You want to determine whether \"noise\" makes a difference in college students\'
ID: 3045315 • Letter: Y
Question
You want to determine whether "noise" makes a difference in college students' ability to memorize new information. You randomly divide 16 college sophomores into two groups with eight participants each. The participants in both groups are asked to memorize a list of 20 nonsense syllables (such as “TSG”, “JMB”, and so on) during a five-minute study period. The participants in Group 1 study the nonsense syllables with a "noisy" background, while those in Group 2 study the syllables with a "no noise" background. After the five-minute study period all participants receive a distracter task wherein they must count backward from 100 by sevens (for example, 100, 93, 85, and so on) for one minutes. At the end of this one minute of counting backward, each participant is asked to recall as many of the original nonsense syllables as possible. The scores below are the number of nonsense syllables each participant correctly recalled.
X1
X2
(“No Noise”, Group 2)
X1
(“Noisy”, Group 1)X2
(“No Noise”, Group 2)
13 9 9 5 7 5 11 9 12 9 10 11 14 12 11 7Explanation / Answer
We have to use here two sample independent t-test for testing whether "noise" makes a difference in college students' ability to memorize new information.
R-code:
x1=c(13,9,7,11,12,10,14,11)
x2=c(9,5,5,9,9,11,12,7)
t.test(x1,x2,alternative="two.sided",var.equal=T,conf.level=1-0.01)
Output:
Two Sample t-test
data: x1 and x2
t = 2.0819, df = 14, p-value = 0.05618
alternative hypothesis: true difference in means is not equal to 0
99 percent confidence interval:
-1.074647 6.074647
sample estimates:
mean of x mean of y
10.875 8.375
Hence, answer of this question is t (14) = 2.08, p > .01 since t=t = 2.0819, df(degrees of freedom ) = 14
and p-value = 0.05618
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