Which statistical test is used to compare the means of two distributions when the population standard deviation is unknown?

The t-test is the appropriate statistical test for comparing the means of two distributions when the population standard deviation is unknown. This test is specifically designed for scenarios where sample sizes are small, and the normality assumption must be considered. The t-test uses the sample means and sample standard deviations to estimate the difference between groups while accounting for the additional uncertainty that comes from estimating the population standard deviation from the sample.

In contrast, the z-test is used when the population standard deviation is known or when the sample size is large (typically over 30), allowing for the approximation of the population parameters. The chi-squared test is employed for categorical data to assess how observed frequencies differ from expected frequencies in a contingency table and is not suited for mean comparisons. ANOVA is used when comparing the means of three or more groups rather than just two, making it irrelevant for this specific question. Thus, the t-test stands out as the proper choice for examining the means between two independent samples when the standard deviation of the population in question is not known.