Two computers are compared by running a collection of different \"benchmark\" pr

Question

Two computers are compared by running a collection of different "benchmark" programs
and recording the time required to complete the program. Six benchmark programs, run
on two computers, produced the following CPU times (in minutes): Use the difference as
(Computer 1 – Computer 2).

If one were asked to test the claim that Computer 1 is faster than Computer 2, the
alternative hypothesis would be?
(a) u1 > u2
(b) u1 < u2
(c) ud > 0
(d) ud < 0

The value of point estimate of the mean di erence in times is
(a) -0.02167
(b) 1.575
(c) 1.5533
(d) 0.16143

If the standard deviation is 0.370784 then the margin of Error for a 95% con dence
interval would be?
(a) 0.389177
(b) 0.305014
(c) 0.370407
(d) 0.426322

Computer Program 1 Program 2 Program 3 Program 4 Program 5 Program 6 1 1.12 1.73 1.04 1.86 1.47 2.10 2 1.15 1.72 1.10 1.87 1.46 2.15

Explanation / Answer

a.
Given that,
mean(x)=1.5533
standard deviation , s.d1=0.4021
number(n1)=6
y(mean)=1.575
standard deviation, s.d2 =0.4143
number(n2)=6
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =2.015
since our test is right-tailed
reject Ho, if to > 2.015
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.5533-1.575/sqrt((0.16168/6)+(0.17164/6))
to =-0.092
| to | =0.092
critical value
the value of |t | with min (n1-1, n2-1) i.e 5 d.f is 2.015
we got |to| = 0.09207 & | t | = 2.015
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > -0.0921 ) = 0.53489
hence value of p0.05 < 0.53489,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: -0.092
critical value: 2.015
decision: do not reject Ho
p-value: 0.53489
we do not have enough evidence to support the claim that Computer 1 is faster than Computer 2

b.
if stanadard deviation = 0.370784 then
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 5 d.f is 2.571
margin of error = 2.571 * 0.151372
= 0.389177

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