Lesson 23: One-Proportion z-Test

Break!

The New York Marlins youth baseball team and two coaches posed at home plate in front of fall trees.

The Turner family selfie in the stands at Williamsport.

Reese winding up to throw a softball while an Army softball player cheers behind her at Williamsport.

Reese in a blue uniform throwing a softball to two college softball players on the outfield grass at Williamsport.

Reese leaping over small hurdles on the outfield grass at Williamsport.

Cal with a glove standing next to a player on the field at Williamsport.

Calendar

Block II calendar with a red box around Tuesday 20 October, Lesson 23, One-Proportion z-Test.

What We Did: Lessons 18 to 22

  • \(\bar{X} \sim N\left(\mu,\ \left(\dfrac{\sigma}{\sqrt{n}}\right)^2\right)\), approximately, for \(n > 40\)
  • Normal population: exact for any \(n\)
  • pnorm(x, mu, sigma/sqrt(n))
  • \(\hat{p} \sim N\left(p,\ \dfrac{p(1-p)}{n}\right)\), approximately, when \(np \ge 10\) and \(n(1-p) \ge 10\)
  • \(\text{estimate} \pm \text{critical value} \times \text{standard error}\)
  • Mean, \(n > 40\): \(\bar{x} \pm z_{\alpha/2}\,\dfrac{s}{\sqrt{n}}\)
  • Proportion, \(n\hat{p} \ge 10\) and \(n(1-\hat{p}) \ge 10\): \(\hat{p} \pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\)
  • qnorm(0.975) for 95%, qnorm(0.95) for 90%, qnorm(0.995) for 99%
  • 95% confident describes the method
  • Small sample, normal population: \(\bar{x} \pm t_{\alpha/2,\,n-1}\,\dfrac{s}{\sqrt{n}}\)
  • \(\text{df} = n - 1\), qt(0.975, n - 1) for 95%
  • \(t_{\alpha/2,\,n-1} > z_{\alpha/2}\), so the \(t\) interval is wider
  • Sample size for width \(w\): \(n = \left(\dfrac{2\,z_{\alpha/2}\,\sigma}{w}\right)^2\), rounded up
  • \(H_0: \mu = \mu_0\) versus \(H_a\): \(\mu < \mu_0\), \(\mu > \mu_0\), or \(\mu \ne \mu_0\)
  • \(z = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}}\), random sample, \(n > 40\)
  • P-value: pnorm(z), 1 - pnorm(z), 2 * pnorm(-abs(z))
  • Reject \(H_0\) if the P-value is at or below \(\alpha\), otherwise fail to reject
  • Type I: reject a true \(H_0\) (\(\alpha\)). Type II: fail to reject a false \(H_0\) (\(\beta\))
  • Fail to reject is not accept
  • \(t = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}}\), \(\text{df} = n - 1\)
  • Random sample, \(n \le 40\), population approximately normal
  • P-value: pt(t, n - 1), 1 - pt(t, n - 1), 2 * pt(-abs(t), n - 1)
  • t.test(x, mu = mu0, alternative = "two.sided")

What We’re Doing: Lesson 23

Objectives

  • Check the large-sample conditions for inference on a proportion. (SLO 5)
  • Carry out a large-sample \(z\) test for a population proportion. (SLO 5)
  • Distinguish statistical significance from practical significance. (SLO 6)

Required Reading

Devore 8.4, 8.5

The Takeaway for Today

NoteKey Concepts from Lesson 23
  • Test statistic: \(z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}}\)
  • Conditions: random sample, \(np_0 \ge 10\) and \(n(1 - p_0) \ge 10\)
  • P-value: pnorm(z), 1 - pnorm(z), or 2 * pnorm(-abs(z))
  • Statistically significant is not the same as practically significant: with a large \(n\), a tiny difference rejects

Warm-Up

Corps Squad

The Corps is told \(25\%\) of cadets are on a corps squad team. In a random sample of \(150\) cadets, \(48\) are on a corps squad team.

If \(p = 0.25\) is true, what is the distribution of \(\hat{p}\)?

\(150(0.25) = 37.5 \ge 10\) and \(150(0.75) = 112.5 \ge 10\).

\[\hat{p} \sim N\left(0.25,\ \frac{0.25(0.75)}{150}\right), \qquad \text{SE} = \sqrt{\frac{0.25(0.75)}{150}} = 0.0354\]

Where does \(\hat{p} = 48/150 = 0.32\) fall?

