Lecture 14 chi-square test, P-valueSome general thoughts about hypothesis testingPowerPoint PresentationBack to the microarray exampleSlide 5Slide 6Confidence intervalLecture 14 chi-square test, P-value•Measurement error (review from lecture 13)•Null hypothesis; alternative hypothesis•Evidence against null hypothesis•Measuring the Strength of evidence by P-value•Pre-setting significance level•Conclusion•Confidence intervalSome general thoughts about hypothesis testing•A claim is any statement made about the truth; it could be a theory made by a scientist, or a statement from a prosecutor, a manufacture or a consumer •Data cannot prove a claim however, because there•May be other data that could contradict the theory•Data can be used to reject the claim if there is a contradiction to what may be expected•Put any claim in the null hypothesis H0•Come up with an alternative hypothesis and put it as H1•Study data and find a hypothesis testing statistics which is an informative summary of data that is most relevant in differentiating H1 from H0.•Testing statistics is obtained by experience or statistical training; it depends on the formulation of the problem and how the data are related to the hypothesis.•Find the strength of evidence by P-value :from a future set of data, compute the probability that the summary testing statistics will be as large as or even greater than the one obtained from the current data. If P-value is very small , then either the null hypothesis is false or you are extremely unlucky. So statistician will argue that this is a strong evidence against null hypothesis.If P-value is smaller than a pre-specified level (called significance level, 5% for example), then null hypothesis is rejected.Back to the microarray example•Ho : true SD denote 0.1 by 0)•H1 : true SD > 0.1 (because this is the main concern; you don’t care if SD is small)•Summary :•Sample SD (s) = square root of ( sum of squares/ (n-1) ) = 0.18•Where sum of squares = (1.1-1.3)2 + (1.2-1.3)2 + (1.4-1.3)2 + (1.5-1.3)2 = 0.1, n=4•The ratio s/ is it too big ?•The P-value consideration:•Suppose a future data set (n=4) will be collected.•Let s be the sample SD from this future dataset; it is random; so what is the probability that s/ will be•As big as or bigger than 1.8 ? P(s/ 0 >1.8)•P(s/ 0 >1.8)•But to find the probability we need to use chi-square distribution :•Recall that sum of squares/ true variance follow a chi-square distribution ;•Therefore, equivalently, we compute•P ( future sum of squares/ 02 > sum of squares from the currently available data/ 02), (recall0 is•The value claimed under the null hypothesis) ;Once again, if data were generated again, then Sum of squares/ true variance is random and follows a chi-squared distributionwith n-1 degrees of freedom; where sum of squares= sum of squared distance between each data point and the sample meanNote : Sum of squares= (n-1) sample variance = (n-1)(sample SD)2For our case, n=4; so look at the chi-square distribution with df=3; from table we see :9.348The value computed from available data = .10/.01=10 (note sum of squares=.1, true variance =.12P-value = P(chi-square random variable> computed value from data)=P (chisquare random variable > 10.0)11.34P-value is between .025 and .01, reject null hypothesis at 5% significance levelConfidence interval•A 95% confidence interval for true variance 2 is•(Sum of squares/C2, sum of squares/C1)•Where C1 and C2 are the cutting points from chi-square table with d.f=n-1 so that•P(chisquare random variable > C1)= .975•P(chisquare random variable>C2)=.025•This interval is derived from•P( C1< sum of squares/ 2 <C2)=.95For our data, sum of squares= .1 ; from d.f=3 of table, C1=.216, C2=9.348; so the confidence interval of 2 is 0.1017 to .4629; how about confidence interval of
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