EPPP Part 1 (Examination for Professional Practice in Psychology) — All Questions

A researcher concludes that a new therapy reduces symptoms (p < .05) when in truth it has no effect. This is:

• a.A lack of statistical power
• b.A Type I error✓
• c.A violation of homogeneity of variance
• d.A Type II error

Concluding there is an effect when the null hypothesis is in fact correct is a Type I error, whose probability is set by alpha. A Type II error is failing to reject a false null, low power raises the risk of Type II rather than Type I errors, and homogeneity of variance is an ANOVA assumption, not a decision error.

All else equal, which change would increase the statistical power of a study?

• a.Increasing the sample size✓
• b.Lowering alpha from .05 to .01
• c.Switching from a one-tailed to a two-tailed test
• d.Adding more variability within groups

Power rises with larger samples, larger effects, a more lenient alpha, a one-tailed test when justified, and less error variance (Cohen, 1988). Lowering alpha to .01, moving to a two-tailed test, and increasing within-group variability all reduce power.

Students with the most extreme test-anxiety scores are selected for a workshop, and their scores are lower at retest. With no comparison group, which threat to internal validity is most likely to account for part of the drop?

• a.Regression toward the mean✓
• b.Instrumentation
• c.Differential attrition
• d.Diffusion of treatment

When participants are selected for extreme scores, their retest scores tend to move toward the mean because of measurement error alone (Campbell & Stanley, 1963). The scenario gives no change in the measure (instrumentation), there is no control group that could receive the treatment (diffusion), and differential attrition requires groups losing members at different rates.

Two variables correlate r = .60. What proportion of the variance in one is accounted for by the other?

• a.6%
• b.60%
• c.64%
• d.36%✓

The coefficient of determination is r squared: .60 x .60 = .36, or 36% of the variance shared. Using r itself (60%) overstates it, 6% misplaces the decimal, and 64% is 1 - r squared, the variance not shared.

A researcher records how many men and how many women choose each of three treatment options and wants to know whether choice is related to gender. The appropriate test is:

• a.A chi-square test✓
• b.A Pearson correlation
• c.A one-way analysis of variance
• d.An independent-samples t-test

Both variables are nominal and the data are frequencies, so a chi-square test of independence is used. ANOVA and the t-test compare means of an interval or ratio outcome across groups, and Pearson correlation requires two continuous variables.

A behavior analyst introduces a token system in one classroom, then two weeks later in a second classroom, then two weeks after that in a third, while collecting data continuously in all three. Which design is this?

• a.Alternating-treatments design
• b.Multiple-baseline design✓
• c.ABAB reversal design
• d.Changing-criterion design

Staggering the start of the same intervention across settings, with continuous measurement in each, is a multiple-baseline design; control comes from change occurring only when the intervention reaches each setting, with no withdrawal needed (Baer, Wolf & Risley, 1968). A reversal design withdraws the intervention, alternating-treatments rapidly alternates conditions, and changing-criterion shifts a performance criterion step by step.

A treatment group averages 60 on an outcome measure and a control group averages 54, with a pooled standard deviation of 12. What is Cohen's d, and how is it conventionally described?

• a.0.20, a small effect
• b.0.50, a large effect
• c.0.50, a medium effect✓
• d.2.00, a large effect

d = (60 - 54) / 12 = 6 / 12 = 0.50 (check: 0.50 x 12 = 6). Cohen (1988) suggested 0.20 as small, 0.50 as medium and 0.80 as large, so 0.50 is medium. Dividing 12 by 6 gives 2.00, which inverts the formula.

The Solomon four-group design adds groups that receive no pretest. What does this allow the researcher to evaluate?

• a.Whether dropouts differed between groups
• b.Whether assignment to groups was random
• c.Whether pretesting interacts with the treatment✓
• d.Whether the outcome measure is stable over time

Solomon (1949) crossed pretest versus no pretest with treatment versus control so that researchers can detect whether taking the pretest sensitizes participants and changes the treatment's effect, a threat to external validity. The design does not test randomization, attrition or measure stability.

A study randomly assigns volunteers from one clinic to treatment or wait-list. The random assignment mainly strengthens:

• a.Construct validity of the measures
• b.Internal validity✓
• c.Test-retest reliability
• d.External validity

Random assignment equates groups on expected pre-existing differences, which supports causal conclusions: internal validity (Shadish, Cook & Campbell, 2002). Generalizing to other people or settings (external validity) depends on how the sample was drawn, here volunteers from one clinic, and assignment does not affect what a measure captures or its reliability.