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# section6.3posted - STOR 155, Section 2 Tuesday, April 13,...

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STOR 155, Section 2 Tuesday, April 13, 2010 Section 6.3

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Population has unknown mean μ, known standard deviation σ . Population is assumed Normal. Sample of size n has mean x. To test the null hypothesis H 0 : μ = μ 0 against an alternative hypothesis H a : Compute the test statistic z = ( x - μ 0 ) / ( σ / n ). Then compute the P-value , depending on H a : If H a : μ > μ 0 , P -value = P ( Z > z ). If H a : μ < μ 0 , P -value = P ( Z < z ). If H a : μ μ 0 , P -value = 2 P ( Z < – | z |). Reject H 0 if the P -value is below your chosen significance level α . (“Result is significant at level 6.3 Use and abuse of tests Summary of 6.2 _ _
Use and abuse of tests: P- value is more informative than significance level The decision to reject or accept H 0 depends on the arbitrary choice of a significance level α . Example 6.21 : Suppose we are doing a 2-sided test and the test statistic is z = 1.95. Then the P- value is 0.0512. If we have decided on a significance level α of 5% and we simply report “the result is not significant at the 5% level,” we are not giving all the information we have. Reporting the P-value is more informative than just stating significance or non-significance at a given level. In Example 6.21, because the P -value is close to .05, it shows that the test may not be conclusive; the decision is not so clear. If the

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## section6.3posted - STOR 155, Section 2 Tuesday, April 13,...

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