Question

The data below shows the pre and post test scores of a random sample of students who
took part in a memory enhancement seminar. The data records the number of items
correctly recalled from a list of nonsense syllables and the differences in the scores (D)
Pretest Post test D
Alice 5 10 -5
Bert 8 16 -8
Charlie 6 10 -4
David 10 20 -10
Emma 12 20 -8
If you wanted to determine if the seminar works to improve memory, you should
propose the following null hypothesis
A.☐  < 0
B.☐ D > 0
C. ☐  D= 0
D.☐ π= 0
E. ☐ none of the above
The logical alternative hypothesis (given the direction of the subtraction above)
would be
A. ☐  D= -7
B. ☐ pre - post > 0
C. ☐   0
D.☐ D < 0
E. ☐ pre - post  0
The standard deviation of the difference scores (SD) for the above data is
A. ☐-7
B. ☐ 6
C.☐ 2.449
D.☐ - 1.75
E. ☐ none of the above
. Assuming normal distributions of pre and post test scores and alpha=.01, what critical
t score(s) would you use to determine whether you should accept or reject the null?
A. ☐+/- 1.533
B. ☐+/- 4.032
C. ☐+3.747
D. ☐–4.604
E. ☐–3.747
. Forget the real data for now. If your computed t = -2.857 and your critical 1 tail
t =-2.834, what would you determine?
A. ☐reject the null
B. ☐not reject the null
. Forget what you actually decided. For alpha = .01, what is the probability that you
would determine that the memory course may not improve memory when in fact it
does?
A. ☐.01
B. ☐.99
C. ☐it depends on how much the course improves memory
D. ☐your p value
E. ☐1-beta
. D X represents
A. ☐the mean of a sampling distribution of differences between two independent
sample means
B. ☐the standard error of the difference between two matched samples
C. ☐the standard error of a sampling distribution of sample means
D. ☐the mean of a sampling distribution of matched sample mean differences
E. ☐the average of your sample difference scores

Answer 1