Unformatted text preview: Introduction
Still do not know the population parameters Dependent ttest: compare two groups of tscores from the "same" subjects same" Independent ttest: compares two groups of tscores from two different groups of subjects
Example: Experimental and control group
Scores are independent of each other Independent ttest
BetweenSubjects Design Independent tTest
Experimental Group Introduction
Key Focus: Difference between the means of two samples
Comparison distribution = distribution of differences between means SAMPLE Differences due to manipulation, not an extraneous variable because mood randomly determined. Control Group 1 Assumptions
1. 2. 3. Normality Independence of observations Homogeneity of variance
  Each of the population distributions follows a normal curve The two populations have the same variance Assumptions
3. When to use:
Independent tTest tInterval/ratio data No information on population parameters (, ) ( Comparing 2 sample means (M) (M Two groups of different individuals
Usually a control group and a test/experimental group Homogeneity of Variance (cont') (cont'
 3X's rule 3X'  SPSS: Levene's test Levene'
  H0: Variances are similar H1: Variances are not similar (i.e., violate assumption)
 p < 0.05 = violation of assumption 1 IV
Only 2 levels 2 Hypotheses
If the null hypothesis is true, then 1 = 2 If the null is true, then the mean of the distribution of differences tells us that 1  2 = 0 Therefore: H0: M1  M2 = 0 H1: M1  M2 0 Independent tTest: Equal n's Independent tTest Equal n's
Quick note watch `n' values df = n1 + n2  2 Standard Error of the Difference Between the Means
sM1M2 = SEDiff btwn means = SDsampling distribution of the diff btwn means t ( N  2) = (M1  M 2 ) sM1  M 2 sM1  M 2 2 s12 s2 = ( )+( ) n1 n2 3 Independent tTest Equal n's
s2 = ( X  Mean)
N 1 2 = SS df
df = N  1 Independent tTest: Unequal n's s = s2 Independent tTest Unequal n's
Quick note watch `n' values df = n1 + n2  2 Standard Error of the Difference Between the Means
sM1M2 = SEDiff btwn means = SDsampling distribution of the diff btwn means t ( N  2) = (M1  M 2 ) sM 1  M 2 1 1 sM1  M 2 = s 2 [( ) + ( )] p n1 n2 4 Pooled estimated variance
Weightedaverage variance Weighted Independent tTest Unequal n's 2 [(n1  1)( s12 ) + (n2  1)( s2 )] 2 sp = (n1 + n2  2) = d MEI nH 2 Power with unequal n's
Harmonic n (nH): gives the equivalent sample size for what you would have with two equal samples Independent tTest Unequal n's (2)(n1 )(n2 ) nH = (n1 + n2 ) = d MEI nH 2 5 Class Example
Researchers are interested in learning which form of treatment is a better option for those suffering from a specific phobia; exposure therapy or antiantianxiety drug, A. Class Example
A total of 128 people (N=128) participated in the study. All participants suffered from a specific phobia. It was determined that all participants had similar anxiety ratings before the start of the study. Participants were separated into two groups exposure therapy (n=64) and drug therapy (n=64). Class Example
After six weeks of treatment, participants were asked to rate their anxiety on a scale of 1 10 (with 1 being not at all anxious and 10 being very anxious). Here are the descriptive statistics... statistics... Class Example
Exposure Therapy n = 64 M = 4.00 SD = 1.95 Drug Therapy n = 64 M = 6.00 SD = 2.50 6 Step One: Hypotheses
H0 : M1 M2 = 0 When comparing exposure therapy to drug A, there will be no difference in anxiety ratings for the participants of each group. H1 : M1 M2 0 There will be a difference... difference... Step Two: Nature of the DV
Anxiety ratings on a scale from 1 10 = Interval Step Three: Test Statistic
Independent tTest tWhy? Do not know population parameters Comparing 2 sample means (comparing two different groups) One IV with 2 levels = 0.05 Step 4: Error Rates = ?? either 0.20 and then power is 0.80 OR if you already know your sample size you need to find power first 7 Step 5: Sample Size
Already know N = 128 and n = 64 for both groups = 0.05 = 0.20 Step 4: Error Rates = dMEI n/2
= 0.50 64 / 2 = 2.83 Power = 0.80 Step 6: Collect Data
Done!
Exposure Therapy n = 64 M = 4.00 SD = 1.95 Drug Therapy n = 64 M = 6.00 SD = 2.50 Step 7: Run Statistical Test
t (N2) = [(M1 M2) / sM1 M2 ] (N[(M = (4.00  6.00) / [((1.95) 2 / 64) + ((2.50) 2 / 64)] = 2 / 0.39632097 = 5.046 8 Step 7: Run Test
t(126)=5.046, p < 0.05 t(126)=Reject the Null
5.046
1.984 1.984 Step 8: Practical Significance
dOBS = [M1 M2] / [(s1 + s2) / 2 ] [M
= (4.00  6.00) /[(1.95 + 2.50) / 2] = 2 / 2.225 = 0.899 Step 9: Decision
We have both practical and statistical significance. Therefore, we reject the null hypothesis. Participants in exposure therapy score reported a greater reduction in their anxiety compared to those who took Drug A, t(126) = 5.046, p <0.05, dOBS = 0.899. Compare dOBS to dMEI = YES practical significance 9 How do we know about directionality?
Exposure Therapy n = 64 M = 4.00 SD = 1.95 Drug Therapy n = 64 M = 6.00 SD = 2.50 When to use:
Independent tTest tInterval/ratio data No information on population parameters (, ) ( Comparing 2 sample means (M) (M Two groups of different individuals
Usually a control group and a test/experimental group Only 1 IV
Only 2 levels Design Comparison: Within Subjects
W/S: Advantages
Fewer subjects for more power Studying learning/development Reduces individual differences Design Comparison: Between Subjects
Btwn: Advantages
No carryover effects No progressive errors W/S: Disadvantages
Carryover effects Progressive error
Tired subjects Btwn: Disadvantages
Need more subjects Individual differences More difficult to study development/change Solution = Counterbalance 10 Main point from comparison
Better (i.e., more efficient) to run a within subjects study as opposed to a between subjects study
But, sometimes not possible to use this design due to study constraints Problems with ttests
Running multiple tests increases your Type I Error Rates Suppose you are running a study. You change the number of ttests you plan to run from 1 test to 100 ttests.
How does this change your Type I Error Rates? What does this change mean? Too many ttests
The problem of too many t tests
Multiple t tests in the same study Possibility any one of them turns out significant at .05 level by chance is greater than .05 Type I Error Rates How do we adjust for this problem? 11 ...
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 Spring '08
 Dicorcia
 Normal Distribution, Statistical hypothesis testing, exposure therapy, independent ttest

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