Question

In: Statistics and Probability

In the course of the thesis work, a student develops a new approach for the solution...

In the course of the thesis work, a student develops a new approach for the solution of a problem (here referred to as method B). The current state-of-the-art approach, method A, is well published in the literature and has been applied to a large standard problem set where its average performance was discovered to be (and published in the main paper by the developers as) 7 with a standard deviation of 3 across the different problems in the problem set. In addition to the publication, the developers of method A also provide their code for anyone to be able to experiment with and the student decides to pick a random set of 15 problems from the standard problem set and apply both methods to these problems, resulting in the following performance numbers for method A: {8, 3, 10, 8, 11, 4, 6, 4, 12, 4, 5, 10, 6, 2, 10}, and the following performance numbers for the student’s method B: {9, 5, 9, 10, 15, 4, 7, 4, 12, 7, 8, 10, 6, 4, 12}. Looking at this data, the student discovers that it seems that method B outperforms method A and sets out to prove this using significance testing with a two-tailed 5% significance threshold. Given that both published performance results as well as the student’s experimental results are available, a number of tests can be performed.

Evaluate the results in terms of the hypothesis that method B has a higher performance than method A. List all the steps (and formulas) involved in the test and what the result implies for the significance of the hypothesis.

Solutions

Expert Solution

a.
Given that,
mean(x)=6.866
standard deviation , s.d1=3.204
number(n1)=15
y(mean)=8.133
standard deviation, s.d2 =3.313
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.145
since our test is two-tailed
reject Ho, if to < -2.145 OR if to > 2.145
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =6.866-8.133/sqrt((10.26562/15)+(10.97597/15))
to =-1.0647
| to | =1.0647
critical value
the value of |t α| with min (n1-1, n2-1) i.e 14 d.f is 2.145
we got |to| = 1.0647 & | t α | = 2.145
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.0647 ) = 0.305
hence value of p0.05 < 0.305,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.0647
critical value: -2.145 , 2.145
decision: do not reject Ho
p-value: 0.305
we do not have enough evidence to support the claim that difference of means between
method A and B.
b.
Given that,
mean(x)=8.133
standard deviation , s.d1=3.313
number(n1)=15
y(mean)=6.866
standard deviation, s.d2 =3.204
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.761
since our test is right-tailed
reject Ho, if to > 1.761
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =8.133-6.866/sqrt((10.97597/15)+(10.26562/15))
to =1.0647
| to | =1.0647
critical value
the value of |t α| with min (n1-1, n2-1) i.e 14 d.f is 1.761
we got |to| = 1.0647 & | t α | = 1.761
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:right tail - Ha : ( p > 1.0647 ) = 0.15251
hence value of p0.05 < 0.15251,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 1.0647
critical value: 1.761
decision: do not reject Ho
p-value: 0.15251
we do not have enough evidence to support the claim that method B has a higher performance than method A


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