Notes for 4/6/15-4/10/15
Notes for 4/6/15-4/10/15 PSYC 3301
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This 13 page Class Notes was uploaded by Rachel Marte on Monday April 13, 2015. The Class Notes belongs to PSYC 3301 at University of Houston taught by Dr. Perks in Fall. Since its upload, it has received 174 views. For similar materials see Introduction to Psychological Statistics in Psychlogy at University of Houston.
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Date Created: 04/13/15
4615 TTests for Related Samples Dependent Samples TTests Repeated Measures Design A repeated measures study is one in which a single sample of individuals is measured more than once on the same dependent variable in all treatment conditions In other words two sets of data are obtained from the same sample of individuals for example we might measure patients depression before and after receiving therapy to see if they improved Because both data sets are from the same sample and are related we cannot do an independent ttest These are dependent samples so we must do a dependent samples ttest Advantages 0 Same subjects in all conditions 0 Results cannot be biased just due to individual differences 0 Compared to an independentmeasures design a repeatedmeasures study is more likely to find a significant effect if it exists because it reduces the contribution of variance due to individual differences 0 More power to find a difference if it exists Matched Subjects Design In a matched subjects design individuals in one sample are matched with individuals in the other sample so that the two individuals are almost equivalent on variables researchers want to control It simulates a repeated design by matching individuals in the two samples on the basis of certain characteristics This reduces overall variation due to factors we aren t studying and don t want affecting our results Difference Scores In repeated measures designs the data consists of two sets of scores so we will use difference scores Dscores instead of raw scores In other words we will subtract each score in one sample from the corresponding score in the other sample to get Dscores to use for our tstatistic Ignoring the obvious aws in performing a study like the one above we can see how difference scores are setup You can see that the participants S OlS 10 all participated in two treatments samples Crest and Colgate Each participant has a score for each treatment The last two columns of the table D and D2 are where we will find the Dscores In column D we will find the actual Dscores by subtracting the scores in one set of data from the scores in the other set CrestColgate or alternatively ColgateCrest Which set you subtract from which doesn t really matter although you may prefer one way over the other because of the research question So the D column would read 2 1 1 2 1 3 etc The D2 column simply requires you to square the D values you just found it would read 4 1 1 4 1 9 etc Summing the D column and dividing it by the sample size n would give you the mean of the di erence scores D this is similar notation to the mean being i Summing the D2 column would give you the sums of squares SS value You will need both of these to complete a dependent samples ttest The sample of difference scores that you will create is then used as the sample for the hypothesis test ie we are taking two related samples and subtracting them from one another to get one sample Hypothesis Testing When we perform a dependent samples ttest we want to know if there is a difference between the two treatment conditions for the general population In other words we are interested in the mean for the population of difference scores We denote the population mean di erence as up The subscript D indicates that we are dealing with difference scores Step I 39 State a hypothesis about a population null and alternative hypotheses 0 Ho up 0 0 H1 mi 0 0 H1 predicts that the treatment has an effect on the scores of one of the conditions Step 2 Set the criteria for a decision Just like other hypothesis tests you need to choose a significance alpha level of between 01 and 05 and decide whether to use a onetailed or twotailed test Determine the tcrit value using the tdistribution table the same way we always have Df and n are always based on the number of difference scores or the number of subjects Step 3 Collect data and compute sample statistics You compute the tstatistic using the difference scores as your data Basically you replace X with D in all the formulas for ttests The df is still n1 and is based on the number of difference scores Main tstatistic formula 5 t MD 55 where 2D 32 SS D s 2 n D n n 1 Note 5 is the mean of the di erence scores 35 is the standard deviation of the di erence scores and s2 is the variance Recall that we talked about how to find the sums of squares SS value above with the Crest vs Colgate example table Step 4 Make a decision Like in any other ttest you will evaluate the tstatistic in comparison to the critical region In other words if the tobt value you calculated in step 3 falls within the critical region you determined in step 2 you will reject the null hypothesis The Four Equations for Finding the TStatistic Find each of these components in the order listed so that you know their value before you need to plug it into the equation in the next step The main goal here is to calculate the tstatistic 1 Variance 2 55 E 2 Standard Deviation of DScores 2 3 Mean of DScores 5 2 Tl 4 TStatistic 5up 55 0 t Advantages and Disadvantages of TTests Disadvantages of Independent TTests 0 They need more subjectsparticipants 0 Changes over time are more difficult to detect more error variance is introduced 0 Individual differences are not controlled for higher error variance Advantages of Dependent TTests 