In a study of teachers, 150 were surveyed and 72% volunteered to a local charity in the past year. Which inference method should be used to test whether the population proportion of volunteers is less than 0.75?

In a study of teachers, 150 were surveyed and 72% volunteered to a local charity in the past year. Which inference method should be used to test whether the population proportion of volunteers is less than 0.75?

• A. One-Proportion Z-Test
• B. Two-Sample T-Test
• C. One-Sample T-Test
• D. Chi-Square Test for Independence

Answer: (A)

When you want to test a claim about a population proportion using binary data from a single sample, you use a one-proportion z-test if the sample is large enough for the normal approximation. Here, you have one sample of 150 teachers with a observed proportion of volunteers of 0.72, and you want to know if the true population proportion is less than 0.75. The null hypothesis is p = 0.75 and the alternative is p < 0.75. Because this is a test of a single proportion and the sample size is large enough (np0 and n(1−p0) are well above 5), the normal approximation is appropriate. The test statistic would be z = (p_hat − p0) / sqrt(p0(1 − p0) / n). With p_hat = 0.72, p0 = 0.75, and n = 150, the standard error is sqrt(0.75 × 0.25 / 150) ≈ 0.035, giving z ≈ (0.72 − 0.75) / 0.035 ≈ −0.85. The corresponding left-tailed p-value is around 0.20, which is not statistically significant at common significance levels, though the key takeaway here is the method itself. The other methods aren’t appropriate for this question: a two-sample t-test compares means between two groups, a one-sample t-test compares a sample mean to a value using the t distribution, and a chi-square test for independence looks at relationships between two categorical variables in a table. The one-proportion z-test is the right choice for testing a claim about a single population proportion.

When you want to test a claim about a population proportion using binary data from a single sample, you use a one-proportion z-test if the sample is large enough for the normal approximation.

Here, you have one sample of 150 teachers with a observed proportion of volunteers of 0.72, and you want to know if the true population proportion is less than 0.75. The null hypothesis is p = 0.75 and the alternative is p < 0.75. Because this is a test of a single proportion and the sample size is large enough (np0 and n(1−p0) are well above 5), the normal approximation is appropriate.

The test statistic would be z = (p_hat − p0) / sqrt(p0(1 − p0) / n). With p_hat = 0.72, p0 = 0.75, and n = 150, the standard error is sqrt(0.75 × 0.25 / 150) ≈ 0.035, giving z ≈ (0.72 − 0.75) / 0.035 ≈ −0.85. The corresponding left-tailed p-value is around 0.20, which is not statistically significant at common significance levels, though the key takeaway here is the method itself.

The other methods aren’t appropriate for this question: a two-sample t-test compares means between two groups, a one-sample t-test compares a sample mean to a value using the t distribution, and a chi-square test for independence looks at relationships between two categorical variables in a table. The one-proportion z-test is the right choice for testing a claim about a single population proportion.