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Sagot :
95% confidence interval is wider than 90% confidence interval.
Given as that unknown mean and a known variance.
let's suppose parameter for mean as µ and
for standard deviation σ.
we have sample of size n as x1,x2,x3,x4.......xn.
Confidence means what degree of assurance do we desire that the interval estimate contains the populace mean µ.
The likelihood that the interval falls within the level of confidence c
The population parameter is contained in estimate.
If the level of confidence is 90%, this means that we are 90%
confident that the interval contains the population mean, µ
The corresponding z-scores for 90% confidence interval are ± 1.645.
If the level of confidence is 95%, this means that we are 95%
confident that the interval contains the population mean, µ
The corresponding z-scores for 95% confidence interval are ± 1.96.
So In comparison to a 95% Confidence Interval, a 90% Confidence Interval would be narrower. This happens because the reliability of an interval containing the actual mean declines as the confidence interval's precision rises (i.e., CI width decreases) (less of a range to possibly cover the mean). Consider the case of 100% confidence; the interval must contain all possible values in order to guarantee that the mean is captured with 100% certainty.
To know more about normal distribution here,
https://brainly.com/question/13906025
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