Question 21(1 point)

The data from a repeated-measures research study show that the difference between treatment I and treatment II is large and positive for some subjects, large and negative for some subjects, and near zero for some subjects. These data will produce

Question 21 options:

a)	a small sample variance and a large t statistic.

b)	a small sample variance and a small t statistic.

c)	a large sample variance and a large t statistic.

d)	a large sample variance and a small t statistic.

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Question 22(1 point)

A research report describing the results from a repeated-measures study includes the following: "The data show no significant difference between the two treatments, t(10) = 1.65, p > .05." Based on this report, you can conclude that a total of _______ individuals participated in the research study.

Question 22 options:

a)

9

b)

10

c)

11

d)

12

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Question 23(1 point)

For which of the following situations would a repeated-measures design have a substantial advantage over an independent-measures design.

Question 23 options:

a)	few subjects are available and individual differences are large

b)	few subjects are available and individual differences are small

c)	many subjects are available and individual differences are large

d)	many subjects are available and individual differences are small

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Question 24(1 point)

A researcher obtains t = 2.10 for a repeated-measures study using a sample of n = 16 participants. Based on this t value, what is the correct decision?

Question 24 options:

a)	reject the null hypothesis with= .05 but not with= .01

b)	reject the null hypothesis with either= .05 or= .01

c)	fail to reject the null hypothesis with either= .05 or= .01

d)	cannot make a decision without additional information

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Question 25(1 point)

A researcher obtains t = 3.00 for a repeated-measures study using a sample of n = 10 participants. If the researcher measures effect size using the percentage of variance accounted for, what value will be obtained for?

Question 25 options:

a)

9

b)	3/12 = 0.25

c)	9/18 = 0.50

d)	9/19 = 0.47