Question

A cognitive psychologist is testing the theory that short-term memory is mediated by subvocal rehearsal. This theory can be tested by reading aloud a string of letters to a participant, who must repeat the string correctly after a brief delay. If the theory is correct, there will be more errors when the list contains letters that sound alike (e.g., G and T) than when the list contains letters that look alike (e.g., P and R). Each participant gets both types of

letter strings, which are randomly mixed in the same experimental session. The number of errors for each type of letter string for each participant are shown in the following table:

Conduct the appropriate t-test (independent or paired) to test the hypothesis that more errors occur when the list contains letters that sound alike than when the list contains letters that look alike using a one-tailed α = .05.

Participant                    Letters sound alike                   Letters look alike

1	8	4
2	5	5
3	6	3
4	10	11
5	3	2
6	4	6
7	7	4
8	11	6
9	9	7

Calculate Cohen’s d as a measure of effect size, and state whether the effect is small, medium, or large.

Calculate the 99% confidence interval to determine whether or not the result is significant at the the α = .01 level.

Report the your results in APA style as you would do in a journal article.

Answer 1

Solution:

Here, we have to use paired t test.

H₀: µ_D = 0 versus H_a: µ_D > 0 (upper/right/one tailed test)

From given data, we have

Dbar = 1.6667

S_D = 2.3452

n = 9

df = n – 1 = 8

Level of significance = 0.05

Critical t value = 1.8595 (by using t-table or excel)

Test statistic = t = (Dbar - µ_D) /[S_D /sqrt(n)]

t = (1.6667 – 0) / [2.3452/sqrt(9)]

t = 1.6667/(2.3452/3)

t = 2.132056967

P-value = 0.0328

P-value < α = 0.05

So, we reject the null hypothesis H₀

There is sufficient evidence to conclude that more errors occur when the list contains letters that sound alike than when the list contains letters that look alike.

Cohen’s d = Dbar/S_D = 1.6667/2.3452 = 0.710686

Effect is medium.

Effect is small if d≤0.2, medium if d lies between 0.2 to 0.8, large if d≥0.8.

Confidence interval = Dbar ± t*[S_D /sqrt(n)]

Confidence level = 99%, df = 8, t = 3.3554 (by using t-table or excel)

Confidence interval = 1.6667 ± 3.3554*[2.3452 /sqrt(9)]

Confidence interval = 1.6667 ± 3.3554* 0.7817

Confidence interval = 1.6667 ± 2.62291618

Lower limit = 1.6667 - 2.62291618 = -0.95621618

Upper limit = 1.6667 + 2.62291618 = 4.28961618

The difference ‘0’ is lies within above confidence interval, so we conclude that result is not significant at α = 0.01 level.