# Quick question about means/confidence intervals

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#1
If I am given the standard deviation, the mean and the number of observations of a random variable, how do I work out the confidence level of the true mean being within 10% of the mean. i.e. the confidence that the true mean = estimated_mean 0.1*estimated_mean

Help would be appreciated.

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6 years ago
#2
(Original post by The Racist Dragon)
If I am given the standard deviation, the mean and the number of observations of a random variable, how do I work out the confidence level of the true mean being within 10% of the mean. i.e. the confidence that the true mean = estimated_mean 0.1*estimated_mean

Help would be appreciated.
It depends on the sample size.

A large sample size you can use the normal distribution, a small one and you use the t-distribution.

For instance, for a normal distribution, the confidence interval is:

For a t-distribution, the confidence interval is:

being the significance level, i.e. in this case, .
0
#3
(Original post by CTArsenal)
It depends on the sample size.

A large sample size you can use the normal distribution, a small one and you use the t-distribution.

For instance, for a normal distribution, the confidence interval is:

For a t-distribution, the confidence interval is:

being the significance level, i.e. in this case, .
The confidence level is what I'm calculating, I want to know the confidence level that the true mean is within 10% of the estimated mean, i don't want the 10% confidence interval of the estimated mean.

I know it's a little confusing, I had to read the question a few times myself...!

So far I have calculated that:

=

But I don't know how to find from that, I'd probably need some crazy formula...

I think I should make a new thread asking how to find alpha when I already know what Z is, this thread has too much unnecessary info in it.
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