School

Carleton UniversityDepartment

PsychologyCourse Code

PSYC 3000Professor

Bruce HutcheonStudy Guide

FinalThis

**preview**shows pages 1-3. to view the full**11 pages of the document.**-For Levene’s test, if you find p < .05 (less than) it means you have failed the test. True: H0

that populations samples are drawn from have the same variance. p < .05 =chance hypothesis

true is less than 5% (populations have different variances, treatment affects variance.

-What null hypothesis is tested using Levene's test? Treatment= no effect on variance, not

about sample variances or sampleing distributions

-Study: (F(1,12) = 3.9, p = .072, (t(12) = 2.3, p = .021, 2-tailed) for the positive-image group (M

= 23.0, sX2 = 20) compared with control (M = 27.6, sX2 = 36).” The samples have passed

Levene's test: 7.2% chance samples from populations with identical variances. low, but not

low enough to reject H0. H0 true.

- pooled variance? 20/2 + 36/2: sample variances are 12 and 16. design is balanced and

therefore each weight is 1/2. The pooled variance is therefore 1/2 of the sample 1 variance

plus 1/2 of the sample 2 variance.

standard error?2 : pooled variance [20/2 + 36/2] = 28. standard error= sqrt(pooled

variance)/sqrt(n). Df freedom say total number of observations is N=14. balanced design we

must have n=7. s.e. = sqrt(28)/sqrt(7) = 2.

-d=large, corresponding t value will have p<.05. False. Even large effect may not be significant

if not enough power (sample too small).

-Population Mean= 50, SD=15, sample N=35, M=42.5. cohens d. Single sample. d= (50-

42.5)/15= 0.5. Sample size not needed.

-1975, 8.4 hours on hmwk +/- 3.2 (SD). 2003= 7.1 +/- 3.2. d= 8.4-7.1/3.2= 0.41

-Repeated measures, matched pairs, mean difference 5. SD before =8, after =12.

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Pool SD by converting to variance, take average and convert back to SD. Variances= 64, 144.

Average- 64=144/2=104 (pooled variance). Pooled SD about 10. D= 5/10=0.5

-Cohen’s d for ind. samples: N= 11 M=15 SD=5, N=6 M=20 SD=4. Mean diff=5, SD= 5,4 (pool).

Convert to variances= 25, 16. Total df= 11+6 -2=15, 10 (2/3) and 5 (1/3) for samples. Pooled

variance= 2/3*25+ 1/3*16= 22/ Square to get pooled SD= 4.69. d=5/4.69= 1.07

balanced design? Each sample ½ df. Pooled variance is average of ind. Sample variance=

25+16/2= 4.53. d= 5/4.53

-Ind. Samples. M1-M2=6, t(18)=3, p=.008, equal sample sizes. T= MD/SE. T=3 therefore SE=

6/3=2. SE= s polled * square root (2/n) , 2= spooled * square root (2/10). S pooled=

2/0.45=4.44. d= 6/4.44= 1.35

- You can increase the power of a test by decreasing the alpha level. F D Alpha=D rejection

regions= D power

- All else being equal, the larger the sample size the greater the value of the noncentrality

parameter T Depends on sqrt of sample size so greater n= greater NCP= greater power

- Power is the probability of rejecting the null hypothesis when it is false. T

- Beta=probability rejecting H0 when it is false. F (chance commit type 2 error/ accept H0

when false).

- When are you susceptible to criticism that you do not have enough power in a t-test? When

accept H0 (always chance you’re wrong, T2E). Power addresses, tells chance of T2E, low

power= high chance T2E

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- When would you have the best chance of demonstrating that there really is a difference due

to a treatment? sampling distribution of the mean is narrow (associated large power). Small

power= SD of mean wide, small sample, hyp. Mean close to true mean.

- anger-management treatment decreases episodes of violent behaviour from a mean of 4.3

per month to a mean of 3.7 per month in a population with a standard deviation of 1. What is

the post hoc power of an independent-samples t-test to detect a treatment effect of this size if

N = 100? D= (4.3-3.7)/1 = 0.6. N=100 then n=50, noncentrality parameter = 0.6*sqrt(50/2)=

0.6*5 = 3. Look up in table to find 0.85

- A priori for 95% power? n= 2*(noncentrality_parameter/d)^2, n = 2*(3.6/0.6)^2 = 2*36 = 72,

N = 2*n = 2*72 = 144

- cued fusion times must be at least 7 seconds faster than uncued. population, fusion times

have a standard deviation of 10 seconds. What sample size, n for 80% power in an

independent-samples t-test, balanced design. D= 7/10 = 0.7. n = 2*(2.8/0.7)squared = 2*(4

squared) = 2*16 = 32

- elevated dietary protein increases anxiety in rats. Want show that mean latency to enter

open space in rats high protein diet greater by at least 1 second than in control rats. SD both

groups is 2 seconds and the sample size for both groups is n = 50. What is the power of a 2-

tailed test if risk of a Type I error set at 5%? Ind. Sample test.

d = (M1 – M2)/s = ½, NCP= 1/2 x sqrt(50/2) = 1/2 x 5 = 2.5, power table shows power of this

test is .71.

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