11 Comparison of Two Means

Tests involving two samples – comparing variances, F distribution TOH - x A = x B ? Step 1 - F-test  s A 2 = s B 2 ? Step 2 - t-test  use different formula for (i) s A 2 = s B 2. (ii) s A 2 ≠s B 2 Goal – whether a given gene is expressed differently between patients and healthy subjects This involves comparing the mean of the two samples To answer this question one must first know whether the two samples have the same variance The method used to compare variances of two samples – F distribution Then we use t-test to test whether the mean of the gene is expressed differently between patients and healthy subjects

Tests involving two samples – comparing variances, F distribution The values measured in controls are: 10, 11, 11, 12, 15, 13, 12 The values measured in patients are: 12, 13, 13, 15, 12, 18, 17, 16, 16, 12, 15, 10, 12. Is the variance different between the controls and the patients at a 5% significant level ? H 0 : s A 2 = s B 2, H 1 : s A 2 ≠s B 2 Need to find a new test statistics, Two-tail test Notation: assume A = controls, B = patients in the following calculation Controls sample A has d.o.f and variance = 6 and 2.66 Patients sample B has d.o.f and variance = 12 and 5.74 Consider the ratio F = 2.66/5.74 = , Significant level for two-tail test = 5%/2 = 2.5% F-distribution (right tail) F 0.025(6,12) = (from Excel) F 0.975(6,12) = (from Excel) F- distribution (right tail)

F distribution – right tail see next page

Tests involving two samples – comparing variances, F distribution F 0.025(6,12) = F 0.975(6,12) =

Tests involving two samples – comparing variances, F-distribution Usually we have F-distribution table for 0.01, 0.025, 0.05 but not !! Given F 0.025(6,12) = , how to find F 0.975(6,12) ??? The F distribution has the interesting property that : left tail for an F with 1 and 2 d.o.f. is = the reciprocal of the right tail for an F with the d.o.f reversed: F[Left tail( A, B )]  = 1/F[right tail( B, A )]  F 0.975(6,12) = 1/ F ( )(12,6) F 0.975(6,12) = 1/ F 0.025(12,6) = 1/ = back to our null hypothesis test Since < < Since the F-statistics is in between and , we will accept the null hypothesis  there is no difference between controls and patients

Tests involving two samples – comparing variances, F-distribution Now, let us consider the ratio The two different choices should lead to same conclusion, since the conclusion should not depend which variance we put on the numerator or denominator Controls sample A has d.o.f and variance = 6 and 2.66 Patients sample B has d.o.f and variance = 12 and 5.74 F = 5.74/2.66 = F-distribution (right tail) F 0.025(12,6) = (from Excel) F 0.975(12,6) = (from Excel) Since < < Since the F-statistics is in between and 5.366, we will accept the null hypothesis  there is no difference between controls and patients REMARK The two F-tests are reciprocal to each other That is < < Reciprocal  1/ > 1/ >1/  > >

Tests involving two samples – comparing means The gene expression level of the gene AC is measured for the patients, P and controls, C are given in the following: geneIDP1P2P3P4P5P6 AC geneIDC1C2C3C4C5C6 AC F-test: H 0 : s P 2 = s C 2, H 1 : s P 2 ≠s C 2 T-test: H 0 : x P = x C, H 1 : x P ≠ x C Mean of gene expression level of patients, X P = Mean of gene expression level of controls, X C = s P 2 = 0.059, s C 2 = To test whether the two samples have the same variance or not, we perform the F-test at a 5% level F = 0.059/0.097 = 0.60, d.o.f. = 10 F 0.025(5,5) = 7.146, F 0.975(5,5) = In between and  accept the null hypothesis  the patients and controls have the same variances

Tests involving two samples – comparing means t-statistic of two independent samples with equal variances The t-score is where the p-value, or the probability of having such a value by chance is This value is smaller than the significant level 0.05, and therefore we reject the null hypothesis, the gene AC is expressed differently between cancer patients and healthy subjects.

Tests involving two samples – comparing means t-statistic of two independent samples with unequal variances The modified t-score is The degree of freedom need to be adjusted as This value is not an integer and needs to be rounded down

Chapter11 p259

Chapter11 p264

Chapter11 p2268