Finding Margin of Error and Mean

[rkslperez]

Hey there,

I would appreciate a push in the right direction.

We are doing confidence interval in our Statistics class. I just got the hang of constructing an interval but now they want me to go backwards... looks like...

question:

A biologist reports a confidence interval of (2.1, 3.5) when estimating the mean height of a sample of seedlings. Find the Margin of Error and then the Mean.


answers given in book .7 and 2.8


Looks like (2.1, 3.5) are the left and right tails and somewhere in the middle of that is the mean, (Xbar).

Book shows Xbar = sum of all X variables / n -- so is it 2.1 + 3.5 / 2 = 2.8 for mean?

then the margin of error - E = Zc(o/ sqr root of n)
but it doesn't list a Zc, o, nor n ??

Confused?

[HallsofIvy]

Sure. You know that the true value lies somewhere between 2.1 and 3.5, a "total error" of 3.5- 2.1= 1.4 (That is, if you we to pick one of the extremes, either 2.1 or 3.5, as your estimate you might be off by as much as 1.4. If you are clever you might notice that if you choose the "middle" point as your estimate, your error couldn't be more that 1.4/2= 0.7- and that is as small as you can make it, your "margin of error". Now you can compute that the estimate that will give you that is 2.1+ 0.7= 3.5- 0.7= 2.8, your "mean".

I stated it this way because you said your problem specifically required you to find the "margin of error" 'and then the Mean'. Personally, I would have done it the other way: knowing that the true value is between 2.1 and 3.5, I would have immediately take the mean as their average (2.1+ 3.5)/2= 5.6/2= 2.8 and then calculated that the margin of error is 3.5- 2.8= 2.8- 2.1= 0.7.