8.3 One-Sample Test for Proportions

One-sample tests for proportions help us determine if a population's characteristic matches a claimed value. We use these tests to check if survey results or sample data align with expectations, like verifying if 50% of students prefer online classes.

To conduct these tests, we need a random sample, sufficient sample size, and enough successes and failures. We set up hypotheses, calculate a test statistic, and interpret the p-value to draw conclusions about the population proportion.

One-Sample Test for Proportions

Conditions for one-sample proportion tests

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• Randomly select the sample from the population to ensure representativeness and avoid bias
• Ensure the sample size is sufficiently large (typically $n \geq 30$) or confirm the population is at least 10 times the size of the sample ($N \geq 10n$) to justify using the normal distribution approximation
• Verify the sample size is less than 10% of the population size ($n < 0.1N$) to avoid applying the finite population correction factor, which adjusts for the impact of sampling without replacement from a relatively small population
• Check that the expected number of successes and failures in the sample are both at least 10 ($np \geq 10$ and $n(1-p) \geq 10$, where $p$ is the hypothesized population proportion) to ensure the normal approximation is appropriate (e.g., testing the proportion of defective products in a manufacturing process)

Hypotheses in proportion testing

• State the null hypothesis ($H_0$) as the population proportion ($p$) being equal to a specific value ($p_0$): $H_0: p = p_0$, representing the status quo or the claim being tested (e.g., $H_0: p = 0.5$, the proportion of students who prefer online classes is 50%)
• Formulate the alternative hypothesis ($H_a$) based on the research question or claim, choosing from one of the following:
  • Two-tailed: $H_a: p \neq p_0$, indicating the population proportion differs from the hypothesized value in either direction (e.g., $H_a: p \neq 0.5$, the proportion of students who prefer online classes is not 50%)
  • Left-tailed: $H_a: p < p_0$, suggesting the population proportion is less than the hypothesized value (e.g., $H_a: p < 0.5$, the proportion of students who prefer online classes is less than 50%)
  • Right-tailed: $H_a: p > p_0$, proposing the population proportion is greater than the hypothesized value (e.g., $H_a: p > 0.5$, the proportion of students who prefer online classes is greater than 50%)

Test statistics and p-values

• Calculate the test statistic ($z$) using the formula: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$, where:
  • $\hat{p}$ represents the sample proportion (e.g., the proportion of students in the sample who prefer online classes)
  • $p_0$ is the hypothesized population proportion (e.g., 0.5, the claimed proportion of students who prefer online classes)
  • $n$ is the sample size (e.g., the number of students surveyed)
• Determine the p-value, which represents the probability of observing a test statistic as extreme as or more extreme than the calculated value, assuming the null hypothesis is true
  • For a two-tailed test, find the area under the standard normal curve beyond $|z|$ in both tails (e.g., the probability of observing a $z$-score as extreme as $\pm 2.5$ or more extreme)
  • For a left-tailed test, find the area under the standard normal curve to the left of $z$ (e.g., the probability of observing a $z$-score of $-1.8$ or less)
  • For a right-tailed test, find the area under the standard normal curve to the right of $z$ (e.g., the probability of observing a $z$-score of $2.1$ or greater)

Interpreting proportion test results

• Compare the calculated p-value to the predetermined significance level ($\alpha$, commonly 0.05) to make a decision:
  • If the p-value is less than $\alpha$, reject the null hypothesis and conclude there is sufficient evidence to support the alternative hypothesis (e.g., if $p$-value $< 0.05$, conclude that the proportion of students who prefer online classes is significantly different from 50%)
  • If the p-value is greater than or equal to $\alpha$, fail to reject the null hypothesis and conclude there is insufficient evidence to support the alternative hypothesis (e.g., if $p$-value $\geq 0.05$, conclude that there is not enough evidence to suggest the proportion of students who prefer online classes differs from 50%)
• Interpret the results in the context of the problem, considering the practical implications and significance of the findings (e.g., discuss how the results might influence decisions related to course offerings or delivery methods based on student preferences)