 18 + 18 = 54 students)
n_day <- 54; xbar_day <- 80.1; s_day <- 11.2

Both samples are large (\(n \geq 30\)), so by Devore we can use the two-sample \(z\)-test:

\[z = \frac{(\bar{x}_{T} - \bar{x}_{D}) - (\mu_T - \mu_D)}{\sqrt{\dfrac{s_{T}^2}{n_{T}} + \dfrac{s_{D}^2}{n_{D}}}}\]

Under \(H_0\): \(\mu_T - \mu_D = 0\), so:

\[z = \frac{(82.3 - 80.1) - 0}{\sqrt{\dfrac{10.5^2}{52} + \dfrac{11.2^2}{54}}} = \frac{2.2}{\sqrt{2.120 + 2.323}} = \frac{2.2}{2.108} = 1.04\]

se <- sqrt(s_turner^2 / n_turner + s_day^2 / n_day)
z_stat <- (xbar_turner - xbar_day) / se
z_stat
[1] 1.043703

Step 3: Find the \(p\)-Value

This is right-tailed, so we want \(P(Z \geq z)\):

p_value <- 1 - pnorm(z_stat)
p_value
[1] 0.1483114

Step 4: Decide and Conclude

At \(\alpha = 0.05\): \(p = 0.1483 > 0.05\). We fail to reject \(H_0\). At the 5% significance level, there is not sufficient evidence that LTC Turner’s students scored higher than MAJ Day’s students on the WPR.

The Setup (General)

We have two independent groups:

Group 1 Group 2
Population mean \(\mu_1\) \(\mu_2\)
Sample size \(n_1\) \(n_2\)
Sample mean \(\bar{x}_1\) \(\bar{x}_2\)
Sample SD \(s_1\) \(s_2\)

We want to test whether \(\mu_1 - \mu_2 = 0\) (i.e., are the two means equal?).

The Two-Sample \(z\)-Test (Large Samples)

ImportantTwo-Sample z-Test for Means

Hypotheses:

\(H_0: \mu_1 - \mu_2 = 0\) vs. 

\(H_a: \mu_1 - \mu_2 < 0\)
\(H_a: \mu_1 - \mu_2 > 0\)
\(H_a: \mu_1 - \mu_2 \neq 0\)

Test statistic:

\[z = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}}\]

Conditions:

Large samples: \(n_1 \geq 30\) and \(n_2 \geq 30\)

Since both samples are large, the CLT kicks in and we use the standard normal \(z\) — no need to worry about degrees of freedom.

\(p\)-values: Same as any \(z\)-test — use pnorm().

Example 1: PT Run Times

A brigade S3 suspects that Soldiers at Fort Bragg have slower 2-mile run times than Soldiers at Fort Drum. Random samples from each installation:

Fort Bragg Fort Drum
\(n\) 45 38
\(\bar{x}\) (min) 16.8 15.9
\(s\) 2.1 1.9

Step 1: State the Hypotheses

Let \(\mu_1\) = Fort Bragg, \(\mu_2\) = Fort Drum. The S3 suspects Liberty is slower (higher times):

\[H_0: \mu_1 - \mu_2 = 0 \qquad \text{vs.} \qquad H_a: \mu_1 - \mu_2 > 0\]

Step 2: Compute the Test Statistic

\[z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}}\]

Under \(H_0\): \(\mu_1 - \mu_2 = 0\), so:

\[z = \frac{(16.8 - 15.9) - 0}{\sqrt{\dfrac{2.1^2}{45} + \dfrac{1.9^2}{38}}} = \frac{0.9}{\sqrt{0.098 + 0.095}} = \frac{0.9}{0.439} = 2.05\]

n1 <- 45; n2 <- 38
xbar1 <- 16.8; xbar2 <- 15.9
s1 <- 2.1; s2 <- 1.9

se <- sqrt(s1^2 / n1 + s2^2 / n2)
z_stat <- (xbar1 - xbar2) / se
z_stat
[1] 2.048632

Step 3: Find the \(p\)-Value

Right-tailed, so \(P(Z \geq z)\):

p_value <- 1 - pnorm(z_stat)
p_value
[1] 0

Step 4: Decide and Conclude

At \(\alpha = 0.05\): \(p = 0.020 \leq 0.05\). We reject \(H_0\). At the 5% significance level, there is sufficient evidence that Fort Bragg Soldiers have slower 2-mile run times than Fort Drum Soldiers.

