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Power

Power. We all know what P(Type I Error) =  is:  = P (rej. H 0 / H 0 true)  = P(Type II Error) = P (acc. H 0 / H 0 false). Power is : “the ability to detect out a small deviation from the hypothesized value.” “The test which can detect a smaller deviation is the more powerful one.”

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Power

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  1. Power We all know what P(Type I Error) =  is:  = P (rej. H0 / H0 true)  = P(Type II Error) = P (acc. H0 / H0 false)

  2. Power is : “the ability to detect out a small deviation from the hypothesized value.” “The test which can detect a smaller deviation is the more powerful one.” What is “power” in terms of , b ?

  3. H0 true H0 false Type II error No Error 1- we accept H0 Type I error No Error 1- we reject H0  = P (acc. H0 / H0 false) 1-  = P (rej. H0 / H0 false) = a correct conclusion 1- = “POWER” (of a test)

  4. 1   R (• j - ••)2 C  For ANOVA (with one factor) we have, POWER = f ( , 1, 2,   = significance level 1 = df of numerator (of Fcalc) 2 = df of denominator (of Fcalc) “Non-Centrality Parameter” = A MEASURE OF HOW DIFFERENT THE ’s ARE FROM ONE ANOTHER (includes “sample size” by including R & C)

  5. 1 -  = Function of a distribution called “non-central F dist’n” (which is NOT the same as the regular F distribution) However, typically, we don’t know  or the  values. So — in practice, you need to “estimate”  from data and “suppose” a value of for purposes of finding POWER

  6. 10(52 + 02+ 52) 3 500 3 12.9 5 Example: One factor ANOVA with  = .05, R = 10, C = 3. We decide that a working estimate of  = 5 and that we (arbitrarily) choose to calculate (1-) assuming that the ’s are each 1• apart. (eg., 17 22 27). Then, 1/2 ( 1 5 ) = 1/2 ( ) 1 5 = 2.58 = =

  7. We have  = 2.58 1 = 2 (3 col) 2 = 27 (30-3)  = .05 Now go to “Power Tables”: eg., the one for 1 = 2 (they are “indexed” by the value of 1): This table is for  = .05 and for  = .01, and for various values of 2 and.

  8. POWER ( 2) at 1 = 2 See next page for chart ) ( 2 = 30,  = 2.6 POWER  .976  = .05, Our Answer:

  9. Figure 3.8 on p.77 (Paul)

  10. Accept H0 = All column means are the same. All column means are each “<5” apart with “97.6%” confidence.

  11. FINDING “REQUIRED”SAMPLE SIZE Given C, , 1-, and “” we can find the minimum sample size required, in terms of R, the number per column.  = Max (•j) – min (•j) = Range of ’s

  12.  Sometimes one estimates  and  separately; sometimes one estimates the ratio and not necessarily the separate  values. A common choice of is 2 (i.e., the range of ’s is 2 standard deviations).      C = 3  = .05 = .90 = 2 Example: R = 8

  13. Table 3.3 on p.81 (Paul)

  14. (1-a) Confidence Interval (C.I.) • For • j: • For • j- • k:

  15. If s is unknown, • Replace s by • Replace Z1-a/2 by t1-a/2 where the df of t is the df of Error.

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