= Solution
For $V=\lambda\phi^4$,
$$
\epsilon_V=\frac{8M_{\rm Pl}^2}{\phi^2},
\qquad
\eta_V=\frac{12M_{\rm Pl}^2}{\phi^2}=\frac32\epsilon_V.
$$
Thus $n_s-1=-6\epsilon_V+2\eta_V=-3\epsilon_V$. Combining this with the <tensor-to-scalar ratio> $r=16\epsilon_V$ gives the <quartic-inflation tilt-tensor relation>
$$
\boxed{n_s-1=-\frac3{16}r}.
$$
The measured tilt $n_s-1=-0.04$ would require $r\simeq0.213$, whereas $r=0.001$ predicts $n_s-1\simeq-1.88\times10^{-4}$. Therefore \b[quartic slow-roll inflation is inconsistent with the stated measurements].
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