Normal curve for p-hat centered at 0.25. The observed p-hat of 0.32 is marked about two standard errors to the right, and the area to its right is shaded.

\[z = \frac{0.32 - 0.25}{0.0354} = 1.98\]

About two standard errors above \(0.25\). Unusual if \(p = 0.25\).

This is the test from the end of last lesson.

One-Proportion \(z\) Test

ImportantOne-Proportion \(z\) Test

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

  • Hypotheses: \(H_0: p = p_0\) versus \(H_a\): \(p > p_0\), \(p < p_0\), or \(p \ne p_0\)
  • Conditions: random sample, \(np_0 \ge 10\) and \(n(1 - p_0) \ge 10\)
  • If \(H_0\) is true, \(Z \sim N(0, 1)\), approximately
\(H_a\) P-value
\(p > p_0\) 1 - pnorm(z)
\(p < p_0\) pnorm(z)
\(p \ne p_0\) 2 * pnorm(-abs(z))
WarningThe Test Uses \(p_0\), the Interval Uses \(\hat{p}\)
  • Test: we assume \(H_0\) is true, so the standard error and the conditions use \(p_0\)
  • Interval: there is no null value, so the standard error and the conditions use \(\hat{p}\)

The Steps

NoteHypothesis Test Steps
  1. Hypotheses: \(H_0\) and \(H_a\), in terms of the parameter
  2. Conditions
  3. Test statistic
  4. P-value
  5. Decision: reject \(H_0\) if the P-value is at or below \(\alpha\), otherwise fail to reject
  6. Conclusion in context

Land Nav

Historically \(70\%\) of cadets pass land navigation on the first attempt. CST tried a new pre-course block. In a random sample of \(120\) cadets, \(93\) passed on the first attempt. Test at \(\alpha = 0.05\).

1. Hypotheses: State \(H_0\) and \(H_a\).

\[H_0: p = 0.70\]

\[H_a: p > 0.70\]

\(p\) is the true proportion of all cadets who pass land navigation on the first attempt with the new block.

2. Conditions: What conditions allow us to do this?

Random sample. \(np_0 = 120(0.70) = 84 \ge 10\) and \(n(1 - p_0) = 120(0.30) = 36 \ge 10\).

3. Test statistic: Compute it.

\[\hat{p} = \frac{93}{120} = 0.775\]

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

\[z = \frac{0.775 - 0.70}{\sqrt{0.70(0.30)/120}} = \mathbf{1.79}\]

4. P-value: Find it.

Standard normal curve. The test statistic z equals 1.79 is marked and the area to its right is shaded.

\[P(Z \ge 1.79) = \mathbf{0.0365}\]

1 - pnorm(1.793)

5. Decision: Reject or fail to reject?

\(0.0365 \le 0.05\). Reject \(H_0\).

6. Conclusion: State it in context.

Since the P-value \(0.0365 \le \alpha = 0.05\), reject \(H_0\). At the 5% level there is sufficient evidence that the true proportion of all cadets who pass land navigation on the first attempt with the new block is greater than 0.70.

Lefties

About \(10\%\) of people are left-handed. Is the Corps different? In a random sample of \(200\) cadets, \(16\) are left-handed. Test at \(\alpha = 0.05\).

1. Hypotheses: State \(H_0\) and \(H_a\).

\[H_0: p = 0.10\]

\[H_a: p \ne 0.10\]

\(p\) is the true proportion of all cadets who are left-handed.

2. Conditions: What conditions allow us to do this?

Random sample. \(np_0 = 200(0.10) = 20 \ge 10\) and \(n(1 - p_0) = 200(0.90) = 180 \ge 10\).

3. Test statistic: Compute it.

\[\hat{p} = \frac{16}{200} = 0.08\]

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

\[z = \frac{0.08 - 0.10}{\sqrt{0.10(0.90)/200}} = \mathbf{-0.94}\]

4. P-value: Find it.

Standard normal curve. The test statistic z equals negative 0.94 is marked. The area to the left of negative 0.94 and the area to the right of 0.94 are both shaded.

\[2\,P(Z \ge \lvert -0.94 \rvert) = \mathbf{0.3458}\]

2 * pnorm(-abs(-0.943))

5. Decision: Reject or fail to reject?

\(0.3458 > 0.05\). Fail to reject \(H_0\).

