Yes, when you have a smaller resistance, your RC product decreases and so does the time constant. This means that the ripple will be greater, and the smoothing will be less if your capacitor is in parallel with the resistor.
Unfortunately, there are some words (like "resolution", "smoothing") for which "greater" can mean one of two things. If I say "greater resolution", does it mean it went from 1 mm to 2 mm (the number is greater, but the resolution is poorer) or does it mean "better" (it went from 2 mm to 1 mm)? It is possible that the confusion in your book came about from a similar unfortunate phrasing.
In general, if your resistance is less, the current is greater and you need a bigger "reservoir" (capacitance) in order to ride through the bumps of the supply.
Finally - you really need to look at the entire circuit to answer questions like this properly: there is usually more than one resistor value, and so there is the "charging" and "discharging" current of the capacitor that needs to be taken into account. If the resistor in question is in series with the capacitor, then the capacitor will be charged more quickly - and so as a percentage of the voltage, it might give you a smaller ripple.
This is why it's not easy to answer this question "definitively" without seeing the actual circuit diagram you are using.