Example 2: Ruck March Times

Two companies complete a 12-mile ruck march. The standard is to finish in under 180 minutes on average. The S3 suspects Alpha Company finishes more than 5 minutes faster than Bravo Company.

Alpha Company Bravo Company
\(n\) 35 40
\(\bar{x}\) (min) 172.4 181.2
\(s\) 12.3 14.1

Let \(\mu_A\) = Alpha’s mean, \(\mu_B\) = Bravo’s mean. The S3 claims Alpha is more than 5 minutes faster:

\[H_0: \mu_A - \mu_B = -5 \qquad \text{vs.} \qquad H_a: \mu_A - \mu_B < -5\]

\[z = \frac{(\bar{x}_A - \bar{x}_B) - (-5)}{\sqrt{\dfrac{s_A^2}{n_A} + \dfrac{s_B^2}{n_B}}} = \frac{(172.4 - 181.2) - (-5)}{\sqrt{\dfrac{12.3^2}{35} + \dfrac{14.1^2}{40}}} = \frac{-8.8 + 5}{3.049} = \frac{-3.8}{3.049} = -1.25\]

n1 <- 35; n2 <- 40
xbar1 <- 172.4; xbar2 <- 181.2
s1 <- 12.3; s2 <- 14.1
delta_0 <- -5

se <- sqrt(s1^2 / n1 + s2^2 / n2)
z_stat <- ((xbar1 - xbar2) - delta_0) / se
z_stat
[1] -1.24655

\(p\)-Value: Left-tailed, so \(P(Z \leq z)\):

p_value <- pnorm(z_stat)
p_value
[1] 0.1062812

Decide and Conclude

\(p = 0.106 > 0.05\). We fail to reject \(H_0\). At the 5% significance level, there is not sufficient evidence that Alpha Company finishes more than 5 minutes faster than Bravo Company.

Preview: The Paired \(t\)-Test

When are samples NOT independent?

Sometimes the two “groups” are really the same individuals measured twice — before and after a treatment, or under two different conditions. These are paired observations.

Examples:

  • Same Soldiers’ AFT scores before and after a training program
  • Same cadets’ exam scores in MA376 vs. MA476
  • Same vehicles’ maintenance time with old vs. new procedure

When data are paired, we don’t compare two separate groups — we compute the difference for each pair and analyze those differences as a single sample.

The Paired \(t\)-Test

ImportantPaired t-Test

Given \(n\) paired observations \((x_{1i}, x_{2i})\), define \(d_i = x_{1i} - x_{2i}\).

Hypotheses:

\(H_0: \mu_d = 0\)

\(H_a: \mu_d \neq, <, > \ 0\)

Test statistic:

\[t = \frac{\bar{d} - \mu_{d_0}}{s_d / \sqrt{n}}, \qquad df = n - 1\]

where \(\bar{d} = \frac{1}{n}\sum d_i\) and \(s_d = \sqrt{\frac{1}{n-1}\sum(d_i - \bar{d})^2}\)

Conditions: Random sample of pairs, differences are approximately normal (or \(n \geq 30\))

\(p\)-values: Use pt() with \(df = n - 1\).

Example: AFT Scores Before and After a PT Program

A company commander puts 10 Soldiers through a new 6-week PT program. She records their AFT scores before and after.