6. Conclusion: State it in context.

Since the P-value \(0.3458 > \alpha = 0.05\), fail to reject \(H_0\). At the 5% level there is insufficient evidence that the true proportion of all cadets who are left-handed differs from 0.10.

In R


    1-sample proportions test without continuity correction

data:  93 out of 120, null probability 0.7
X-squared = 3.2143, df = 1, p-value = 0.0365
alternative hypothesis: true p is greater than 0.7
95 percent confidence interval:
 0.7066342 1.0000000
sample estimates:
    p 
0.775
  • correct = FALSE matches the hand calculation
  • X-squared is \(z^2\): \(1.79^2 = 3.214\)
  • Same P-value

Statistical vs. Practical Significance

ACFT Pass Rate

The Army’s first-time ACFT pass rate has been \(90\%\). After a new training program, every sample below shows \(\hat{p} = 0.905\). Test \(H_0: p = 0.90\) versus \(H_a: p > 0.90\) at \(\alpha = 0.05\).

What happens as \(n\) grows?

\(n\) \(\hat{p}\) \(z\) P-value Decision
500 0.905 0.37 0.3547 Fail to reject
5,000 0.905 1.18 0.1193 Fail to reject
50,000 0.905 3.73 0.0001 Reject

Same half a percentage point every time. Only \(n\) changed.

Does the commander care?

Probably not. At \(n = 50{,}000\) the 95% interval is

\[0.905 \pm 1.96\sqrt{\frac{0.905(0.095)}{50000}} = (0.9024,\ 0.9076)\]

The program adds well under one percentage point. Real, but too small to matter.

ImportantStatistical vs. Practical Significance
  • Statistically significant: the P-value is at or below \(\alpha\). The difference is probably not chance.
  • Practically significant: the difference is large enough to matter in context.
  • With a large enough \(n\), any difference from \(p_0\) is statistically significant.
  • A small P-value does not mean a big effect. Look at \(\hat{p}\) or the interval to judge size.
  • The same holds for tests about \(\mu\).

Look Ahead: Comparing Two Means

What happens if we want to compare means?

New Run Program

CST randomly assigns \(22\) cadets to two groups. \(12\) train with a new run program and \(10\) train with the standard program. Does the new program lower the mean 2 mile run time? Test at \(\alpha = 0.05\).

Group \(n\) \(\bar{x}\) (min) \(s\) (min)
1: New 12 14.8 1.1
2: Standard 10 16.1 1.4

1. Hypotheses

\[H_0: \mu_1 - \mu_2 = 0\]

\[H_a: \mu_1 - \mu_2 < 0\]

\(\mu_1\) and \(\mu_2\) are the true mean 2 mile run times of all cadets on the new and standard programs.

2. Conditions

Two independent random samples. Both samples are small, so both populations must be approximately normal.

3. Test statistic

\[t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}} = \frac{14.8 - 16.1}{\sqrt{\dfrac{1.1^2}{12} + \dfrac{1.4^2}{10}}} = \mathbf{-2.39}\]

\[\nu = \frac{\left(\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}\right)^2}{\dfrac{(s_1^2/n_1)^2}{n_1 - 1} + \dfrac{(s_2^2/n_2)^2}{n_2 - 1}} = 16.97, \text{ round down to } 16\]

4. P-value

t curve with 16 degrees of freedom. The test statistic t equals negative 2.39 is marked and the area to its left is shaded.

\[P(T \le -2.39) = \mathbf{0.0149}\]

pt(-2.386, 16)

5. Decision

\(0.0149 \le 0.05\). Reject \(H_0\).

6. Conclusion

Since the P-value \(0.0149 \le \alpha = 0.05\), reject \(H_0\). At the 5% level there is sufficient evidence that the true mean 2 mile run time of cadets on the new program is less than that of cadets on the standard program.

Same six steps. New parameter, new standard error, new degrees of freedom. Lesson 24.

