Question

Math meets chemistry...

Answer 1

After completing a titration (and various calculations), I get two final values which refer to a percentage of mass... These are : 0.109g and 0.116g.
The actual values should be 0.137, for both.
In my analysis I am asked to discuss the significance of my values with respect to the "confidence interval" ... What is that, re: statistics? And how do I calculate it?
Thanks to any help!

Answer 2

ok your confindence interval is a value that you are resonably sure that the value of your statistic falls in the interval of your findings

Answer 3

so the eventual calculated 'error'? +-0.1 (or something) after getting through all the calcs with the error of each calculation etc?

Answer 4

does it say anything about the distubition of the population?

Answer 5

No, it doesn't .... But I CAN calculate the 'associated error' - maybe this is just a case of being unfamiliar with the terminology?

Answer 6

ummm not really it's more like if you do the sampling x times you will get your goal mark a x percentage of the time.

Answer 7

Oh, right..... Ummm OK? Hahaha so if the only values I have are an actual, two test results, and their error... Do you think my answer will just be a hypothetical response?

Answer 8

So your interval is (.109g,.116) and you are looking for .137g so in this case the interval doesn't hold the .137

Answer 9

So this would indicate that you should redo your titration in an attempt to get with in your range

Answer 10

What if my answers are \[0.109 \pm 0.1\]
etc? then they hold, right?

Answer 11

Confidence interval = (lower bound, upper bound)

Lower bound = point estimate - margin of error

Upper bound = point estimate + margin of error


The point estimate is the true value you're trying to measure (in this case, it's the percentage of mass)

The margin of error is determined by the sample size and the level of confidence desired.

Answer 12

Oh I should clarify that the point estimate (in this case) should be the average percentage of mass and it should be close to (but not exactly equal to) the true value.

Answer 13

i don't think you should add or subtract the .1 i'm guessing thats your standard devation and is taken in account to find your C.I.

Answer 14

Thanks guys!