A random sample of 20 beat cops yields a mean salary of 51,300 with a standard deviation of 1,900. Which inferential test is appropriate to assess whether the true mean salary is less than 52,000?

A random sample of 20 beat cops yields a mean salary of 51,300 with a standard deviation of 1,900. Which inferential test is appropriate to assess whether the true mean salary is less than 52,000?

• A. 1-sample t-test with 19 degrees of freedom
• B. 1-sample z-test
• C. Paired t-test
• D. Two-sample t-test

Answer: (A)

When you’re testing whether a single population mean is below a fixed value from one sample, you compare the sample mean to the hypothesized mean using a one-sample t-test if the population standard deviation is unknown and the sample is not large. In this case, the population standard deviation isn’t known (you have only the sample standard deviation, 1,900), and the sample size is 20, so a t-test is appropriate. Compute the t statistic: t = (Xbar − μ0) / (s/√n) = (51,300 − 52,000) / (1,900/√20) ≈ −700 / 425 ≈ −1.65. With 19 degrees of freedom (n − 1), this is a one-sided test (mean < 52,000). The critical value for a 0.05 level one-sided test is about −1.73, so the observed t is not enough to declare significance at 0.05, though the estimate points in the expected direction. Other tests don’t fit here: a z-test would be used only if the population standard deviation were known or the sample were very large; a paired t-test is for related measurements on the same subjects or matched pairs; a two-sample t-test is for comparing means of two independent groups.

When you’re testing whether a single population mean is below a fixed value from one sample, you compare the sample mean to the hypothesized mean using a one-sample t-test if the population standard deviation is unknown and the sample is not large. In this case, the population standard deviation isn’t known (you have only the sample standard deviation, 1,900), and the sample size is 20, so a t-test is appropriate.

Compute the t statistic: t = (Xbar − μ0) / (s/√n) = (51,300 − 52,000) / (1,900/√20) ≈ −700 / 425 ≈ −1.65. With 19 degrees of freedom (n − 1), this is a one-sided test (mean < 52,000). The critical value for a 0.05 level one-sided test is about −1.73, so the observed t is not enough to declare significance at 0.05, though the estimate points in the expected direction.

Other tests don’t fit here: a z-test would be used only if the population standard deviation were known or the sample were very large; a paired t-test is for related measurements on the same subjects or matched pairs; a two-sample t-test is for comparing means of two independent groups.