HYPOTHESIS TESTING SAMPLING DISTRIBUTION. the sampling distribution for di erences of means is. 2 is known. normal if.

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1 Introduction to Statistics in Psychology PSY Professor Greg Francis Lecture 5 Hypothesis testing for two sample case Why do we let people die? H : µ = a H a : µ 6= a H : = a H a : 6= a always compare one-sample to a hypothesized population parameter sometimes we want to compare two (or more) population parameters H : µ = µ H a : µ 6= µ TWO-SMPLE SE FOR THE MEN useful when you want to compare means of two groups di erent teaching methods surial with and without drug depression with and without treatment height of males and females the null hypothesis is that there is no di erence between the means H : µ = µ H a : µ 6= µ or another way to say the same thing H : µ µ = H a : µ µ 6= 3 STTISTI since we want to compare the di erence of two population means our statistic should be the di erence of two sample means X X and we will compare that statistic to the hypothesized alue of the parameter H : µ µ = if the statistic is much di erent from the hypothesized parameter, we will reject H same approach as before, di erent sampling distribution the sampling distribution for di erences of means is normal if is known a t distribution if estimated by s is unknown and (central limit theorem again) two restrictions. the two samples drawn from the respectie populations are independent. the ariances of the two populations are equal INDEPENDENE drawing a sample with a particular alue of X should not a ect the probability of drawing a sample with any other particular alue of X remember statistical independence P (X and Y )=P (X) P (Y ) same idea here 4 5 6

2 INDEPENDENE in practice this means we need to be careful about how we sample if comparing treatments, randomly diide a random sample into an experimental group and a control group if comparing fixed populations, take random samples from each (no oerlap, so no risk of dependence) aoid situations like repeating subjects: e.g. comparing depression for the same subjects before and after treatment (there are ways to test this situation, but not with these techniques) VRINE to carry out hypothesis testing we need to calculate standard error to get standard error we need to estimate (or know) the standard deiation since we sample two groups, we need a pooled estimate of to get a pooled estimate we need to be certain that = note this is a statement about the populations we would not expect the sample ariances to be identical we want to compare population means from two populations H : µ = µ = = we hae Independent samples of size n and n although we draw two random samples (one from each population), we are only interested in one statistic X X but we need to know the sampling distribution for this statistic OF DIFFERENES it turns out that the sampling distribution is familiar. Shape: s sample sizes get large, distribution becomes normal.. entral tendency: The mean of the sampling distribution equals µ µ. 3. Variability: The standard deiation of the sampling distribution (standard error of the di erence between means) is X X = u t + n as before, the sampling distribution is normal only if you know the population standard deiation, when you must estimate the population standard deiation, the sampling distribution becomes a t distribution our estimate is called the pooled estimate because we use scores from both samples FORMULS deiation formula s = (X i X ) + (X i X ) n + n deiations relatie to the sample mean of each sample! s = raw score form: " X i ( X i ) # " /n + X i ( X i ) # /n n + n X i refers to the ith score from sample X i refers to the ith score from sample n refers to the number of scores in sample n refers to the number of scores in sample

3 FORMULS standard deiations s = (n )s +(n )s n + n where s is the ariance among scores in sample s is the ariance among scores in sample sum of squares s = SS +SS n + n where, the sum of squares is SS = (X i X ) SS = (X i X ) STNDRD ERROR we use the estimate of s to calculate an estimate of standard error for the sampling distribution of di erences u s X X = + n n this gies us an estimate of the standard deiation of the t distribution we need to know one more thing DEGREES OF FREEDOM we hae two samples with (possibly) di erent numbers of scores the degrees of freedom in sample df = n from sample df = n added together gies the result (depends on independence!) d.f. = n + n (same as in denominator of s estimate) now we hae eerything we need to apply the same techniques of hypothesis testing as before. State the hypothesis.. Set the criterion. 3. ompute the test statistic. 4. onstruct the confidence interal. 5. Interpret the results. EXMPLE neurosurgeon beliees that lesions in a particular area of the brain, called the thalamus, will decrease pain perception. If so, this could be important in the treatment of terminal illness accompanied by intense pain. s a first attempt to test this hypothesis, he conducts an experiment in which 6 rats are randomly diided into two groups of 8 each. nimals in the experimental group receie a small lesion in the part of the thalamus thought to be inoled in pain perception. nimals in the control group receie a comparable lesion in a brain area belieed to be unrelated to pain. Two weeks after surgery each animal is gien a brief electrical shock to the paws. The shock is administered with a ery low intensity leel and increased until the animal first flinches. In this manner, the pain threshold to electric shock is determined for each rat. The following data are obtained. Each score represents the current leel (milliamperes) at which flinching is first obsered. The higher the current leel, the higher is the pain threshold. Note that one animal died during surgery and was not replaced. HYPOTHESIS () State the hypothesis. Directional hypothesis because we expect the lesion will increase the threshold. H : µ = µ or µ µ = (lesion makes no di erence) H a : µ <µ or µ µ < (lesion increases pain threshold, less sensitiity) 6 7 8

4 RITERION () Step : Set the criterion. we will set =.5 for a one-tailed test We expect a negatie t alue (see H a ) DT now we consider the data from the experiment the researcher gets the following ontrol Group Experimental Group (False lesion) (Thalamic lesion) X X OMPUTING TEST STTISTI Step 3. we hae n =8,n =7 from the data we calculate X =.875 X =.443 X X =.568 s =.394 (using any formula you want), so that the estimate of standard error is s X X = u + n n s X X = u 8 + = OMPUTING THE TEST STTISTI Statistic Parameter Test statistic = Standard Error of the Statistic t = (X X ) (µ µ ) s X X t = (.568).93 =.94 we need to calculate the degrees of freedom df = n + n =5 =3 We use the t Distribution alculator to compute p =.57 we reject H! INTERPRET RESULTS Step 4. you could create confidence interals for the di erence of means, but they would not be ery useful here Step 5. our interpretation of the test is that the di erence between the calculated sample means would hae occurred by chance less than 5% of the time if the null hypothesis were true in practice, this means that the study supports the theory that lesions to the thalamus decrease pain perception ONLUSIONS comparing means of two samples independent samples much more powerful than one-sample case many more experiments can be tested same basic technique 3 4

5 NEXT TIME robustness of tests homogeneity of ariance iolations of assumptions Worry or don t worry? 5