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The confidence interval can be expressed in terms of statistical significance, e.g.: "The 95% confidence interval represents values that are not statistically significantly different from the point estimate at the .05 level." [20] Interpretation of the 95% confidence interval in terms of statistical significance.
a) The expression inside the square root has to be positive, or else the resulting interval will be imaginary. b) When g is very close to 1, the confidence interval is infinite. c) When g is greater than 1, the overall divisor outside the square brackets is negative and the confidence interval is exclusive.
Classically, a confidence distribution is defined by inverting the upper limits of a series of lower-sided confidence intervals. [15] [16] [page needed] In particular, For every α in (0, 1), let (−∞, ξ n (α)] be a 100α% lower-side confidence interval for θ, where ξ n (α) = ξ n (X n,α) is continuous and increasing in α for each sample X n.
4.2 Confidence intervals. ... Download as PDF; Printable version; In other projects ... and simplifies formulas in some contexts, ...
The probability density function (PDF) for the Wilson score interval, plus PDF s at interval bounds. Tail areas are equal. Since the interval is derived by solving from the normal approximation to the binomial, the Wilson score interval ( , + ) has the property of being guaranteed to obtain the same result as the equivalent z-test or chi-squared test.
In the social sciences, a result may be considered statistically significant if its confidence level is of the order of a two-sigma effect (95%), while in particle physics and astrophysics, there is a convention of requiring statistical significance of a five-sigma effect (99.99994% confidence) to qualify as a discovery. [3]
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The rule can then be derived [2] either from the Poisson approximation to the binomial distribution, or from the formula (1−p) n for the probability of zero events in the binomial distribution. In the latter case, the edge of the confidence interval is given by Pr(X = 0) = 0.05 and hence (1−p) n = .05 so n ln(1–p) = ln .05 ≈ −2