Question

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Topic: Margin of Error
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Answer 1

Question:
Eight trials are simulated. The results are shown in the table.
112 106 105 110
109 108 111 105
What is the estimated margin of error, using standard deviation? Round to two decimal places.

What I DO know.

Standard deviation of the trails is 2.71241 / BUT I’m rounded so 2.71

I was thinking we could take the lower limit and the upper limit from the data table. 112 - 105 / 2 = 3.5 (Margin of Error) BUT I think it depends on how many deviations we are going from the average, no?

For example, for 95% coverage in a bell curve we would multiply 2.71 X 2 = 5.42

I am a little up in the air on this and I would appreciate an opinion. Thank you.

Answer 2

It doesn't explicitly say ANYTHING about deviations, but I can't help but assume...

Answer 3

Typically, \(z\dfrac{\sigma}{\sqrt{n}}\) for the margin of error of the mean. Remember that there is likely a units problem. +/- 3 potato chips does not constitute the percentage that you need.

Answer 4

OK, I know that formula @tkhunny , BUT we haven't covered that at ALL in my Algebra II class. It would seem odd to use it if we haven't learned it, that's why I'm trying to simplify it and find another way.

Answer 5

I do find this problem different in relation to the fact that most of the problems I have seen thus far have given me the margin of error and asked...

Answer 6

For the interval, for the point estimate, etc...

Answer 7

It is sometimes acceptable to use (max - min)/6 in place of \(\sigma\), but that's only deliberately introducing error that is not anticipated by the margin of error.

Answer 8

This may sound really elementary to you, but where does the six come from?

Answer 9

3 standard deviations each direction. It assumes noting landed farther away than that.

Answer 10

Ahhh! Ok! So based on your previous example,
112 -105 = 7

7/6 = 1.1666...

Answer 11

Someone that doesn't look right. I apologize this is relatively new to me.

Answer 12

@tkhunny What are your thoughts?

Answer 13

I almost feel like we need to 'dumb it down' for a lack of better words...

Answer 14

It may actually be as simple as +-

Answer 15

That is ONLY an estimate for \(\sigma\). It does not spare you from the rest of the formula.

Answer 16

Alternatively, if we ASSUME these are the absolute maximum and minimum, we might be able to state that half the range is the +/- margin of error.

112 - 105 = 7
7/2 = 3.5
3.5 / 108.25 = 0.03233256351039260969976905311778
MOE = +/- 3.3%

I guess that might be more meaningful than nothing.

Answer 17

Ah! Ok, I see what you're saying now. Thank you for the visual!