0 They can be used even if few subjects are available you get two scores for each participant so twice as much data 0 Individual di erences are controlled for by using the same subjects 0 Variance and standard error are drastically reduced because you are dealing with difference scores 0 Related samples ttests are useful for studies of learning memory or other changes Disadvantages of Dependent TTests 0 Factors other than treatment e ects can cause scores to change from one treatment to the next 0 Fatigue boredom or memory may play roles 0 Subjects may do well on the first test but then get bored or fatigued while doing it for a second time resulting in lower scores on the second test due to boredom or fatigue not due to treatment effects 0 Subjects may learn from the first test and use that to their advantage while doing it for a second time resulting in higher scores on the second test due to memory or learning effects not due to treatment effects 0 Carryover e ects function of first treatment 0 The first treatment may have effects that actually carry over to the second treatment so that the scores of the second treatment are affected by the scores of the first treatment 0 A good example of this is the testing of drugs The first drug you test may have longerlasting effects than you anticipate which would affect your results if you test the second drug too soon 0 Progressive error function of time passing 0 Things tend to deteriorate over time which could affect your results 0 Counterbalance in order to avoid problems o In counterbalancing you would alter the order of testing so that half of the subjects get the first treatment first followed by the second one while the other half would get the second treatment first followed by the first one This prevents boredom fatigue memory and carryover effects or at least makes it obvious if they are present so that you know to adjust your study accordingly Assumptions of Dependent TTests Assumptions 0 Observations are independent within the treatments 0 Population distribution of difference scores must be normal not a concern unless sample size is smaller than 30 Example Dependent TTest Example We want to know if a treatment has a positive effect at an alpha level of 05 given the following data Subject Before After Treatment D D2 Treatment AfterB efore A 3 12 9 8 l B 5 l 0 5 25 C 7 8 l l D 1 14 l 3 l 69 zD28 SS276 Note The first 3 columns would be given to you in the problem The last two D and D2 you will need to add yourself and calculate their values Remember that summing the D column is needed to find the mean of the dijference scores 5 later and summing the D2 column gives you the SS value which will also be used later List information ZD28 0t05 SS276 positive onetailed test Step I 39 State a hypothesis about a population null and alternative hypotheses H0 No difference 5 0 or AfterBefore0 H1 The treatment has a positive effect After gt Before Step 2 Set the criteria for a decision df3 There are 4 sets of dijference scores 4 3 A tcrit Step 3 Collect data and compute sample statistics Step I 39 Find the variance ss 2 11 1 276 o 32 4 1 276 o 32 3 o 2 92 Choose the proper equation Plug in known quantities Simplify denominator Simply Step 2 Find the standard deviation of Dscores 0 S n Choose the proper equation Plug in known quantities Simplify inside of radical Simply Choose the proper equation Plug in known quantities SimplifY Choose the proper equation Plug in known quantities 5 2 SD 4 0 5 V23 0 5 480 Step 3 Calculate the mean of Dscores 5 2 Tl 5 E 4 o 5 7 Step 4 Calculate the tstatistic t D PD 13 0 t E 48 tobt Step 4 Make a decision There is not a significant difference because tobt doesn t fall within the critical main SimplifY 146 2353 tobt tcrit I We Will retain Ho Our data suggest that the treatment has no effect PF p f 39n in HE Taii 125 11 It 11 E15 L E L 39T L Frupum39 n in Twin T3515 Eumbined riff ELSE HE quot H135 an an 1 1mg llm 311321 39Ezlf E1 1amp3 E El 1 3315 4313 g 9925 lt 3 gt 365 1633 H32 ELEM SEEM I UnM 1 25 iqT T diam Effect Size We will still use Cohen s d and the r2 value to calculate effect sizes The formula for r2 is the same as it has always been but the formula for Cohen s 1 looks a little different Cohen s d d 4815 amp 41015 Introduction to the Analysis of Variance ANOVA Why Use ANOVAS Although ttests are very useful when you are comparing two groups they are far less helpful when you have more than two groups or more than one independent variable Say that you are testing the effect on leadership skills for three different types of trainings You want to know which training has the best outcomes AND if the effect of the training differs by gender e g women might benefit most from training A while men find training C most effective In order to do a ttest you would have to compare each group to each other group two at a time This would take forever and require several different ttests Instead of doing a series of ttests you can just use an analysis of variance test or ANOVA ANOVAs are useful if we have more than two groups e g three types of training or more than one independent variable e g training types and gender ANOVA Terms and Design Know the difference between factors and levels 0 the variables that designate the groups being compared the independent variables 0 E g In the above example training types and gender are both factors 0 individual conditions or values that make up a factor 0 E g In the above example training types has three levels training A B and C and gender has two levels male and female We use the number of levels per factor to describe the design of an AN OVA For our extended example we would have a 3x2 design The number of numbers tells us how many factors there are there are two numbers in the example 3 and 2 The individual