Two new mathematics learning techniques are being tested. Twenty students were r
ID: 3150692 • Letter: T
Question
Two new mathematics learning techniques are being tested. Twenty students were randomly selected from a population. nA = 9 of them were randomly assigned to use technique A, and nB = 11 of them were randomly assigned to use technique B. Each student spent 30 minutes learning the technique to which they were assigned, and then were asked to complete a task. The time to complete the task was recorded, in seconds. A shorter time indicates better mastery of the task. The data are below:
Technique A: 23.1, 21.4, 20.6, 15.5, 21.9, 36.0, 30.2, 33.1, 33.4
Technique B: 32.7, 36.8, 39.1, 37.3, 40.3, 46.8, 75.5, 53.0, 55.6, 54.1, 55.7
We wish to test:
H0 : µA µB = 0 vs.
HA : µA µB =/= 0, using = 0.05.
a) Use the bootstrap to perform the test, using B = 10000 resamplings. Compute a p-value, and make a reject or not reject decision. Finally, state your conclusion in the context of the problem.
Explanation / Answer
Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=26.1333
Standard Deviation(s.d1)=7.1453 ; Number(n1)=9
Y(Mean)=47.9
Standard Deviation(s.d2)=12.4473; Number(n2)=11
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =26.1333-47.9/Sqrt((51.05531/9)+(154.93528/11))
to =-4.897
| to | =4.897
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 8 d.f is 2.306
We got |to| = 4.89692 & | t | = 2.306
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -4.8969 ) = 0.001
Hence Value of P0.05 > 0.001,Here we Reject Ho
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