 Solution

A null hypothesis specifies the population comparison to be tested, for example equality of the population means for women and their complement. The alternative hypothesis for a two-sided test is a nonzero difference. A Student's t-test measures an estimated difference in units of its estimated standard error:On the reported calibration, the sampling distribution under the null hypothesis is Student's t-distribution with $1225$ statistical degrees of freedom. Consequently the two-sided test has p-valueWith this many statistical degrees of freedom, the Student t-distribution is close to the standard normal distribution: $|t|=2.5$ exceeds the approximate $5\%$ two-sided critical value $1.96$, but falls short of the $1\%$ value $2.576$. Using the supplied calibration, reject at the $5\%$ significance level, but not at $1\%$. The large statistical degrees of freedom justify this comparison; a value $2.5$ need not be significant for an arbitrary very small number of statistical degrees of freedom.

The p-value is the probability, under the specified null hypothesis and sampling assumptions, of a test statistic at least as discrepant as the observed one. It is not the probability that the null hypothesis is true, nor the probability that the observed difference happened “by chance”. A significance level of $5\%$ controls the long-run type I error rate of the testing procedure when its assumptions hold. Failing to reject at $1\%$ does not prove equality. Statistical significance also does not measure the substantive size or importance of a difference.

The reported sample means place the women's mean $1.4$ scale units above the overall mean. The negative test statistic can reflect the order of subtraction; it does not reverse that observed ordering. An appropriate pre-specified directional one-sided test would have a tail p-value approximately $0.00627$ on the supplied calibration, but one cannot choose that direction after inspecting the results simply to halve a p-value.

There is a design issue behind the numerical calculation: the women are included in the overall sample. Comparing a subgroup mean with its containing sample mean requires accounting for shared observations. If $w$ is the women's sample fraction and ${\overline{x}}_O$ the mean for the remaining respondents, thenThese two estimates are not independent. An ordinary independent-two-sample Student's t-test of women against the entire respondent sample would therefore use an inappropriate standard error. The interpretation above accepts the supplied test statistic and its stated null calibration; a valid analysis of the underlying data should compare disjoint groups, or calculate the shared-sample covariance correctly. The summary alone cannot establish which calculation was used. Simple random sampling supports sampling inference under the stated design, but an observed gender association does not establish a causal effect.