Ho:  = 0 = 120Suppose we change  to 0.01, one-tailedThis can be approximated byOne Sample, One-SidedOne-sample, two-tailedPOWER--Probability of rejecting Ho when it is true--Probability of accepting Ho when it is false—depends on several factorsa. True value of parameterb. Hypothesized value or Hoc. Value of d. Sample size, nFor a fixed  and n, and before conducting the test, we can compute many values for by postulating specific values of parameter if Ho is false.We want to know for a given hypothesis, how well test controls Type II error. If Ho is false, we want probability of rejecting Ho. This is the complement to .1- -- Power = probability of rejecting Ho when it is false. We must specify a specificalternative value of the parameter to calculate power. Assume underlying distribution is normal and 2 is known. Ho:  = oH1:  = 1 < 0For a Type I error of , we reject Ho if z < z and accept Ho if z > z.Power = Pr(reject Ho | Ho is false) = Pr(z < z |  = 1) 1010μμ|n/*zμXμμ|znσ/μXααPrPrUnder H1, /n),σ~N(μX21nσμμzzPrornσμμzΦnσ/μn/*σzμΦααα101010PowerReminder: We are testing Ho, the null hypothesis. Under Ho, the rejection region is .Under H1, the area to the left of this point is POWER = 1- .Important concept—How likely are we to get a significant difference under the alternative hypothesis? If there is low power, there is little likelihood of getting a significant difference even if Ho is false. Example: Suppose we want to study height in a population and we sample 30 individuals. Let 0 = 60 inches and 2 = 8.44 inches2. Let  = 0.05.Ho:  = 0 = 60H1:  = 1 = 58What is the power of this test?    983401256212562770636451477259052226451309052258606451Power10..zPror.Φ..Φ.*..Φ..ΦnσμμzΦαTherefore, the probability of rejecting Ho at the 5% level of significance if n = 30 is about 98%.We can calculate power for a test in the opposite direction.Ho:  = 0H1:  = 1 > 0We will reject Ho if z > z1-We will accept Ho if z  z1-If z > z1-, αznσ/μX10thennσ)μ(μzzPrornσ)μ(μzΦnσ/μn/*σzμΦ)μ|μn/*σzμX(Pr)μ|μn/*σzμX(Prn/*σzμXorαααααα101101110110110101111PowerNote: (-X) = 1- (X) and z = -z1- nσμμzzPrornσμμzΦαα0101PowerExample: Suppose we are studying birth weight of 100 live births from diabetic mothers. Let 0 = 120 ounces,  = 25 ounces. Let  = 0.05Ho:  = 0 = 120H1:  = 1 = 125What is the power of this test?    6403550355026451100251201256451Power01..zPror.Φ.Φ.ΦnσμμzΦαTherefore, the probability of rejecting Ho at the 5% level of significance if n = 100 is about 64%.Power depends on 4 factors1. 2. 1 vs 03. 24. n Suppose we change  to 0.01, one-tailedHo: μ = 0 = 120H1: μ = μ1 = 125 = 0.01σ = 25n = 100    3703260326023262100251201253262Power01010..zPror.Φ.Φ.ΦnσμμzΦ.As  decreases, power decreases.Suppose  = 0.05, 1 = 128, that is further from 0Ho: μ = 0 = 120H1: μ = μ1 = 128 = 0.05σ = 25n = 100    940555155512364511002586451100251201286451Power01..zPror.Φ..Φ.Φ.ΦnσμμzΦαAs 1 gets further from 0, power increasesSuppose 2 increasesHo:  = 0 = 120H1:  = 1 = 125 = 0.05 = 36 ouncesn = 100    400255025503916451100361201256451Power01..zPror.Φ..Φ.ΦnσμμzΦαAs 2 increases, power decreasesSuppose n increasesHo:  = 0 = 120H1:  = 1 = 125 = 0.05 = 25n = 400    9903552355246451400251201256451Power01..zPror.Φ.Φ.ΦnσμμzΦαAs n increases, power increasesReviewFormula for Power when alternative mean is less than hypothesized mean—one-tailed alternativeHo:  = oH1:  = 1 < 0 nσμμzzPrnσμμzΦαα1010PowerFormula for Power when alternative mean is greater than hypothesized mean—one-tailed alternativeHo:  = 0H1:  = 1 > 0 nσμμzzPrnσμμzΦαα0101PowerPOWERTwo-sided alternative:Combine area from two sides. That is, reject Ho for very high or very low values.Ho:  = 0H1:   0 with the specific alternative  = 1Exact formula where the underlying distribution is normal and the population variance, 2, is known.nσ)μ(μzΦnσ)μ(μzΦα/α///01211021PowerThis can be approximated byn/σ|μ|μzzPrn/σ|μ|μzΦα/α/10211021Powersince one of the two areas is usually very, very small. Example: We have a new education program to help patients reduce weight. The effect on blood pressure is unknown. Suppose 30 patients are to be tested. The change in blood pressure after two weeks has  = 8 mm Hg. What powerwould such a study have of detecting a significant difference in blood pressure over two weeks if it is hypothesized that the true mean change in blood pressure from baseline to two weeks could be 4 mm Hg in either direction?n = 30,  = 8, |0 - 1| = 4,  = 0.05   78230780787429613084961Power1021..zPror.Φ..Φ/.Φn/σ|μ|μzΦα/Sample Size DeterminationOne Sample, One-SidedWe are testingHo:  = 0H1:  = 1 < 0Assume both distributions are normal and that  is known. We will conduct the testat the