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Psychology 10. Analysis of Psychological Data April 23, 2014. The plan for today. Finishing the Kelly & Nils example. Assumptions of two-way ANOVA. An example of two-way ANOVA calculated from raw data: Summary statistics; Graphics; Computations; Interpretation; Checking assumptions.
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Psychology 10 Analysis of Psychological Data April 23, 2014
The plan for today • Finishing the Kelly & Nils example. • Assumptions of two-way ANOVA. • An example of two-way ANOVA calculated from raw data: • Summary statistics; • Graphics; • Computations; • Interpretation; • Checking assumptions.
Example (cont.) • Male TA, male student: T = 583.5. • Male TA, female student: T = 532. • Female TA, male student: T = 582.5. • Female TA, female student: T = 590.5. • n = 7. • S X2 = 193391.8.
What are our null hypotheses? • H0 : mNils = mKelly. • H0 : mMales = mFemales. • H0 : There is no interaction between TA sex and student sex.
Calculating SSAfor TAs • For Nils, TA= 583.5 + 532 = 1115.5 • For Kelly, TA= 582.5 + 590.5 = 1173.0 • The grand total G is 1115.5 + 1173.0 = 2288.5 • SSA= 1115.52 / 14 + 1173.02 / 14 – 2288.52 / 28 = 118.0804
ANOVA Table Source SS df MS F ----------------------------------- TA Sex 118.080 1 118.080 Student Sex Interaction Within ------------------------ Total
Calculating SSBfor Student Sex • For male students, TB= 583.5 + 582.5 = 1166.0 • For female students, TB= 532 + 590.5 = 1122.5 • The grand total G is still 2288.5 • SSA= 11662 / 14 + 1122.52 / 14 – 2288.52 / 28 = 67.58036
ANOVA Table Source SS df MS F ----------------------------------- TA Sex 118.080 1 118.080 Student Sex 67.580 1 67.580 Interaction Within ------------------------ Total
Calculating the Interaction SS • SScells= 583.52 / 7 + 5322 / 7 + 582.52 / 7 + 590.52 / 7 – 2288.52 / 28 = 312.0982 • SSA×B = Sscells - SSA – SSB • 312.0982 - 118.0804 - 67.58036 = 125.4374
ANOVA Table Source SS df MS F ----------------------------------- TA Sex 118.080 1 118.080 Student Sex 67.580 1 67.580 Interaction 126.437 1 126.437 Within ------------------------ Total
Calculating the Within-Groups SS directly • In the four cells, SX2= 50311.75 (Nils, males), 42289.5 (Nils, females), 49957.75 (Kelly, males), and 50832.75 (Kelly, females) • 50311.75 - 583.52 / 7 = 1672.857 • 42289.5 - 5322 / 7 = 1857.5 • 49957.75 - 582.52 / 7 = 1485.429 • 50832.75 - 590.52 / 7 = 1019.857 • SSW = 1672.857 + 1857.5 + 1485.429 + 1019.857 = 6035.643
ANOVA Table Source SS df MS F ----------------------------------- TA Sex 118.080 1 118.080 Student Sex 67.580 1 67.580 Interaction 126.438 1 126.438 Within 6035.643 24 251.487 ------------------------ Total
SST • SST may be calculated two ways: • 118.080 + 67.580 + 126.438 + 6035.643 = 6347.741 • 50311.75 + 42289.5 + 49957.75 + 50832.75 - 2288.52 / 28 = 6347.741 • The fact that we get the same answer both ways confirms our other calculations.
ANOVA Table Source SS df MS F ----------------------------------- TA Sex 118.080 1 118.080 .47 Student Sex 67.580 1 67.580 .27 Interaction 126.438 1 126.438 .50 Within 6035.643 24 251.485 ------------------------ Total 6347.741 27
Making a decision • For each of those F statistics, the degrees of freedom are 1 in the numerator and 24 in the denominator. • From the table, the critical value is 4.26. • We fail to reject each one of the null hypotheses.
Interpreting the decision • We have not found evidence that the population mean midterm score of students with the male TA differs from the population mean of students with the female TA. • We have not found evidence that the population means of midterm scores for males and females differ. • We have not found evidence that male/female differences depend on which TA the student has.
Main effects and interactions • It may seem strange to talk about Male/Female differences depending on TA when we haven’t actually found evidence of Male/Female differences. • It is possible for the interaction to be present even when no main effects are present.
Assumptions of two-way ANOVA • Independence between groups. • Independence within groups. • Equal variances in all populations. • Normal distributions in all populations.
Two-way ANOVA from scratch • Anemia is a serious third world nutrition problem. • In Ethiopia, traditional iron pots have mostly been replaced by aluminum pots. • A 1999 study randomly assigned 4 lots of three different types of food (meat, legumes, vegetables) to three types of pot (aluminum, clay, iron). • DV = iron content (milligrams iron per 100 grams of food).
Hypotheses • H0 : mMeat = mLegumes = mVeggies . • H0 : mAluminum = mClay = mIron. • H0 : There is no interaction between pot material and type of food.
Cooperative exercise • Calculate summary statistics. • Create an interaction plot. • Test all three null hypotheses. • Interpret the test results. • Evaluate the assumptions.