Find Confidence Interval w/o Formula: Uniform Distribution

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Homework Help Overview

The discussion revolves around finding a confidence interval for a set of numbers, specifically focusing on uniform and Bernoulli distributions. The original poster has already calculated the mean and standard deviation but is seeking guidance on how to compute the confidence interval without a provided formula.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need for a formula to compute the confidence interval and reference a practical example from Wikipedia. There are mentions of using Z-scores for confidence intervals and the implications of approximating uniform and Bernoulli distributions to a normal distribution.

Discussion Status

Some participants have offered insights into using Z-scores for calculating confidence intervals, while others are questioning the applicability of these methods to uniform and Bernoulli distributions. There is no explicit consensus on the approach to take for these specific distributions, indicating an ongoing exploration of ideas.

Contextual Notes

The original poster has clarified the need for confidence intervals specifically for uniform and Bernoulli distributions, which may not follow the same methods as normal distributions. This introduces additional complexity to the discussion.

lildrea88
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i am given a set of numbers. I have already found the mean, standard deviation, etc.
i am now asked to find the confidence interval. but was not given a formula in order to compute this. does anyone know one?
 
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write it in terms of the mean (that u shaped thing in the normalization eqation), choose the correct value of Z for the confidence interval they give.

95% confidence interval ==> z=1.96 (for 2 tails)

98% confidence is something like , i don't know off by heart, check the normal table
 
cloud360 said:
write it in terms of the mean (that u shaped thing in the normalization eqation), choose the correct value of Z for the confidence interval they give.

95% confidence interval ==> z=1.96 (for 2 tails)

98% confidence is something like , i don't know off by heart, check the normal table

So you want P(-1.96<Z<1.96) and you know that Z = (X-μ)/(σ/√n)

so in all you want the CI to be

P(-1.96<(X-μ)/(σ/√n)<1.96)
 
i need a confidence interval for the uniform and bernoulli distribution...sorry i should have said that before
 
lildrea88 said:
i need a confidence interval for the uniform and bernoulli distribution...sorry i should have said that before

I can't really recall if it is the same for those distribution, but you can approximate them to a normal distribution.
 

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