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final exam.docx

Course Code
PSYC 3000
Bruce Hutcheon
Study Guide

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-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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