Learning Objectives

By the end of this chapter, you will be able to:

  • Set up hypotheses for population proportions
  • Check conditions for the z-test for proportions
  • Calculate the test statistic for proportions
  • Perform complete hypothesis tests for proportions
  • Interpret results in context

When to Use Z-Test for Proportion

Use this test when:

  1. Testing a claim about a single population proportion (p)
  2. Sample size is large enough: np₀ ≥ 10 AND n(1-p₀) ≥ 10
  3. Sample is random from the population
flowchart TD
    A[Testing proportion p?]
    B{np₀ ≥ 10?}
    C{n(1-p₀) ≥ 10?}
    D[Use Z-test for proportion]
    E[Use exact binomial test]

    A --> B
    B -->|Yes| C
    B -->|No| E
    C -->|Yes| D
    C -->|No| E

Hypotheses for Proportions

Type H₀ H₁
Two-tailed p = p₀ p ≠ p₀
Right-tailed p = p₀ p > p₀
Left-tailed p = p₀ p < p₀

Where p₀ is the hypothesized value.

Test Statistic Formula

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

Where:

  • $\hat{p} = \frac{x}{n}$ = sample proportion
  • $x$ = number of successes
  • $n$ = sample size
  • $p_0$ = hypothesized proportion

Standard Error (under H₀)

\[SE = \sqrt{\frac{p_0(1-p_0)}{n}}\]

Note: We use p₀ (not $\hat{p}$) in the standard error because we assume H₀ is true.

Step-by-Step Example 1: Two-Tailed Test

Problem: A government claims that 60% of citizens are satisfied with public services. A survey of 400 citizens finds 220 satisfied. Test at α = 0.05 whether the satisfaction rate differs from 60%.

Solution:

Step 1: State hypotheses

  • $H_0: p = 0.60$
  • $H_1: p \neq 0.60$ (two-tailed)

Step 2: Check conditions

  • np₀ = 400 × 0.60 = 240 ≥ 10 ✓
  • n(1-p₀) = 400 × 0.40 = 160 ≥ 10 ✓

Step 3: Calculate sample proportion \(\hat{p} = \frac{220}{400} = 0.55\)

Step 4: Calculate standard error \(SE = \sqrt{\frac{0.60 \times 0.40}{400}} = \sqrt{\frac{0.24}{400}} = \sqrt{0.0006} = 0.0245\)

Step 5: Calculate test statistic \(z = \frac{0.55 - 0.60}{0.0245} = \frac{-0.05}{0.0245} = -2.04\)

Step 6: Find critical value

  • Two-tailed, α = 0.05: z* = ±1.96

Step 7: Decision

  • |z| = 2.04 > 1.96
  • Reject H₀

Step 8: Conclusion At the 0.05 level of significance, there is sufficient evidence to conclude that the satisfaction rate differs from 60%. The sample suggests it is lower (55%).

Step-by-Step Example 2: Right-Tailed Test

Problem: A politician claims more than 50% of voters support a policy. In a sample of 500 voters, 270 support it. Test at α = 0.05.

Solution:

Step 1: State hypotheses

  • $H_0: p = 0.50$
  • $H_1: p > 0.50$ (right-tailed)

Step 2: Check conditions

  • np₀ = 500 × 0.50 = 250 ≥ 10 ✓
  • n(1-p₀) = 500 × 0.50 = 250 ≥ 10 ✓

Step 3: Calculate sample proportion \(\hat{p} = \frac{270}{500} = 0.54\)

Step 4: Calculate standard error and test statistic \(SE = \sqrt{\frac{0.50 \times 0.50}{500}} = \sqrt{\frac{0.25}{500}} = \sqrt{0.0005} = 0.0224\)

\[z = \frac{0.54 - 0.50}{0.0224} = \frac{0.04}{0.0224} = 1.79\]

Step 5: Find critical value

  • Right-tailed, α = 0.05: z* = 1.645

Step 6: Decision

  • z = 1.79 > 1.645
  • Reject H₀

Step 7: Conclusion At the 0.05 level of significance, there is sufficient evidence to conclude that more than 50% of voters support the policy.

Step-by-Step Example 3: Left-Tailed Test

Problem: A company claims its defect rate is less than 5%. In a sample of 200 items, 6 are defective. Test at α = 0.05.

