t-test for 2 matched/related/ dependent samples

Transcription

1 HUMBEHV 3HB3 two-sample t-tests & statistical power Week 9 Prof. Patrick Bennett Concepts from previous lectures t distribution standard error of the mean degrees-of-freedom Null and alternative/research hypotheses (H vs H1) t-test for 2 matched/related/ dependent samples Experimental Question: - Is family therapy an effective treatment for anorexia? 17 girls participated in study - weighed before & after treatment - weights (in pounds given in Table 13.1) Statistical Question: - Does before/after weight differ?

2 subject before after N=17 subjects weighed before & after family therapy diff = after - before before & after measures are not independent because they come from the same subject before & after measures are correlated - not surprising because measures were taken on same subjects at different times t tests depend, in part, on N (e.g., df) should Ntotal be Nbefore + Nafter? no, because the 2 sets of measures are not independent our analysis must take dependence into account simple solution: analyze difference scores Weight After Therapy (lbs.) r =.54 r s = Weight Before Therapy (lbs.) diff = after - before subject before after diff mean: standard deviation: D = s D =7.16 Difference (after-before) lbs Hand et al. (1994)

3 sd Hypotheses for scores: - H: μbefore = μafter - H1: μbefore μafter µd = D sd p N D = standard deviation: sd = D t= 7.16 p 17 df = N = = t = t = H: (μafter - μbefore) = μd = - H1: (μafter - μbefore) = μd density for difference scores: df=16.3 mean: α =.5 (2-tailed), tcritical = ± t=.1 1 = 17 1 = t df=16 t= 7.16 p 17 = = t = t = density.3.4 α =.5 (2-tailed), tcritical = ± t 2 4 H: (μafter - μbefore) = μd = H1: (μafter - μbefore) = μd tobserved is more extreme than tcritical reject H in favour of H1 is family therapy an effective treatment for anorexia? 17 participants weighed before & after therapy t test used to evaluate H of no change in weight - rejected H in favour of H1 (i.e., weight change zero) - direction of effect (after > before) means weight gain not loss conclude that family therapy is/was an effective treatment? - can you think of an alternative explanation of result?

4 is family therapy an effective treatment for anorexia? - rejected H in favour of H1 (i.e., weight change zero) - but this is weak evidence for an effect of therapy - because simple alternative explanation exists: weight gain was due to normal growth over time experiment needs a control group - in experimental studies, a group that does not receive the treatment/procedure of interest - in correlational studies, a group that is not exposed to or does not experience the variable of interest control group in Hand et al study: - set of individuals who do not receive therapy - ideally, individuals would be assigned randomly to therapy and no-therapy groups. Why? is family therapy an effective treatment for anorexia? original experiment did include a control group - experimental group: N=17, received family therapy - control group: N=26, did not receive family therapy new experimental hypothesis: - was weight gain different in the two groups? - let μft & μc represent mean weight gain i groups H: μft = μc H1: μft μc Hand et al. (1994) Sampling Distributions of Group Means control (N=26): - mean = -.45, therapy (N=17): -, is difference between group means due to chance? Difference (after-before) lbs Scores mean = -.45 Sampling Distribution mean = -.42 sd = /srt(26) Freuency Freuency Scores Sampling Distribution mean = 7.24 sd = /srt(17) control therapy Mean Weight Gain (lbs.) Mean Weight Gain (lbs.) Group distributions of 1, simulated group means

5 Sampling Distribution of Sampling distribution of difference between means mean = -.45 Freuency Mean(Therapy) - Mean(Control) mean = 7.66 sd = 2.34 Each mean has a sampling distribution Difference between means also has a sampling distribution - mean = μ1-μ2; variance = VAR1 + VAR2 - Variance of sum or difference of 2 independent variables euals the sum of their variances - if means are distributed normally, then difference (or sum) is distributed normally Central Limit Theorem distributions of 1, simulated group means Sampling Distribution of estimating population parameters for group difference from 2 samples mean = -.45 Freuency Mean(Therapy) - Mean(Control) mean = 7.66 = (-.42) ˆµ D = X 1 X2 sd = 2.34 = srt{ [(7.99 2) /26 + (7.16 2) /17] } ˆ D = s Is our observed difference between group means unusually large given the null hypothesis that the two groups do not differ? We will try to answer this uestion using a t test.

6 t test for 2 independent samples convert difference between means to a t statistic X2 ) (µ 1 µ 2 ) = ( X 1 X2 ) (µ 1 µ 2 ) s X1 X2 when H is (µ 1 µ 2 )= X2 ) estimating standard error of difference X2 ) (µ 1 µ 2 ) s X1 X2 pooled variance estimate s X1 X2 = ˆ X1 X2 = s X1 X2 = s s 2 p + s2 p when n1 n2 s s 2 p is a weighted average of and s 2 2 s 2 p = ( 1) +( 1)s t test for independent means s 2 p X2 ) + s2 p df = (n1-1) + (n2-1) = n1 + n2-2 Hand et al. (1994) mean = -.45 H: μft = μc H1: μft μc s 2 p X2 ) + s2 p s 2 p = 25(7.992 ) + 16( ) = 58.9 Is the observed value of t=3.22 unusual given that the Null hypothesis is true? when H is true (μ1 - μ2 = ), t statistic follows t distribution with n1+n2-2 degrees of freedom t = (.45) df = = 41 = 7.71/2.39 = 3.22

7 General Strategy reject H if t exceeds critical values of t reject H if t exceeds critical values of t.2 density.3 2-tailed test.1 significance level =.5 2-tailed test df = 41 critical t = ±??..4 General Strategy zt critical t values is family therapy an effective treatment for anorexia? measured weight gain in control and therapy groups difference between groups was significant - t(41) = 3.22, p<.5, 2-tailed reject null hypothesis that weight gain was the same i groups result supports (i.e., is consistent with) hypothesis that family therapy is an effective treatment for anorexia next steps in research program? - replicate findings! - "Without replication, all results should be taken as preliminary." -- Gary Marcus significance level =.5 2-tailed test df = 41 critical t = ± 2.21