Soldier 1 2 3 4 5 6 7 8 9 10
Before 445 478 512 390 467 501 423 489 455 438
After 460 491 520 412 475 498 441 502 470 450
\(d_i\) (After \(-\) Before) 15 13 8 22 8 \(-3\) 18 13 15 12

She wants to know if the program improved scores (i.e., \(d > 0\)):

\[H_0: \mu_d = 0 \qquad \text{vs.} \qquad H_a: \mu_d > 0\]

Compute the differences:

\[\bar{d} = \frac{15 + 13 + 8 + 22 + 8 + (-3) + 18 + 13 + 15 + 12}{10} = \frac{121}{10} = 12.1\]

\[s_d = \sqrt{\frac{\sum(d_i - \bar{d})^2}{n-1}} = 7.17\]

before <- c(445, 478, 512, 390, 467, 501, 423, 489, 455, 438)
after  <- c(460, 491, 520, 412, 475, 498, 441, 502, 470, 450)
d <- after - before
d
 [1] 15 13  8 22  8 -3 18 13 15 12
n <- length(d)
d_bar <- mean(d)
s_d <- sd(d)
d_bar
[1] 12.1
s_d
[1] 6.773314

Test statistic:

\[t = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{12.1}{7.17 / \sqrt{10}} = \frac{12.1}{2.267} = 5.34\]

t_stat <- (d_bar - 0) / (s_d / sqrt(n))
t_stat
[1] 5.649164

\(p\)-Value: Right-tailed with \(df = 9\):

p_value <- 1 - pt(t_stat, df = n - 1)
p_value
[1] 0.0001569676

Decide and Conclude

\(p < 0.001 \leq 0.05\). We reject \(H_0\). At the 5% significance level, there is strong evidence that the PT program improved AFT scores. The average improvement was 12.1 points.

WarningWhy not a two-sample test?

These are the same 10 Soldiers measured twice — the before and after scores are not independent. A two-sample \(z\)-test would ignore the pairing and lose power. The paired \(t\)-test accounts for the natural variability between Soldiers by focusing on each individual’s change.

Board Problems

Problem 1: Graduation Rate

A military academy claims that 95% of cadets who start their senior year will graduate. In a random sample of \(n = 250\) seniors from recent classes, 230 graduated. Is there evidence the true graduation rate is below the claim?

  1. State the hypotheses.

  2. Check the conditions.

  3. Compute the test statistic and \(p\)-value.

  4. At \(\alpha = 0.05\), state your conclusion in context.

  1. \(H_0: p = 0.95\) vs. \(H_a: p < 0.95\)

  2. Random sample (stated). \(n = 250 < 10\%\) of all seniors. \(np_0 = 250(0.95) = 237.5 \geq 10\) and \(n(1-p_0) = 250(0.05) = 12.5 \geq 10\). Conditions met.

n <- 250; x <- 230
p_hat <- x / n; p0 <- 0.95
se <- sqrt(p0 * (1 - p0) / n)
z <- (p_hat - p0) / se
z
[1] -2.176429
p_value <- pnorm(z)
p_value
[1] 0.01476161
  1. \(p = 0.0148 < 0.05\), so we reject \(H_0\). At the 5% significance level, there is sufficient evidence that the graduation rate is below 95%.

Problem 2: Equipment Readiness

A battalion standard requires that 80% of vehicles be fully mission-capable (FMC) at any given time. A random inspection of \(n = 120\) vehicles finds 89 are FMC. Is there evidence the battalion is not meeting the standard?

  1. State the hypotheses.

  2. Check the conditions.

  3. Compute the test statistic and \(p\)-value.

  4. At \(\alpha = 0.05\), state your conclusion in context.

  5. Is the observed difference practically significant? The battalion has 120 vehicles total.

  1. \(H_0: p = 0.80\) vs. \(H_a: p < 0.80\)

  2. Random inspection (stated). \(np_0 = 120(0.80) = 96 \geq 10\) and \(n(1-p_0) = 120(0.20) = 24 \geq 10\). Conditions met. (Note: if this is ALL vehicles, the 10% condition fails — the sample IS the population.)

n <- 120; x <- 89
p_hat <- x / n; p0 <- 0.80
se <- sqrt(p0 * (1 - p0) / n)
z <- (p_hat - p0) / se
z
[1] -1.597524
p_value <- pnorm(z)
p_value
[1] 0.05507446

  1. \(p = 0.078 > 0.05\), so we fail to reject \(H_0\). At the 5% significance level, there is not sufficient evidence that the FMC rate is below 80%.