Inference Procedures

Parameter Conditions Test statistic Confidence interval Critical value P-value by \(H_a\)
\(\mu\)
\(\sigma\) known, or \(\sigma\) unknown and \(n > 40\)		 Random sample
\(\sigma\) known and population normal, or \(n > 40\)		 \(z = \dfrac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}\)
use \(s\) for \(\sigma\) when \(n > 40\)		 \(\bar{x} \pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\) qnorm(1-a/2) \(>\) 1-pnorm(z)
\(<\) pnorm(z)
\(\ne\) 2*pnorm(-abs(z))
\(\mu\)
\(\sigma\) unknown, \(n \le 40\)		 Random sample
Population approx. normal		 \(t = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}} \sim t_{n-1}\) \(\bar{x} \pm t_{\alpha/2,\,n-1}\dfrac{s}{\sqrt{n}}\) qt(1-a/2,n-1) \(>\) 1-pt(t,n-1)
\(<\) pt(t,n-1)
\(\ne\) 2*pt(-abs(t),n-1)
\(p\) Random sample
\(np_0 \ge 10\)
\(n(1-p_0) \ge 10\)		 \(z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}}\)
the test uses \(p_0\)		 \(\hat{p} \pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\)
the interval uses \(\hat{p}\)		 qnorm(1-a/2) \(>\) 1-pnorm(z)
\(<\) pnorm(z)
\(\ne\) 2*pnorm(-abs(z))
\(\mu_1 - \mu_2\)
independent		 Two independent samples
Both normal, or both \(n > 40\)		 \(t = \dfrac{(\bar{x}_1 - \bar{x}_2) - \Delta_0}{\sqrt{s_1^2/n_1 + s_2^2/n_2}} \sim t_{\nu}\) \((\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2,\,\nu}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}\) qt(1-a/2,nu) \(>\) 1-pt(t,nu)
\(<\) pt(t,nu)
\(\ne\) 2*pt(-abs(t),nu)
\(\mu_D\)
paired		 Random sample of matched pairs, \(n_1 = n_2\)
Differences approx. normal when \(n \le 40\)		 \(t = \dfrac{\bar{d} - \Delta_0}{s_D/\sqrt{n}} \sim t_{n-1}\)
on the differences \(d_i = x_i - y_i\)		 \(\bar{d} \pm t_{\alpha/2,\,n-1}\dfrac{s_D}{\sqrt{n}}\) qt(1-a/2,n-1) \(>\) 1-pt(t,n-1)
\(<\) pt(t,n-1)
\(\ne\) 2*pt(-abs(t),n-1)
\(p_1 - p_2\) Two independent samples
\(n_i\hat{p}_i \ge 10\)
\(n_i(1-\hat{p}_i) \ge 10\)		 \(z = \dfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\)
pooled \(\hat{p} = \dfrac{x_1+x_2}{n_1+n_2}\)		 \((\hat{p}_1 - \hat{p}_2) \pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\)
not pooled		 qnorm(1-a/2) \(>\) 1-pnorm(z)
\(<\) pnorm(z)
\(\ne\) 2*pnorm(-abs(z))

Board Problems

Problem 1: Can We Use the \(z\) Test?

For each, can we use the one-proportion \(z\) test? If not, why not?

  1. A random sample of \(40\) cadets, testing \(H_0: p = 0.20\).

  2. A random sample of \(200\) cadets, testing \(H_0: p = 0.05\).

  3. A random sample of \(60\) cadets, testing \(H_0: p = 0.50\).

  4. The first \(300\) cadets through the gym door on a Saturday, testing \(H_0: p = 0.40\).

  1. No. \(np_0 = 40(0.20) = 8 < 10\).

  2. Yes. \(np_0 = 200(0.05) = 10 \ge 10\) and \(n(1 - p_0) = 190 \ge 10\).

  3. Yes. \(np_0 = n(1 - p_0) = 30 \ge 10\).

  4. No. Not a random sample. Saturday gym goers are not representative of the Corps.

Problem 2: Night Vision Goggles

The contract allows \(5\%\) of night vision goggles to arrive defective. The S4 thinks the defect rate is higher. In a random sample of \(300\) goggles, \(24\) are defective. Test at \(\alpha = 0.01\).

  1. State the hypotheses.

  2. Check the conditions.

  3. Compute the test statistic.

  4. State the rejection region. Is \(z\) in it?

  5. Find the P-value.

  6. Decide and conclude.

  1. \[H_0: p = 0.05\]

\[H_a: p > 0.05\]

\(p\) is the true proportion of all delivered goggles that are defective.

