# Chapter 15 Comparing Two Populations: Dependent samples.

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Chapter 15 Comparing Two Populations: Dependent samples

Dependent Samples Dependent Samples - measurements are somehow related to one another either because: –The measurements came from the same subject (Within-subjects design), or –Subjects were paired or matched (Matched-pairs design) May also be called a “Repeated- measures” designs

Why use dependent samples? Advantages –Fewer subjects required with a within- subjects design (each subject is measured twice) –Increased power through decreased variability Disadvantages –“Carry-over” effects

Carry-over Effects Let’s say we are interested in methods to teach reading to pre-schoolers We –introduce the whole-language method, measure, –then use the phonics method, and measure We have a repeated-measures (within- subjects design), but

Carry-over Effects How can we remove the effects of teaching reading with the whole- language method? There is a “carry-over” to the next condition (phonics) condition, because the preschoolers may have learned to read Do we expect to forget how to read once they learn to?

Carry-over Effects Matched-pairs designs will eliminate the carry-over effect problem Matching, though, must be done well, without respect to the obtained scores, –matching is done before the experiment Along a dimension that is relevant to the study

Dependent Samples t-test With pairs of scores, either the result of matching or taking multiple measurements from the same subject Calculate the difference (D) between the two scores (maintaining positive and negative values) Calculate the mean difference, standard error, and then t

Dependent Samples t-test

M D is the mean of the differences s D is the standard deviation of the differences s M D is the estimated standard error of the sampling distribution of M D Δ 0 is the difference between the population means (parameter)

Example of a Dependent Samples t-test Example of a Dependent Samples t-test We want to compare a new teaching method with a traditional method on math 2 groups - 10 12 th grades in each, Group “Old” receives old method, Group “New” gets new method Subjects are matched on their IQ

Hypothesis test of Old vs. New (teaching math) 1. State and Check Assumptions –About the population Normally distributed? - don’t know Homogeneity of variance – we’ll check – About the sample Independent Random sample? – yes Dependent samples –About the sample Interval level

Hypothesis test of Old vs. New (teaching math) 2.Hypotheses H O : μ Old = μ New (the effectiveness of the old and new method are the same) μ Old - μ New = 0 (the difference between the effectiveness of Old and New is 0) H A : μ 1 ≠ μ 2 (the effectiveness of Old and New are not equal) μ 1 - μ 2 ≠ 0 (there is a difference between the effectiveness the Old and New)

Hypothesis test of Old vs. New (teaching math) 3.Choose test statistic –2 groups dependent samples Dependent-sample t-test

Hypothesis test of Old vs. New (teaching math) 4.Set Significance Level α =.05 Critical Value Non-directional Hypothesis with df = n p - 1 = 10 - 1 = 9 From Table C, critical value of α /2 =.025 t crit = 2.262, so we reject H O if t ≤ - 2.262 or t ≥ 2.262

Hypothesis test of Old vs. New (teaching math) 5.Compute Statistic –We need:

Scores on the WAT (Wisconsin Achievement test) OldNewD 78 74 4 Need: Δ 55 45 10 M D 95 88 7 s D 57 65 -8 s M D 60 64 -4 80 75 5 50 41 9 83 68 15 90 80 10 70 64 6

Scores on the WAT OldNewD 78 74 4ΣD = 54 55 45 10ΣD 2 = 712 95 88 7M D = 5.4 57 65 -8SS(D) = 420.4 60 64 -4s 2 D = 46.7111 80 75 5s D = 6.833 50 41 9n p = 10 83 68 15s M D = s D /√n p 90 80 10 = 6.833/ √10 70 64 6 = 2.161

Example (non- directional H A ) t is greater than t crit, reject H O

Hypothesis test of Old vs. New (teaching math) 6. Draw Conclusions –because our t falls within the rejection region, we reject the H O, and –conclude that the old and new method differ in their effectiveness in teaching math, with the old method superior

Comparing Dependent and Independent Samples t-tests Suppose, instead of pairing subjects based on IQ, we compared the mean of the two groups using a 2- independent samples t-test

