Consider the following analogy to water pipes: Wires are like pipes already filled with water; the water resembles the movable charge, in case of metal electrons. Voltage is a pressure difference between two points, e.g. because one is higher up than the other. The pipes are really broad in comparison to the slow flow of water in it, so one can ignore friction/turbulence. Resistors are like chokes, simply a very narrow passage.
From this point of view it should be clear that the $\frac{volume\space of\space water}{time \space interval}$ is the same everywhere in the pipe. If you leave the pressure constant and enlarge the choke point, there will be more flow of water. If you squeeze it almost tight, there will be almost no flow of water, even though the pipes are nice and wide everywhere else.
From this point of view it should be really intuitive and trivial how simple circuits involving Rs and voltage sources behave.
My guess is that you confused current (which is $\frac{charge}{time \space interval}$ ) with the speed the charges move, which is called [drift velocity](see wikipedia).
This site has correct and very intuitive explanations of related concepts, and Ohm's Law and parallel circuits are very important but also easily understood given the analogy I gave above.
UPDATE :
John : Thanks for data. Graph is ok. I note your intercept is E=3.94V but your calculations use E=4.5V. This explains the discrepancy in your results. If you use 3.94V you get r ranging from 1.59 to 1.76, close to slope value of 1.68 Ohms.
ORIGINAL ANSWER :
Your line of best fit gives an average internal resistance r based on all measurements. If data points do not lie exactly on this line then the value of r calculated for individual data points (measured pairs of V and I) will not be exactly the same as the slope of the line of best fit.
If you have drawn the line correctly some points will be above the line and some below, with about as many each side, and with the above and below points distributed randomly.
However, it sounds as though there is a consistent trend in your data points : eg all 'below' points at low current and all 'above' points at high current. This suggests that internal resistance was not in fact constant, within the limitations of experimental error. You do not say how big an effect this is : if small, you may be able to ignore it.
EMF and r should be measured when the current drawn is very small, ideally 0. Possibly you have taken readings at a high current, or you have taken a long time to take them. This can have two effects : (i) depleting the battery, reducing EMF, and (ii) increasing r because the battery is warming up and this increases internal resistance.
Your observation that internal resistance increased as current decreased suggests to me that you may have started readings with a high current then worked down to low current.
You will need to decide for yourself what went wrong, perhaps after consulting your teacher again and explaining how you took the readings.
Best Answer
That's only true if the voltage is fixed (constant).
Because it takes a certain amount of work per unit charge (i.e. voltage) to move the charge through the resistor per unit time. The greater the resistance the greater the work per unit charge (voltage) needed to overcome the resistance and maintain the same charge per unit time through the resistor. Consequently, if the voltage across the resistor is fixed, increasing the resistance will decrease the current.
Hope this helps.