  1. Random sample. \(np_0 = 300(0.05) = 15 \ge 10\) and \(n(1 - p_0) = 285 \ge 10\).

  2. \[\hat{p} = \frac{24}{300} = 0.08\]

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

\[z = \frac{0.08 - 0.05}{\sqrt{0.05(0.95)/300}} = \mathbf{2.38}\]

  1. Reject \(H_0\) if \(z \ge 2.326\). qnorm(0.99). Yes, \(2.38 \ge 2.326\).

  2. \[P(Z \ge 2.38) = \mathbf{0.0086}\]

1 - pnorm(2.384)

  1. Since the P-value \(0.0086 \le \alpha = 0.01\), reject \(H_0\). At the 1% level there is sufficient evidence that the true proportion of all delivered goggles that are defective is greater than 0.05.

Problem 3: Coin Toss

Over \(120\) home football games, the visiting captain called the coin toss correctly \(52\) times. Is the call different from a fair \(50\%\)? Test at \(\alpha = 0.05\).

  1. State the hypotheses.

  2. Check the conditions.

  3. Compute the test statistic and P-value.

  4. Decide and conclude.

  5. Does this prove the call is a fair \(50\%\)?

  1. \[H_0: p = 0.50\]

\[H_a: p \ne 0.50\]

\(p\) is the true proportion of coin tosses the visiting captain calls correctly.

  1. Treat the games as a random sample. \(np_0 = n(1 - p_0) = 60 \ge 10\).

  2. \[\hat{p} = \frac{52}{120} = 0.4333\]

\[z = \frac{0.4333 - 0.50}{\sqrt{0.50(0.50)/120}} = \mathbf{-1.46}\]

\[2\,P(Z \ge \lvert -1.46 \rvert) = \mathbf{0.1441}\]

2 * pnorm(-abs(-1.461))

  1. Since the P-value \(0.1441 > \alpha = 0.05\), fail to reject \(H_0\). At the 5% level there is insufficient evidence that the true proportion of coin tosses the visiting captain calls correctly differs from 0.50.

  2. No. Fail to reject is not accept. The data are consistent with \(0.50\), but also with other values near \(0.43\).

Problem 4: Two Sleep Studies

Currently \(30\%\) of cadets sleep \(7\) or more hours a night. Two studies test a new sleep program, each with \(H_a: p > 0.30\) at \(\alpha = 0.05\).

  • Study A: random sample of \(20{,}000\) cadets, \(6{,}200\) sleep \(7\) or more hours
  • Study B: random sample of \(50\) cadets, \(20\) sleep \(7\) or more hours

  1. Find \(\hat{p}\), \(z\), and the P-value for each study.

  2. Which study is statistically significant?

  3. Which study shows the bigger improvement?

  4. What would you tell the Commandant about each?

  1. Study A: \(\hat{p} = 0.31\)

\[z = \frac{0.31 - 0.30}{\sqrt{0.30(0.70)/20000}} = \mathbf{3.09}, \qquad P(Z \ge 3.09) = \mathbf{0.0010}\]

Study B: \(\hat{p} = 0.40\). Conditions: \(50(0.30) = 15 \ge 10\), \(50(0.70) = 35 \ge 10\).

\[z = \frac{0.40 - 0.30}{\sqrt{0.30(0.70)/50}} = \mathbf{1.54}, \qquad P(Z \ge 1.54) = \mathbf{0.0614}\]

  1. Study A rejects \(H_0\). Study B fails to reject.

  2. Study B: \(10\) percentage points versus \(1\).

  3. Study A: the improvement is real but only one percentage point, statistically significant but probably not practically significant. Study B: a promising improvement, but \(n = 50\) is too small to rule out chance. Run a larger study.

Before You Leave

Today

  • \(z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}}\)
  • Random sample, \(np_0 \ge 10\) and \(n(1 - p_0) \ge 10\)
  • P-value: pnorm(z), 1 - pnorm(z), 2 * pnorm(-abs(z))
  • Statistically significant does not mean practically significant

Any questions?

Next Lesson

Lesson 24: Two-Sample t-Test

  • Distinguish independent-sample from paired designs
  • Carry out the two-sample \(t\) test for \(\mu_1 - \mu_2\) using the approximate degrees of freedom
  • Construct and interpret a confidence interval for \(\mu_1 - \mu_2\)

Reading: Devore 9.2

Upcoming Graded Events

  • WebAssign 8.4: due at the start of Lesson 24
  • WPR II: Lesson 28 (covers Lessons 18-26)
  • TEE: 15-18 Dec 2026