= Solution
For $S=\sum_i t_i$, the <likelihood> is $L(\lambda)=\lambda^ne^{-\lambda S}$, and its log derivative is $n/\lambda-S$. Thus, almost surely,
$$
\widehat\lambda=n/S,\qquad I_{\rm full}(\lambda)=n/\lambda^2.
$$
The large-sample distribution is that of the estimator, rather than of the fixed parameter: $\sqrt n(\widehat\lambda-\lambda)\xrightarrow d N(0,\lambda^2)$. A <Wald confidence interval> is consequently
$$
\boxed{\widehat\lambda\ \pm\ z_{1-\alpha/2}\widehat\lambda/\sqrt n},
$$
intersected with the positive parameter space if necessary. The unadorned parameter in the source's distribution request is interpreted as this estimator.
A single aphid survives to $T$ with probability $r=e^{-\lambda T}$, independently of the others. Hence $N_T\sim\operatorname{Bin}(n,r)$, and
$$
L_T(\lambda)=\binom n{n_T}e^{-\lambda Tn_T}(1-e^{-\lambda T})^{n-n_T},\qquad
\widetilde\lambda=-\frac1T\log(n_T/n)
$$
for $0<n_T<n$. If $n_T=n$, the optimum is at the boundary $\lambda=0$; if $n_T=0$, the <likelihood> increases to its supremum as $\lambda\to\infty$, so there is no finite maximum <likelihood> estimate. <Fisher information> transforms from the binomial probability parameter as
$$
I_T(\lambda)=\frac{n(r'(\lambda))^2}{r(1-r)}=\frac{nT^2}{e^{\lambda T}-1}.
$$
The given strict inequality shows
$$
\boxed{\frac{I_T(\lambda)}{I_{\rm full}(\lambda)}=\frac{(\lambda T)^2}{e^{\lambda T}-1}<1}.
$$
Thresholding discards information about how long each survival time is. The second experiment therefore has a larger information bound on <variance> and larger asymptotic uncertainty for an efficient estimator. This is an expected-information comparison, rather than a claim that every realised finite-sample <confidence interval> must be wider.
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