REASONING Explain what happens to the margin of sampling error when the size of the sample $n$ increases. Why does this happen?
since, $n$ is in the denominator it always divides the whole value and thus the margin of sampling
error decreases when $n$ increases.
this question is asking us to figure out what happens to the margin. Very the m E. When r N r sample size gets bigger. Okay, well, look at the formula for margin of error that I've written out here. OK, we've got the march a very equals two times the square root of our percent p times a quantity of one minus that percent divided by n our sample size. Look where n is. Okay, And is the denominator all right? There's a fraction there and is the denominator in it. We know a fraction bar means the same thing is division. Well, think about how division works, right? We know division is supposed to be splitting things up into smaller pieces. So the larger our denominator is, the bigger the number that we're dividing by meeting the more groups we have to split things up, meaning thes smaller. The number we get right, If you just think about a random example, if I say what's 12 divided by three? Well, that's four. But if we take 12 divided by four, a bigger denominator than what we had, we get a smaller answer, right? And I can keep going if we take 12 divided by six. That's the biggest denominator that we've divided by yet. And that gives us the smallest number of an answer that we've had yet, right? So the idea here is as n is increasing as our denominator is increasing, our margin of air is going to have to get what smaller? Because that's the answer we're getting from the fraction on the right side. So the bigger the denominator, the worst, the smaller the number that fraction gives us, the smaller the margin of error we're gonna happen.