You are correct about the second scenario, for the reason you give, but not about the first scenario. The theory of the finite population correction (fpc) applies only to a random sample without replacement (Lohr (2009) Sec 2.8,pp 51-530. The key word is random. The hallmark of a random sample is that selection is determined by random numbers or the physical equivalent. In your first scenario the 45% of the population who responded were not selected by random numbers. The same would be true if the 45% were part of an even large random sample of the population: response is not governed by random numbers.
Even if you have a sample of a substantial part of the population with (near) 100% response, you should still omit the fpc if the purpose of your study is to develop predictions, estimate odds ratios, or to otherwise test hypotheses or quote p-values. The reasoning is interesting (Cochran, 1977, p.39): For a finite population it is seldom of scientific interest to ask if a null hypothesis (e.g. that two proportions are equal) is exactly true. Except by a very rare chance, it will not be, as one would discover this by enumerating the entire population. This leads to the adoption of a "superpopulation" viewpoint, which is taken by almost all statisticians these days. Your second scenario is a variant of this. See also Deming(1966) pp 247-261 "Distinction between enumerative and analystic studies"; Korn and Graubard (1999), p. 227.
ADDED NOV 26
I should have noted that the finite population correction is a minor concern here. The major problem is the 55% non-response and the subsequent non-response bias. Survey professionals universally agree that it is better to take a smaller manageable sample and to focus on reducing non-response by personalizing the initial contacts and by following-up with non-responders. Post-survey weighting fixes may also help, but will increase standard errors.
In summary, to answer your three questions:
- Your interpretation of the first scenario is incorrect.
- You really don't need to say anything. If your goal is to describe only the finite population from which you drew the sample, then you can mention that you omit the fpc because the effect is miniscule. Otherwise, when you do hypothesis testing or prediction, you could mention that omit the fpc, but I've never seen anyone do it.
- The decision of whether to use the fpc is the assessment you describe in the question.
So the answer is "Yes".
Additional discussion See a related CV discussion here.
References
Cochran, W. G. (1977). Sampling techniques (3rd Ed.). New York: Wiley.
Deming, W. E. (1966). Some theory of sampling. New York: Dover Publications.
Korn, E. L., & Graubard, B. I. (1999). Analysis of health surveys (Wiley series in probability and statistics). New York: Wiley.
Levy, Paul S, and Stanley Lemeshow. 2008. Sampling of populations : methods and applications. Wiley series in survey methodology. Hoboken, N.J: Wiley.
Lohr, Sharon L. 2009. Sampling: Design and Analysis. Boston, MA: Cengage Brooks/Cole.
2nd order inclusion probability is defined as the probability that both item i and item j (with j $ \ne i$) are in the sample.
By use of the inclusion-exclusion principle https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle ,
P(i and j both in the sample) = P(i in the sample) + P(j in the sample) - (1 - P(neither i nor j in the sample))
You already know that P(i in the sample) = P(j in the sample) = $1 - (1- \frac{1}{N})^m$
We have that P(neither i nor j in the sample) = $(1- \frac{2}{N})^m$
Putting it altogether results in P(i and j both in the sample) = $1 - 2(1- \frac{1}{N})^m + (1- \frac{2}{N})^m$
Best Answer
Expanding on the answer of @Scortchi . . .
Suppose the population had 5 members and you have budget to sample 5 individuals. You are interested in the population mean of a variable X, a characteristic of individuals in this population. You could do it your way, and randomly sample with replacement. The variance of the sample mean will be V(X)/5.
On the other hand, suppose you sample the five individuals without replacement. Then, the variance of the sample mean is 0. You've sampled the whole population, each individual exactly once, so there is no distinction between "sample mean" and "population mean." They are the same thing.
In the real world, you should jump for joy each time you have to do the finite population correction because (drumroll . . .) it makes the variance of your estimator go down without you having to collect more data. Almost nothing does this. It's like magic: good magic.
Saying the exact same thing in math (pay attention to the <, and assume sample size is greater than 1): \begin{equation} \textrm{finite sample correction} = \frac{N-n}{N-1} < \frac{N-1}{N-1} = 1 \end{equation}
Correction < 1 means that applying the correction makes the variance go DOWN, 'cause you apply the correction by multiplying it against the variance. Variance DOWN == good.
Moving in the opposite direction, entirely away from math, think about what you are asking. If you want to learn about the population and you can sample 5 people from it, does it seem likely that you will learn more by taking the chance of sampling the same guy 5 times or does it seem more likely that you will learn more by ensuring that you sample 5 different guys?
The real world case is almost the opposite of what you are saying. Almost never do you sample with replacement --- it's only when you are doing special things like bootstrapping. In that case, you are actually trying to screw up the estimator and give it a "too big" variance.