= 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>:
$$
t=\frac{\widehat\Delta-0}{\widehat{\operatorname{SE}}(\widehat\Delta)}.
$$
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-value>
$$
\boxed{p=\Pr\{|T_{1225}|\geq2.5\}\simeq0.01255.}
$$
With 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 $2.576$. \b[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, then
$$
\overline x=w\overline x_F+(1-w)\overline x_O,\qquad
\overline x_F-\overline x=(1-w)(\overline x_F-\overline x_O).
$$
These 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>.
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