Testing population proportion is a hypothesis testing procedure used to assess whether or not a sample from a population represents the true proportion of the entire population. Testing a sample population proportion is a widely used statistical method with various applications across different fields.
Table of Contents
Purpose of Testing Population Proportion (one-sample)
The main purpose of testing a sample population proportion is to make inferences about an entire population based on the sample information. Testing a sample population proportion helps to determine whether an observed sample proportion is significantly different from a hypothesized population proportion.
Common Uses of Testing Population Proportion
The following are some common uses of population proportion:
- Marketing research: To determine if a certain proportion of customers prefer one product compared to another.
- Quality control: In manufacturing, population proportion tests can be used to test/check if the proportion of defective items in a production batch exceeds an acceptable threshold.
- Medical research: To test the efficacy of a new treatment by comparing the proportion of patients who recover using the new treatment versus a standard treatment.
- Political polling: To estimate the proportion of voters supporting a particular candidate or policy.
- Social sciences: To examine the prevalence of certain behaviors or attitudes in a population.
Applications Population Proportion in Various Fields
- Business: Testing customer satisfaction rates, conversion rates in A/B testing for websites, or employee retention rates.
- Public health: Estimating vaccination rates, disease prevalence, or the effectiveness of public health campaigns.
- Education: Assessing the proportion of students meeting certain academic standards or the effectiveness of new teaching methods.
- Psychology: Evaluating the proportion of individuals exhibiting certain behaviors or responses in experiments.
- Environmental science: Measuring the proportion of samples that exceed pollution thresholds.
Types of Testing Population Proportion
There are two types of population proportion tests.
- One-sample z-test for proportion: One-sample proportion tests are used when comparing a sample proportion to a known or hypothesized population proportion.
- Two-sample z-test for proportions: Two-sample proportion tests are used when comparing proportions from two independent samples.
Assumptions and Considerations
The following are assumptions and considerations when testing population proportion:
- The sample should be randomly selected and representative of the population.
- The sample size (number of observations in the sample) should be large enough (typically $np$ and $n(1-p)$ should both be greater than 5, where $n$ is the sample size and $p$ is the proportion).
- For two-sample tests, the samples should be independent of each other.
- Interpretation: The results of these tests are typically interpreted using p-values or confidence intervals, allowing researchers to make statistical inferences about the population based on the sample data.
Data Frive Decisions from Proportion Tests
By using tests for population proportions, researchers and professionals can make data-driven decisions, validate hypotheses, and gain insights into population characteristics across a wide range of fields and applications.
Suppose, a random sample is drawn and the population proportion (say) $\hat{p}$ is measured and $n\hat{p}\ge 5$, $n\hat{q}\ge5$, the distribution of $\hat{p}$ is approximately normal with $\mu_{\hat{p}} =p$ and $\sigma_{\hat{p}}=\sqrt{\frac{pq}{n}}$. Also, suppose that one of the possible null hypotheses of the following form, when testing a claim about a population proportion is:
$H_o: p=p_o$
$H_o:p\ge p_o$
$H_o\le p_o$
For simplicity, we will assume the null hypothesis $H_o:p=p_o$. The standardized test statistics for a one-sample proportion test is
\begin{align*}
Z&=\frac{\hat{p} – \mu_{\hat{p}}}{\sigma_{\hat{p}}}\\
&=\frac{\hat{p} -p_o }{\sqrt{\frac{p_oq_o}{n}}}
\end{align*}
This random variable will have a standard normal distribution. Therefore, the standard normal distribution will be used to compute critical values, regions of rejection, and p-values, as we use it to test a mean using a large sample.
Example 1 (Defective Items): Testing Population Proportion
A computer chip manufacturer tests microprocessors coming off the production line. In one sample of 577 processors, 37 were found to be defective. The company wants to claim that the proportion of defective processors is only 4%. Can the company claim be rejected at the $\alpha = 0.01$ level of significance?
Solution:
The null and alternative hypotheses for testing the one-sample population proportion will be
$H_o:p=0.04$
$H_1:p\ne 0.04$
By focusing on the alternative hypothesis symbol ($\ne$), the test is two-tailed with $p_o=0.04$.
The $\hat{p} = \frac{37}{577} \approx 0.064$.
the standardized test statistics is
\begin{align*}
Z &= \frac{\hat{p} – p_o}{\sqrt{\frac{p_oq_o}{n}}}\\
&=\frac{0.064 – 0.04}{\sqrt{\frac{(0.04)(0.96)}{577}}}\\
&=\frac{0.024}{0.008}\approx 3.0
\end{align*}
Looking up $Z=3.00$ in the standard normal table (area under the standard normal curve), we get a value of 0.9987. Therefore, $P(Z\ge 3.00) = 1-0.9987) = 0.0013$.
Note that the test is two-tailed, the p-value will be twice this amount or $0.0026$.
Since the p-value ($0.0026$) is less than the level of significance ($0.01$), that is $0.0025 < 0.01$ (p-value < level of significance), we will reject the company’s claim. It means that the proportion of defective processors is not 4%, it is either less than 4% or more than 4%.
Example 2 (Opinion Poll): Testing Population Proportion
An opinion poll of 1010 randomly chosen/selected adults finds that only 47% approve of the president’s job performance. The president’s political advisors want to know if this is sufficient data to show that less than half of adults approve of the president’s job performance using a 5% level of significance.
Solution:
The null and alternative hypothesis of the problem above will be
$H_o:p\ge 0.50$
$H_1:p< 0.50$
By focusing on the alternative hypothesis symbol (<), the test is left-tailed with $p_o=0.50$.
The $\hat{p} = 0.47$. The standardized test statistics for one-sample population proportion will be
\begin{align*}
Z &= \frac{\hat{p} – p_o}{\sqrt{\frac{p_oq_o}{n}}}\\
&=\frac{0.47 – 0.50}{\sqrt{\frac{(0.5)(0.5)}{1010}}}\\
&=\frac{-0.03}{0.01573}\approx -1.91
\end{align*}
For a left-tailed test (for $\alpha = 0.05$), the $Z_o=-1.645$. Since $-1.91 < -1.645$, the null hypothesis should be rejected. So the data does support the claim that $p<0.50$ at the $\alpha=0.05$ level of significance.
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