1) In a planned study, here is a known population with a normal distribution, pop mean = 50, pop standard dev = 5. What is the predicted (standardized) effects size (d) if the researchers predict that those given an experimental treatment have a mean of

a) 50

; this is no effect

b) 52

; this is a small effect

c) 54

; this is a large effect

d) 56

; this is a large effect

e) 47

; this is medium (negative) effect

For each part, also indicate whether the effect is approximately small, medium, or large.

2) For each of the following studies, make a chart of the four possible correct and incorrect decisions, and explain what each would mean.

a) A study of whether increasing the amount of recess time improves school children's in-class behavior.

H0 is true / H0 is false
Reject H0 / Incorrectly concluding that increasing recess time does improve children’s in-class behavior, when it really doesn’t / Correctly concluding that increasing recess time does improve children’s in-class behavior, when it really does
Fail to reject H0 / Correctly concluding that increasing recess time does not improve children’s in-class behavior, when it really doesn’t / Incorrectly concluding that increasing recess time does not improve children’s in-class behavior, when it really does

b) A study of whether color-blind individuals can distinguish gray shades better than the population at large.

H0 is true / H0 is false
Reject H0 / Incorrectly concluding that color blind individuals can distinguish gray shades better than the general population, when they really can’t / Correctly concluding that color blind individuals can distinguish gray shades better than the general population, when they really can
Fail to reject H0 / Correctly concluding that color blind individuals can not distinguish gray shades better than the general population, when they really can’t / Incorrectly concluding that color blind individuals can not distinguish gray shades better than the general population, when they really can

c) A study comparing individuals who have ever been in psychotherapy to the general public to see if they are more tolerant of other people's upsets than is the general population.

H0 is true / H0 is false
Reject H0 / Incorrectly concluding that individuals who have been in psychotherapy are more tolerant of others, when they really aren’t / Correctly concluding that individuals who have been in psychotherapy are more tolerant of others, when they really are
Fail to reject H0 / Correctly concluding that individuals who have been in psychotherapy are not more tolerant of others, when they really aren’t / Incorrectly concluding that individuals who have been in psychotherapy are not more tolerant of others, when they really are

3) Here is information about several possible versions of a planned experiment. Figure effect size and power for each; sketch the distributions involved, showing the area for alpha, beta and power (Assume all populations have a normal distribution). (I haven’t included sketches because that takes a lot longer to do in the digital realm)

Population / Predicted Mean / n / alpha-level / 1 or 2 tailed
Mean / Std dev / treatment
A / 90 / 4 / 91 / 100 / 0.05 / 1
B / 90 / 4 / 92 / 100 / 0.05 / 1
C / 90 / 2 / 91 / 100 / 0.05 / 1
D / 90 / 4 / 91 / 16 / 0.05 / 1
E / 90 / 4 / 91 / 100 / 0.01 / 1
F / 90 / 4 / 91 / 100 / 0.05 / 2

a) ; this is a small effect

  • critical Z for  = 0.05 & 1-tailed test is 1.645
  • convert this z-score into a raw score (using the information from the null distribution)
  • convert this raw score into a z-score (using the information from the treatment distribution)
  • using the unit normal table look up the area under the curve in the tail for that z-score: z(-0.85) = 0.1977 = 
  • Power = 1- = 1 – 0.1977 = 0.8123 or 81.23%

b) ; this is a medium effect

  • critical Z for  = 0.05 & 1-tailed test is 1.645
  • convert this z-score into a raw score (using the information from the null distribution)
  • convert this raw score into a z-score (using the information from the treatment distribution)
  • using the unit normal table look up the area under the curve in the tail for that z-score: z(-3.35) < 0.001
  • 1977 = 
  • Power = 1- = 1 – 0.001 = 0.999 or approximately 100%

c) ; this is a medium effect

  • critical Z for  = 0.05 & 1-tailed test is 1.645
  • convert this z-score into a raw score (using the information from the null distribution)
  • convert this raw score into a z-score (using the information from the treatment distribution)
  • using the unit normal table look up the area under the curve in the tail for that z-score: z(-1.7) = 0.0446
  • 1977 = 
  • Power = 1- = 1 – 0.0446 = 0.9554 or 95.54%

d) ; this is a small effect

  • critical Z for  = 0.05 & 1-tailed test is 1.645
  • convert this z-score into a raw score (using the information from the null distribution)
  • convert this raw score into a z-score (using the information from the treatment distribution)
  • using the unit normal table look up the area under the curve in the tail for that z-score: z(-0.34) = 0.3669
  • 1977 = 
  • Power = 1- = 1 – 0.3669 = 0.6331 or 63.31%

e) ; this is a small effect

  • critical Z for  = 0.01 & 1-tailed test is 2.33
  • convert this z-score into a raw score (using the information from the null distribution)
  • convert this raw score into a z-score (using the information from the treatment distribution)
  • using the unit normal table look up the area under the curve in the tail for that z-score: z(-0.175) = 0.4286
  • 1977 = 
  • Power = 1- = 1 – 0.4286 = 0.5714 or 57.14%

f) ; this is a small effect

  • critical Z for  = 0.05 & 2-tailed test is ±1.96
  • convert this z-score into a raw score (using the information from the null distribution)
  • convert this raw score into a z-score (using the information from the treatment distribution)
  • using the unit normal table look up the area under the curve in the tail for that z-score: z(-0.55) = 0.2912
  • 1977 = 
  • Power = 1- = 1 – 0.2912 = 0.7088 or 70.88%

4) On a particular memory task in which words are learned in a random order, it is known that people can recall a mean of 11 words with a standard deviation of 4, and the distribution follows a normal curve. A cognitive psychologist, to test a particular theory, modifies that task so that the words are presented in a way in which words that have a related meaning are presented together. The cognitive psychologist predicts that under these conditions, people will recall so many more words that there will be a large effect size. She plans to test this with a sample of 20 people, using the alpha level = 0.01 and a two tailed test.

a) What is the power of this study?

  • she is expecting a large effect size, so we will assume Cohen’s d = 0.8
  • critical Z for  = 0.01 & 2-tailed test is ±2.58
  • convert this z-score into a raw score (using the information from the null distribution)
  • convert this raw score into a z-score (using the information from the treatment distribution)
  • using the unit normal table look up the area under the curve in the tail for that z-score: z(-1.00) = 0.1587
  • 1977 = 
  • Power = 1- = 1 – 0.1587 = 0.8413 or 84.13%

b) Sketch the distributions involved, showing the area for alpha, beta, and power.

(I haven’t included sketches because that takes a lot longer to do in the digital realm)

c) Explain your answer to someone who understands hypothesis testing involving means of samples but has never learned about effect size or power.

Using her experience with the research area, the researcher expects to have a large effect. Using this expectation, she is able to predict the approximate raw score mean of the treatment group. Using this information (and information about the population standard deviation and the number of participants in her sample), she was able to predict the likelihood of being able to detect an effect of that size if it is present.