Solution:

Step 1: State hypotheses

  • $H_0: p = 0.05$
  • $H_1: p < 0.05$ (left-tailed)

Step 2: Check conditions

  • np₀ = 200 × 0.05 = 10 ≥ 10 ✓ (barely)
  • n(1-p₀) = 200 × 0.95 = 190 ≥ 10 ✓

Step 3: Calculate sample proportion \(\hat{p} = \frac{6}{200} = 0.03\)

Step 4: Calculate test statistic \(SE = \sqrt{\frac{0.05 \times 0.95}{200}} = \sqrt{0.0002375} = 0.0154\)

\[z = \frac{0.03 - 0.05}{0.0154} = \frac{-0.02}{0.0154} = -1.30\]

Step 5: Find critical value

  • Left-tailed, α = 0.05: z* = -1.645

Step 6: Decision

  • z = -1.30 > -1.645 (not in rejection region)
  • Fail to Reject H₀

Step 7: Conclusion At the 0.05 level of significance, there is insufficient evidence to conclude that the defect rate is less than 5%.

Step-by-Step Example 4: Complete Problem with p-Value

Problem: A hospital claims 80% of patients are satisfied. A survey of 250 patients finds 185 satisfied. Test at α = 0.01 and find the p-value.

Solution:

Step 1: State hypotheses

  • $H_0: p = 0.80$
  • $H_1: p \neq 0.80$ (two-tailed)

Step 2: Check conditions

  • np₀ = 250 × 0.80 = 200 ≥ 10 ✓
  • n(1-p₀) = 250 × 0.20 = 50 ≥ 10 ✓

Step 3: Calculate sample proportion \(\hat{p} = \frac{185}{250} = 0.74\)

Step 4: Calculate test statistic \(SE = \sqrt{\frac{0.80 \times 0.20}{250}} = \sqrt{\frac{0.16}{250}} = 0.0253\)

\[z = \frac{0.74 - 0.80}{0.0253} = \frac{-0.06}{0.0253} = -2.37\]

Step 5: Calculate p-value For two-tailed test: \(p\text{-value} = 2 \times P(Z < -2.37) = 2 \times 0.0089 = 0.0178\)

Step 6: Decision

  • p-value = 0.0178 > α = 0.01
  • Fail to Reject H₀

Step 7: Conclusion At the 0.01 level of significance, there is insufficient evidence to conclude that the satisfaction rate differs from 80%.

Note: At α = 0.05, we would reject H₀ since 0.0178 < 0.05.

Confidence Interval for Proportion

\[\hat{p} \pm z^* \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]

Note: For CI, we use $\hat{p}$ in the standard error (not p₀).

Example 5: 95% CI for Proportion

Using Example 1: $\hat{p}$ = 0.55, n = 400

\[SE_{CI} = \sqrt{\frac{0.55 \times 0.45}{400}} = \sqrt{0.000619} = 0.0249\]

\(95\% \text{ CI} = 0.55 \pm 1.96 \times 0.0249 = 0.55 \pm 0.049\) \(= (0.501, 0.599)\)

Since 0.60 is outside this interval, we reject H₀: p = 0.60.

Summary Table

Component Formula
Sample Proportion $\hat{p} = \frac{x}{n}$
Test Statistic $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$
Standard Error (test) $SE = \sqrt{\frac{p_0(1-p_0)}{n}}$
Standard Error (CI) $SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
Conditions np₀ ≥ 10 and n(1-p₀) ≥ 10

Common Applications in Public Administration

Scenario Example H₁
Voter support p > 0.50 (majority supports)
Satisfaction rate p ≠ 0.80 (differs from claim)
Compliance rate p < 0.05 (below standard)
Participation rate p ≠ 0.70 (changed from previous)
Error rate p < 0.02 (below threshold)

Practice Problems

Problem 1

A politician claims 55% support a policy. Sample: n = 300, x = 150. Test at α = 0.05 if support differs from 55%.

Problem 2

Test if more than 70% of employees are satisfied:

  • n = 200, 156 satisfied
  • α = 0.05

Problem 3

A quality standard requires less than 3% defects. Sample: n = 500, 10 defects. Test at α = 0.01.

Problem 4

For Problem 1: (a) Find the p-value (b) Construct a 95% CI for the true proportion

Problem 5

In a sample of 400, 180 prefer option A. Can we conclude at α = 0.05 that fewer than 50% prefer option A?

Summary

Aspect Key Point
Purpose Test claims about population proportion
Conditions np₀ ≥ 10 and n(1-p₀) ≥ 10
Test Statistic Uses p₀ in standard error
CI Uses $\hat{p}$ in standard error
Interpretation Conclude about true population proportion

Next Topic

In the next chapter, we will study Large Sample Test for Two Proportions - comparing proportions from two independent populations.