Question

I'm willing to do the work. I want to learn!
Topic: Margin of Error
I will upload the question and attempted work when thread becomes open.

Answer 1

The standard deviation for the data in the table is 2.71.

For a confidence level of 95%, what is margin of error.

Round to two decimal places.

Two Standard Deviations:

2 X 2.71 = 5.42

X +- 2SD

X = 112 + 106 + 105 + 110 + 109 + 108 + 111 + 105 = 866

X = 866/8 = 108.25

108.25 +- 5.42

108.25 + 5.42

108.25 - 5.42

Would the two prior statements be a way to get the interval estimates, it seems like the interval estimates would be far to 'large'?

Answer 2

Here is the data table as well...

Answer 3

Alternatively, I was think we could take the lower limit and the upper limit from the data table.

112 - 105 / 2 = 3.5 (Margin of Error)

I would greatly appreciate any feedback you may have to offer.

Answer 4

If the prior statement is the case we would still have the same mean in which,

108.25 +- 3.5

108.25 + 3.5 = 111.75

108.25 - 3.5 = 104.75

Those intervals are more reasonable in my opinion, no?

Answer 5

If we used a 5.42 margin of error for our interval estimates it would be (102.83, 113,67).

Answer 6

Compared to (104.75, 111.75). They're both close, but...

Answer 7

@mathmate It's me again. I don't want it to appear like I'm taking advantage of your skills here, but I'm generally befuddled on this question. If you could kindly contribute your opinion once more I would truly appreciate it. I recognize you have other students, so no problem if you can't OR simply don't want to. Thanks again!

Answer 8

I apologize if it's a little confusing, I suppose I'm a little confused as well, so it might translate...

Answer 9

For almost all math problems, the way the question is phrased has a vital impact on the solution, and its related assumptions. In statistics, it is even more so.
So please post the \(original\) question, (preferably a screen shot).
For the question of margin of error, it is important to know the sample size (n=8), population standard deviation (if known), population mean (if known), any known distribution (normal, Poisson, ...). In this case, n=8 is too small to use the central limit theorem.... etc etc. BTW, have you done the t-distribution?
Also, someone else may be more knowledgeable than I am on this subject.

Answer 10

Firstly, thank you for your detailed response. I do know what a t-distribution is, but because this is basic Algebra II we haven't gone into it yet. I truly feel that they wouldn't make it this complicated. I can provide the exact screenshot of the question though. Looking at it I realized that I misphrased the problem. It does NOT ask for interval estimates, but just the margin of error. Does that make more sense now? My apologies!

Answer 11

Oh and @mathmate thank you for returning once again! I appreciate your help up till this point!

Answer 12

In this case, I think what you did was as close as you can get, except I would use 1.96 instead of \(2\sigma\) if you want to split hairs. lol

Answer 13

1.96 as the margin of error?