Estimates of Percentage Errors in Cosomology

[resurgance2001]

Homework Statement

Suppose that there is some hypothetical cosmological parameter called C, and suppose that the real value of C is C(real) = 1. Also suppose that everyone in the universe is able to measure C.

Our value of C is C(ours) = 1.15 +/- .15 at confidence level 95 %

Make an estimate for the percentage of measurements that would appear as discrepant as ours.

Homework Equations

$$ \mu = \bar X \pm Z^* \left( \frac \varsigma {\sqrt n} \right)$$

The Attempt at a Solution

I don't know how to proceed from here because I don't know if I have enough information or if I am supposed to make up some of the information myself. I am not 100% sure that I have the right equation any way. Thanks

 

[BvU]

Any idea what the symbols in the relevant equation stand for ?
PF guidelines don't really allow assistance if the attempt at solution comes down to "I don't know"
but what the heck:
You do understand what confidence level 95% means in

Our value of C is C(ours) = 1.15 +/- .15 at confidence level 95 %

I do miss something in the problem statement, namely a clear definition of 'discrepant with ours' . What if someone measures 1.60 +/- .15 at confidence level 95 % ? Is that a discrepancy or agreement ? What does your textbook (or notes) have to say ?

 

[resurgance2001]

Thanks for looking at this. When they say as discrepant as ours they mean that our value of 1.15 versus the real value of 1.0 is considered discrepant. So they are asking what percentage of other observers who make the measurement are likely to come up with measurements that are as far away from the real value as ours.

The textbook does not cover this. It is an issue that has been taken up with the course team because it is a question which seems to be asking for some knowledge/skill that has not been taught anywhere in the course. Notwithstanding whatever the answer is that they are expecting is only very simple because the question is only worth 2 mark out of 100, and it does say we are only making an estimate.

However my knowledge and understanding of statistics is very limited, hence being stuck.

 

[resurgance2001]

The equation I have given is one I have found which is said to be the standard equation for confidence levels.

$$\mu = real value $$
$$\bar X = average experimental value$$

n = sample number
Z* = confidence level
the funny sigma means standard deviation

 

[BvU]

Ah, I can interpret the

Make an estimate for the percentage of measurements that would appear as discrepant as ours

litterally, too. The probability that someone else finds exactly the same answer is as good as zero. That is definitely not intended by the exercise composer. (sorry for wrongly(?) quoting 'discrepant with ours').
And 'as discrepant as ours' is probably also not intended; perhaps 'as discrepant as ours or worse'? Even then:
I'm still short:

suppose that everyone in the universe is able to measure C

but if you don't know how accurate they can do so, then what ? Is someone who reports 1.02 $\pm$ 0.02 (95% CL) as discrepant (namely 1.96 sigma away) as we are ?

I don't think this can be answered without making more assumptions - which may be fine.

The equation I have given is one I have found which is said to be the standard equation for confidence levels.
$$\mu ={\rm real\ value} $$
$$\bar X = {\rm average\ experimental\ value}$$
n = sample number -- you mean sample size: the number of measurements that is being averaged over.
Z* = confidence level
the funny sigma means standard deviation

Sort of, yes. 95% CL means you expect that if the experiment is repeated infinitely -- but with the same accuracy --, about 5% of the results will be outside the interval [1, 1.3]

So perhaps they simply want 5% as an answer: the chance that you are more than 1.96 sigma away (either below or above) from the true value

 

[resurgance2001]

Thanks - yes I am coming to the same conclusion - that what they are looking for is 5%. It seems to me though at really daft question - even possibly a fake question! Thanks for your help. It is very appreciated.

 

[BvU]

You're welcome. Hope it helps...