Scores on the WAT Old 78 ΣX O = 718 55ΣX 2 O = 53772 95M O = 71.8 57SS(X O ) = 2219.6 60s 2 O = 246.622 80s O = 15.70 50n O = 10 83 90 70 New 74ΣX N = 664 45 ΣX 2 N = 45992 88M N = 66.4 65 SS(X N ) = 1902.4 64 s 2 N = 211.377 75 s N = 14.54 41 n N = 10 68 80 64

Independent Samples t-test Cannot reject H O

Comparison (cont.) By matching subjects, a priori (prior to the experiment) the variance is reduced substantially, thereby Allowing us to find a difference that was obscured

Ranking When completing a Wilcoxon, ranking must be done with the smallest difference receiving the lowest rank (1) The biggest difference gets a rank of n p

Hypothesis Test with non- parametric alternative to t-test Nicotine patches supposedly reduces smoking A psychologist records the number of cigarettes each of 9 patients smokes per day without a patch and with a nicotine patch.

OOPS!

Hypothesis Test 1.State and Check assumptions –Random Sample - yes –Dependent Sample (Within-subjects design) measures come from the same person (wearing the patch, and not wearing the patch) –Number of cigarettes smokes normally distributed? - not sure –Homogeneity of Variance - nope

Hypothesis Test 2.State Hypotheses H O : μ BEFORE = μ AFTER H A : μ BEFORE > μ AFTER

Hypothesis Test 3.Choose Test –2 dependent samples –H of V - no Wilcoxon T m (Wilcoxon signed-ranks test)

Hypothesis Test 4.Set Significance level α =.05 with n p > 8, the Wilcoxon T m approaches a z directional alternative at.05, using Table A z = 1.65 if obtained z ≥ 1.65, Reject H O

Hypothesis Test 5.Compute Test Statistic

Wilcoxon Test Non-parametric alternative to dependent sample t-test –Calculate the differences between paired scores (D) –Calculate the absolute value of each (|D|) –Rank |D| –“Sign” the ranks using the sign from D –Add all positive Ranks (T + )

Hypothesis Test 6.Draw conclusions –Since the obtained z was greater that the critical z of +1.65, H O is rejected, and –Conclude that the distributions are different with the before values greater than the after values

Review Procedure for calculating t Independent samples Compute (for each group):

Calculating t - Independent Samples To find t, the estimated standard error of the sampling distribution is calculated:

Computing t The t statistic ( a family of distributions which varies with df) is computed using:

Review Procedure for finding t Dependent Samples –matched-pairs, correlated samples, repeated-measures, etc. (2 samples) –Compute D (the difference between pairs of scores) n p (the number of pairs of scores M D (the mean of the differences)

Dependent Samples t (cont.) –Compute s D (the standard deviation of the differences) s M D (the estimated standard error)

Estimated Standard Error and t

Conceptually, what is the t statistic? In both tests, the numerator is a measure of the observed difference between scores (either unpaired or paired) The denominator, of both tests, is an estimate of the standard deviation of the sampling distribution of M 1 - M 2 (we call it “standard error”)

Error Terms When a sample is taken from a population, statistics computed from the sample are estimates of parameters, however, A certain amount of variability is expected (no two sample means are exactly alike) This variability from sample to sample is called “error”

Error and t-tests The amount of error obtained from a sample is used to estimate the “standard” error that might be expected in the situation (given n, etc.) This error can produce large (significant) differences between sample statistics, just by chance, because no two sample means are exactly alike

t statistic as a ratio Therefore, obtained difference t = ———————————————— difference expected by chance (“error”) We will return to this type of conception again

Can’t tell the difference between independent and dependent samples? If measurements are related to one another - dependent samples You will not be told when to use independent or dependent samples t- tests You have to decide

Example A psychologist believes that environment is more important than genetics in influencing intelligence. She locates 12 pairs of identical twins that have been reared apart, one twin in each pair in an enriched environment and the other in an impoverished one. Independent or Dependent Samples?

Another A new experimental drug, ABZ, is thought to have beneficial effects on AIDS. 20 AIDS patients are randomly selected and assigned to one of two conditions: ABZ for 90 days or Placebo for 90 days. Independent or Dependent Samples?

A third example A researcher suspects that increasing the level of lighting during the winter months will increase mood. The researcher selects 36 students, tests their mood, then replaces the light bulbs in the dorm from 75-W to 100-W for 1 month, then tests each student again Independent or Dependent Sample?

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