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Confidence Intervals and Hypothesis Testing, Exams of Probability and Statistics

How to calculate confidence intervals for population means and proportions, as well as the relationship between confidence intervals and hypothesis testing. It covers the formulas for confidence intervals for single and paired samples, as well as independent samples, and discusses the use of z-scores and t-scores. The document also mentions the importance of assuming normal distribution for using t-scores and the differences between testing hypotheses using confidence intervals and other methods.

Typology: Exams

Pre 2010

Uploaded on 07/23/2009

koofers-user-itm
koofers-user-itm 🇺🇸

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Download Confidence Intervals and Hypothesis Testing and more Exams Probability and Statistics in PDF only on Docsity! Confidence intervals are rather straightforward. You don’t know the population mu or sigma so you estimate them by xBAR and s respectively and the CI is [xBAR] +/- (z or t) [sx / root(n)]. Likewise, you don’t know the population p so you estimate it by pHAT and the CI is [pHAT] +/- z [root(pHAT qHAT) / root(n)]. Similarly, for paired data [Dbar] +/- (z or t) [sd / root(n)]. while for independent samples [x1bar – x2bar] +/- z [root( s12 / n1 + s22 / n2 )] and [p1HAT – p2HAT] +/- z [root( p1HAT q1HAT / n1 + p2HAT q2HAT / n2)]. In every case the bold expression is our estimator of the sd of our estimator, e.g. [sx / root(n)] is our estimator of the sd of our estimator [xBAR]. We know that t is not to be used unless the population distribution of X or D is at least close to normal. Tests can be based on CI. For example, if we wish to test H0: mu = 16 ounces versus H1: mu is not 16 ounces at level alpha = 0.05 we could simply create a 95% CI for mu [xBAR] +/- (z or t) [sx / root(n)] and reject H0 if this CI fails to cover 16 ounces. If, truly, mu = 16 ounces then the chance we will (falsely) reject H0 is 5% (after all, we are using a 95% CI). We do not approach testing in this way, by linking it to CI, because there are many exceptions to doing so. For example, if the boundary point is p0 then the usual test statistic is not A: (pHAT – p0) / [root(pHAT qHAT) / root(n)]. Instead, we use B: (pHAT – p0) / [root(p0 q0) / root(n)]. The reasons for this are touched upon in your textbook so I will not repeat them here except to say that form B is widely adopted and (the main point)
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