Statistics and finding margin of error:
Given that you have calculated a 99% confidence interval of 13.4 to 21.16. Determine the margin of error?

M(error) = Z_(alpha/2) * variance/sqrt(sample)

Question

Statistics and finding margin of error:
Given that you have calculated a 99% confidence interval of 13.4 to 21.16. Determine the margin of error?

M(error) = Z_(alpha/2) * variance/sqrt(sample)

anonymous · Answer

@jdoe0001 @terenzreignz @Compassionate

anonymous · Answer

P( -2.575 < $$(\bar{x} - \mu ) \over{ \sigma\over \sqrt(n)}$$ < 2.575)

terenzreignz · Answer

so far everything I've googled says the margin of error is just half of the difference of the two endpoints...

terenzreignz · Answer

Does this involve a normal distribution, by any chance?

anonymous · Answer

yes it does - sorry for the belated response.

anonymous · Answer

(21.16 - 13.4)/2 was the correct answer.

anonymous · Answer

Thanks a bunch @terenzreignz

terenzreignz · Answer

That was it? Just half the difference of the endpoints after all?
XD

terenzreignz · Answer

You and your formulas -_-
and normally, when someone tags me like that I'd say 'please call me TJ'
but that'd be rather weird in your case, so... bleh... 

Have a nice day ^_^

anonymous · Answer

Vice versa, have a great day!