Comparing Computer Chip Speeds: Fifty specimens of a new computer chip were test
ID: 3310814 • Letter: C
Question
Comparing Computer Chip Speeds: Fifty specimens of a new computer chip were tested for speed in a certain application, along with 50 specimens of chips with an old design. The average speed in MHz, for the new chips was 495.6, and the standard deviation was 19.4. The average speed for the4 old chips was 481.2, and the standard deviation was 14.3 1. Can you conclude that the mean speed for the new chips is greater than that of the old chips? State the appropriate null and alternative hypotheses, and then find the p-value. What do you conclude? 2. A sample of 60 even older chips had an average speed of 391.2 MHz with a standard deviation of 17.2 MHz. Someone claims that the mean speed of the new chips is over 100 MHz faster than these very old ones. Do the data provide convincing evident for this claim? State the appropriate null and alternative hypotheses, and then find the p-value. What do you conclude?Explanation / Answer
Q1.
Given that,
mean(x)=495.6
standard deviation , s.d1=19.4
number(n1)=50
y(mean)=481.2
standard deviation, s.d2 =14.3
number(n2)=50
null, Ho: u1 < u2
alternate, H1: u1 > u2
level of significance, = 0.1
from standard normal table,right tailed t /2 =1.299
since our test is right-tailed
reject Ho, if to > 1.299
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =495.6-481.2/sqrt((376.36/50)+(204.49/50))
to =4.2249
| to | =4.2249
critical value
the value of |t | with min (n1-1, n2-1) i.e 49 d.f is 1.299
we got |to| = 4.22489 & | t | = 1.299
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 4.2249 ) = 0.00005
hence value of p0.1 > 0.00005,here we reject Ho
ANSWERS
---------------
null, Ho: u1 < u2
alternate, H1: u1 > u2
test statistic: 4.2249
critical value: 1.299
decision: reject Ho
p-value: 0.00005
the mean speed for the new chip is greater than that of the old chips
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