  2. The observed rate is \(89/120 = 74.2\%\), which is 5.8 percentage points below 80%. That’s about 7 vehicles short of the standard — arguably a meaningful gap in readiness, even though the test wasn’t statistically significant. This is a case where practical significance exists but statistical significance doesn’t — likely because the sample size wasn’t large enough to detect the difference.

Problem 3: Screen Time

A wellness officer claims that fewer than 30% of cadets spend more than 4 hours per day on their phones. In a random sample of \(n = 80\) cadets, 30 report spending more than 4 hours daily.

  1. State the hypotheses.

  2. Check the conditions.

  3. Compute the test statistic and \(p\)-value.

  4. At \(\alpha = 0.05\), state your conclusion in context.

  1. \(H_0: p = 0.30\) vs. \(H_a: p > 0.30\) (the data suggests the officer’s claim might be wrong — the sample proportion is above 30%)

  2. Random sample (stated). \(n = 80 < 10\%\) of all cadets. \(np_0 = 80(0.30) = 24 \geq 10\) and \(n(1-p_0) = 80(0.70) = 56 \geq 10\). Conditions met.

n <- 80; x <- 30
p_hat <- x / n; p0 <- 0.30
se <- sqrt(p0 * (1 - p0) / n)
z <- (p_hat - p0) / se
z
[1] 1.46385
p_value <- 1 - pnorm(z)
p_value
[1] 0.07161745
  1. \(p = 0.0285 < 0.05\), so we reject \(H_0\). At the 5% significance level, there is sufficient evidence that more than 30% of cadets spend over 4 hours daily on their phones.

Problem 4: Marksmanship Scores

Two companies use different rifle zeroing techniques. After qualification, Company A (\(n = 35\)) has \(\bar{x}_1 = 31.4\) with \(s_1 = 3.2\), and Company B (\(n = 40\)) has \(\bar{x}_2 = 28.7\) with \(s_2 = 3.8\).

  1. State the hypotheses to test whether there is a difference in mean scores.

  2. Compute the test statistic and \(p\)-value.

  3. Construct and interpret a 95% CI for \(\mu_1 - \mu_2\).

  4. State your conclusion in context.

  1. \(H_0: \mu_1 - \mu_2 = 0\) vs. \(H_a: \mu_1 - \mu_2 \neq 0\)

n1 <- 35; n2 <- 40
xbar1 <- 31.4; xbar2 <- 28.7
s1 <- 3.2; s2 <- 3.8

se <- sqrt(s1^2 / n1 + s2^2 / n2)
z <- (xbar1 - xbar2) / se
p_value <- 2 * (1 - pnorm(abs(z)))
z
[1] 3.339775
p_value
[1] 0.0008384623
diff <- xbar1 - xbar2
me <- qnorm(0.975) * se
c(diff - me, diff + me)
[1] 1.115491 4.284509

We are 95% confident that Company A scores between 1.12 and 4.28 points higher than Company B on average.

  1. \(p = 0.0008385 \leq 0.05\). We reject \(H_0\). There is sufficient evidence of a difference in mean marksmanship scores between the two zeroing techniques.

Before You Leave

Today

  • Reviewed the one-proportion \(z\)-test
  • Two-sample \(z\)-test compares means from two independent groups (large samples)
  • Test statistic: \(z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}}\), use pnorm() for \(p\)-values
  • CI for \(\mu_1 - \mu_2\): \((\bar{x}_1 - \bar{x}_2) \pm z_{\alpha/2} \cdot SE\) — if it doesn’t contain 0, reject \(H_0\)

Any questions?

Next Lesson

Lesson 23: Paired t-Test

  • Identify paired versus independent designs
  • Test a mean difference using paired \(t\)
  • Construct a CI for a mean difference

Upcoming Graded Events

  • WebAssign 8.4, 9.2 - Due before Lesson 23
  • WPR II - Lesson 27