Variances UnequalRound d down to the nearest integer, dTesting for Equality of Two VariancesIn previous test, assumed variances from two populations were the same. Sometimes we want to test thisH0: 12 = 22H1: 12  22Assume the 2 samples are independent and come from N(1, 12) and N(2, 22) distributions.Usually, we don’t know 12 or 22. We will base test on sample estimates, s12, s22.Best test is based on ratio s12 / s22. We reject H0 if the ratio is very small or very large. To do this, we first have to describe the sampling distribution of s12 / s22. This ratio follows an F-distribution under H0. This is a family of distributions dependent on 2 parameters called the numerator degrees of freedom and the denominator degrees of freedom. If samples are of size n1 and n2, then s12 / s22 ~ F(n1-1, n2-1).1211 ,nnFMost F distributions are positively skewed and the amount of skew depends on the relative magnitudes of the two degrees of freedom. The only exceptions are when the numerator degrees of freedom are either 1 or 2. The mode is then at 0. The F distribution takes positive values only. The F distribution is well tabled (See Table 9 for selected degrees of freedom). This table only gives the upper percentage pointsp)Fss(p,,nn 112112221PrWe can also show that p,,nn,p,nnFF1111212111Therefore, we can also compute the lower percentiles because of this symmetry property.Example: Suppose n1 = 9, n2 = 13.284905131123810204112045139750128025812975081202512897508129750128..FF..FF.F.F.,,,.,.,,,.,.,,.,,Practice using Table 9.num df = 8, denom df = 16Critical values: F.95 = 2.59, F.975 = 3.12, F.995 = 4.52num df = 6, denom df = 16Critical values: F.95 = 2.74, F.975 = 3.34, F.995 = 4.91To return to the F-test:Ho: 12 = 22H1: 12  22We will simplify test. We will always conduct test by putting larger sample variancein the numerator. Therefore, we will only reject H0 for large values of F. Reject H0 if 211211α/,,nnFFFor example, if n1 = 13 and n2 = 21,  = 0.056829752012211211.FF,.,α/,,nnIf s12 is the larger variance, reject H0 if 6822221.ssNote if  = 0.01, critical value is 6839952012.F,.,As numerator degrees of freedom increases, the critical value decreasesAs denominator degrees of freedom increases, the critical value decreasesExample: Behr investigated alterations of thermoregulation in patients with certainpituitary adenomas. The standard deviation of the weights of a sample of 13 patients was 21.4 kg. The weights of a sample of 5 control subjects yielded a standard deviation of 12.4 kg. We wish to know if we may conclude that the weights of the two populations have the same variability. H0: 12 = 22H1: 12  22Let  = 0.05Reject H0 if F > F12,4,0.975 = 8.75Test Statistic:982412421222221...ssFDecision: Do not reject H0. The weights of the two populations do not appear to have different variability. One-tailed alternatives:H0: 12 = 22Ho: 12 = 22H1: 12 > 22H1: 12 < 22Ratio: s12 / s22Ratio: s22 / s12Note that the larger variance is always in the numerator. For a one-tailed alternative, the critical value for F is obtained for 1- , not 1-/2.Two-Sample t-test for Independent SamplesVariances UnequalAssume 2 normally distributed samples:First sample of size n1 from N(1, 12)Second sample of size n2 from N(2, 22)with 12  22H0: 1 = 2H1: 1  2Called Behrens-Fisher problemThe significance test will be based on 21XX  where)nσnσ,μN(μ~XX)/nσ,N(μ~X)/nσ,N(μ~X22212121212222212111Under H0,)nσnσN(0,~XX22212121If 12 and 22 are known, then the test statistic is  N(0,1)~nσnσ0XXz22212121usually, 12, 22 are not known but are estimated by s12, s22We will not pool s12, s22 as we did when we assumed that 12 and 22 were equal.Test Statistic to consider: 222121210nsnsXXtUnfortunately, the exact distribution of t is difficult to derive. We will use an approximation called Satterthwaite’s approximation. This will allow us to use the t statistic defined above. We have to adjust degrees of freedom. Compute d = approximate degrees of freedom.1122222121212222121nnsnnsnsnsd'Round d down to the nearest integer, dReject H0 if t > td,1-/2 or t < t d,/2 (= -td, 1-/2)p-value: If t  0, p-value is 2 * area to left of t under a td distribution.If t > 0, p-value is 2 * area to right of t under a td distribution.Example: Protoporphyrin levels were measured in 2 samples of subjects. Sample 1 consisted of 50 adult male alcoholics. Sample 2 consisted of 40 adult healthy male non-alcoholics. Are protoporphyrin levels the same in the 2 groups?H0: 1 = 2H1: 1  2Let  = 0.01Data: Sample Mean St. Dev. n1 340 250 502 45 25 40     freedomofdegrees5022500151231894625126539625154912506251512503940254950250402550250112222222222222222121212222121d''or.....nnsnnsnsnsd'Reject H0 if t > 50, 0.995 = 2.70 or t < t50, 0.005 = -2.70  298575635395625126529562515125029540255025004534002222212121....nsnsXXtReject H0p-value: Use d = 50 degrees of freedomp < 0.0005 *2 = 0.0012-sided 100% * 1 -  Confidence Interval for 1 - 2 when 12  22= 2221212121nsns*tXXα/d'',For the example,