Large Sample Test for Single Proportion

Definition

A large sample test for a single proportion is a hypothesis test used to test a claim about one population proportion $p$ based on a sample proportion $\hat{p}$ , when the sample size is large enough for the normal approximation to the binomial distribution to be valid.

The test is typically based on the test statistic:

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

where:

$\hat{p} = \frac{x}{n}$ is the sample proportion,
$x$ is the number of successes,
$n$ is the sample size,
$p_0$ is the hypothesized population proportion under the null hypothesis.

The test checks whether the difference between the sample proportion and the hypothesized proportion is large enough to be considered statistically significant.

Main Content

1. Concept of Population Proportion and Sample Proportion

Population proportion $p$

This is the true proportion of individuals/items in the entire population with a certain characteristic. It is usually unknown and is the parameter being tested.

Sample proportion $\hat{p}$

This is the proportion calculated from the sample and is used as an estimate of the population proportion.

For example, if a company claims that 60% of its customers are satisfied, then:

population proportion claim: $p = 0.60$
if a sample of 200 customers shows 118 satisfied customers, then: $\hat{p} = \frac{118}{200} = 0.59$

The sample proportion is only an estimate, so statistical testing is needed to determine whether the observed difference from the claim is due to random sampling variation or a real effect.

2. Hypothesis Testing for a Single Proportion

Null hypothesis $H_0$

This represents the assumed value of the population proportion, usually written as: $H_0: p = p_0$

Alternative hypothesis $H_1$

This represents the research claim or suspicion and may be:
Two-tailed: $H_1: p \neq p_0$
Right-tailed: $H_1: p > p_0$
Left-tailed: $H_1: p < p_0$

The choice of alternative hypothesis depends on the problem statement.

Example: If an examiner wants to test whether the pass rate is different from 70%: $H_0: p = 0.70,\quad H_1: p \neq 0.70$

If a manufacturer wants to check whether the defect rate is more than 5%: $H_0: p = 0.05,\quad H_1: p > 0.05$

The hypotheses form the foundation of the test and determine the rejection region.

3. Test Statistic and Normal Approximation

Test statistic

The large sample z-test uses the standard normal distribution to measure how far the sample proportion is from the hypothesized proportion.

Normal approximation condition

The approximation is valid when the sample size is sufficiently large, usually when: $np_0 \ge 5 \quad \text{and} \quad n(1-p_0) \ge 5$ Some books use 10 instead of 5 as a stricter requirement.

The test statistic is: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

Interpretation:

A z-value near 0 means the sample proportion is close to the hypothesized proportion.
A large positive z-value suggests the sample proportion is greater than $p_0$ .
A large negative z-value suggests the sample proportion is less than $p_0$ .

Example: Suppose $n=100$ , $x=56$ , and $p_0=0.50$ . Then: $\hat{p}=0.56$ $z=\frac{0.56-0.50}{\sqrt{\frac{0.50(0.50)}{100}}} =\frac{0.06}{0.05}=1.20$ This z-value can then be compared with the critical value or used to find a p-value.

Working / Process

1. State the hypotheses

Identify the claimed proportion and write the null hypothesis.
Decide whether the test is left-tailed, right-tailed, or two-tailed based on the wording of the problem.

2. Choose the significance level and compute the test statistic

Select a significance level such as $\alpha = 0.05$ or $\alpha = 0.01$ .
Find the sample proportion $\hat{p}$ .
Check the large sample condition.
Calculate the z-test statistic: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

3. Make the decision and interpret the result

Compare the calculated z-value with the critical value, or determine the p-value.
If the z-value falls in the rejection region or if p-value $< \alpha$ , reject $H_0$ .
Otherwise, fail to reject $H_0$ .
State the conclusion in the context of the problem.

Example process: Suppose a survey of 400 people shows 220 support a proposal. Test the claim that support is 50%.

$H_0: p = 0.50$
$H_1: p \neq 0.50$
$\hat{p} = 220/400 = 0.55$
$z = \dfrac{0.55-0.50}{\sqrt{0.5(0.5)/400}} = \dfrac{0.05}{0.025} = 2.0$

At 5% significance level for a two-tailed test, the critical values are approximately $\pm 1.96$ . Since $2.0 > 1.96$ , reject $H_0$ . This suggests the true proportion is significantly different from 50%.

For visual understanding of a two-tailed decision region:

      Reject H0          Fail to Reject H0          Reject H0
---------|----------------------|----------------------|---------
       -1.96                    0                    +1.96

Advantages / Applications

Simple and practical

The test is straightforward to apply and interpret, especially when the sample size is large.

Widely applicable

It is used in opinion polls, medical research, quality control, marketing surveys, and education.

Decision-making tool

It helps determine whether an observed proportion supports or contradicts a claim with statistical evidence.

Example in quality control

A factory may test whether the proportion of defective products exceeds an acceptable level.

Example in public health

Researchers may test whether the proportion of vaccinated individuals in a community meets a target rate.

Example in surveys

A political analyst may test whether the proportion of voters favoring a candidate is significantly greater than 50%.

Summary

The large sample test for a single proportion is used to test a claim about one population proportion.
It uses the sample proportion and the normal approximation to make statistical decisions.
Important terms to remember: population proportion, sample proportion, null hypothesis, alternative hypothesis, z-test, significance level, rejection region, p-value, normal approximation