numbers tell us how many levels each factor has the first factor has 3 levels training A B and C and the second factor has two levels male and female Therefore a 3x2 design would have two factors one of which has three levels and one of which has two levels A 2x3x2 design would have three factors two of which have two levels and one of which has three levels The chart below is an illustration of our example problem a 3x2 design Factor 1 Training Type Lecture Based Interactive 0n the ilob Factor 1 Training lecture based Training has 3 Factor 2 ender Women levels training Women who Women who Women who received iectu re received interactive received on the joh hased training iectu re hasxed training training Men who received Men who received Men who received lecture hased interactive lecture on the ioh training training based training Factor 2 has 2 levels For now we are going to focus on the simplest ANOVAs with only one factor and independent levels Hypothesis Testing with ANOVAs Hypotheses We are starting outwith ANOVAS that have three populations groups you would use a t test if you had two populations 0 H0 There are no differences between the three groups 0 H0 uD O 0 H1 Populations have different means and sample differences are indicative of population mean differences 0 H1 HD O 0 H1 predicts that the training has a systematic effect on the scores of one condition Note The hypotheses only predict that there is or is not a di erence somewhere They do not predict where this di erence is ie which groups are di erent perhaps two groups are the same and only one is dijj erent ANOVAs are only used to tell if there is a difference somewhere Other tests must be performed to determine where this dijj erence is Logic of ANOVA Scores vary and we want to measure the amount of the variability and explain where it comes from To do this we divide the variance into two parts 0 BetweenTreatments Variance 0 Looks at the dijj erences between group means 0 WithinTreatment Variance 0 Looks at dijj erences within groups Goal of ANOVA We split the variance in two because we want to determine which part best explains our data There are two possibilities o The differences between levels are simply due to chance and are not caused by a treatment effect 0 The differences between levels are significantly greater than can be explained by chance alone and are instead caused by treatment e ects In other words there are two possible explanations for the difference variance that exists 0 Treatment e ects 0 Chance 0 Individual differences 0 Experimental error Di erences within groups can only be due to chance or error while di erences between groups can be due to either treatment e ects or chanceerror Notice that means that betweentreatments variance should be greater than withintreatment variance and that we need to determine how much of the betweentreatments variance is due to chance and how much is due to treatment effects Total Variability Between Within Treatment Treatment Variance Variance Treatment Effects Chance Chance FRatio We use the Fratio to isolate the differences due to the treatment effect The theoretical definition of the F ratio treatment e f f ectdi f f erences due to chance 0 F Differences due to chance 01 Variance between treatments Variance within treatments If the treatment has no e ect the F ratio will be I The farther away the F ratio is from I the greater the treatment e ect MSbetween SSbetween and We calculate the Fratio using the formula F Where M S between within df between 55 MSWthin dfww MS stands for Mean Square and 15 used 1n the place of the term var1ance within In this class you Will be given the SS sums of squares values but it may also be useful to know T2 2 how to calculate them yourself SS between 2 Z 7 G Where T treatment total sum of all the scores in a particular treatment N total number of all scores in a study and G sum of all scores in a study grand total and SSWmm Z SSlnSlde each treatment In addition dfbetween k 1 and dfwithm N k Where k number of treatments samples or levels To get the total SS you can add the values of SSbetween and SSWmm SStotal SSbetween 2 SSWmm or use the equation SStotal Z x2 Note A clear list of steps for calculating the F ratio can be found on page 13 Example Given the following table calculate SS between and SSWmm A B C 2 2 3 4 4 5 SSA0 SSB5 SSC5 G20 TA4 TB7 Tc9 N n2 n2 n2 G22344520 GA1A2B1B2C1C2 N2226 NnAnBnc TA224 TB347 Tc459 SSZ x f2 ssA2 22 2 22 o ssB3 352 4 352 5 ssc4 452 5 452 5 dfbetween k 1 dfbetween 3 1 2 dfwithin N k dfwithin 6 3 3 SSbetween Z SSbetween y ISSbetween 633 I SSWlthm SSA SSE 55C SSWithin O Example Research question What is the effect of three different trainings on leadership Given SSbetween300 SSwithin100 5 participants per group Calculate Fobt List information SSbetween300 k3 SSwithin100 N15 k3 because there are 3 groups N15 because there are 5 participants in each of 3 groups 55515 Step I 39 Calculate a f s dfbetween k 1 O dfb 3 1 0 dfwithin N 0 df 15 3 0 Step 2 Calculate MS s SSbetween 0 MSbetween dfbetween 300 O MSbetween 2 O S 150 SSwi in MSwithin th dfwithin 100 12 0 ithm 83 0 M Swithin Step 3 Calculate F rati0 F MSbetween MSwithin 150 833 Fobt Steps to Calculate FRatio Step 0 Step 3 Calculate SS s T2 62 SSbetween 2 g W SSwithin Z SSinside each treatment Note Usually this step will be unnecessary In this class you will be given the SS values Calculate df s dfbetween k 1 dfwithin N k Calculate MS s 55 t MSbetween df e ween between MS Within thhm df within Calculate F Rati0 F MSbetween MSwithin Total 62 2 SS 2X N df N 1 Between treatments Within treatments 88 n 83 for the means SS 2 ZSseach Treatment 72 92 df N k or 88 3 W n M3 53 at k 1 df M5 55 at MS between treatments Fratto MS within treatments
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