Among students who did not withdraw

[manhhiep2509]

Hello.

Please clarify the choice A in the question 31 (diagnostic test in OG 13). The below is my attempt to explain the choice and where I am stuck.

Choice A says "Among students who did not withdraw, students enrolled in online courses got higher grades, on average, than students enrolled in classroom-based courses."

I know that the average grade for online courses matches that for class-based courses.

I know that the rate of withdrawal is higher for online courses than for class-based courses. -- Here, I did not really understand how the rate can be related to the formula to calculate the average grades but I guess the rate can decrease the average grades.

I know that the rate of withdrawal decreases the average grade for online classes, so there must be something compensating for the decrease so that the two average grades could be equal.

However, since the number of people who attend courses could affect the average grades, the grades of people who do not withdraw is not the only factor that decides the average grades. That is why I cannot explain the choice A.

Thank you.

[RonPurewal]

The rate of withdrawal is proportional to the number of students in the class. It would be expressed as a percent, or as a fraction, or as a number per 1000, or whatever. (This is not specialized knowledge; the word "rate" ALWAYS refers to a quantity that's proportional to some total, or, in other words, something divided by something else.) So, your objection doesn't fly.

(If the problem had mentioned the relative numbers of withdrawals, then you'd have a valid point here.)

[manhhiep2509]

Is the grade of students who withdraw zero?

Even though I think it may be zero, since the stimulus does not say so I consider the grade as a variable in the calculation of average grade. However, with the calculation, I cannot come up with the result like that in choice A.

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is there another way to solve the problem and similar ones, besides using math formula?

Thank you.

[tim]

The problem does say that a withdrawal counts as a failure, so you can assume that is a zero. You don't really need a math formula though, just the intuition that if two sets have the same average and one has more zeros, then it's non-zeros must be higher on average than those in the other set.