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     Confidence Interval Calculator for the Mean (Unknown Population Standard Deviation)

This calculator computes a confidence interval for the population mean \(\mu\), in the case that the population standard deviation \(\sigma\) is not known, and we use instead the sample standard deviation. Please type the sample mean, the sample standard deviation, the sample size and the confidence level, and the confidence interval will be computed for you:

Sample Mean (\(\bar X\))
Sample St. Dev. (\(s\))
Sample Size (\(n\))
Confidence Level
(Ex: 0.99, 0.95, or 99, 95 without "%", etc)

More about the confidence intervals so you can better interpret the results obtained by this calculator: A confidence interval is an interval (corresponding to the kind of interval estimators) that has the property that is very likely that the population parameter is contained by it (and this likelihood is measure by the confidence level). In this case the population parameter is the population mean (\(\mu\)). Confidence intervals have several properties:

  • They correspond to an interval that is very likely to contain the population parameter being analyzed

  • Such likelihood is measured by the confidence level, that is set at will

  • The higher the confidence level, the wider the confidence interval is (if everything else is equal)

  • For confidence intervals for \(\mu\), they are symmetric with respect to the sample mean, this is the sample mean is the center of the interval.

The formula for a confidence interval for the population mean \(\mu\) when the population standard deviation is not known is

\[CI = (\bar x - t_{\alpha/2, n-1} \times \frac{ s }{ \sqrt n }, \bar x + t_{\alpha/2, n-1} \times \frac{ s }{ \sqrt n })\]

where the value \(t_{\alpha/2, n-1}\) is the critical t-value associated with the specified confidence level and the number of degrees of freedom df = n -1. For example, for a confidence level of 95%, we know that \(\alpha = 1 - 0.95 = 0.05\) and a sample size of n = 20, we get df = 20-1 = 19 degrees of freedom, and using a t-distribution table table (or Excel) we find that \(t_{0.025, 19} = 2.093\).

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