= Solution
{c}
Let $F_{n-1}$ be the <cumulative distribution function> of the <Student t-distribution>. The two-sided <p-value> is $2[1-F_{n-1}(|t_0|)]$. If it is less than $\alpha$, strict monotonicity gives $|t_0|>t_{n-1,1-\alpha/2}$, hence
$$
|\overline x|>t_{n-1,1-\alpha/2}\frac s{\sqrt n}.
$$
The magnitude of the interval's centre is larger than its half-width, so \b[zero lies outside the confidence interval]. This is the <duality between confidence intervals and hypothesis tests>.
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