Fifty specimens of a new computer chip were tested for speed ina certain applica
ID: 2954121 • Letter: F
Question
Fifty specimens of a new computer chip were tested for speed ina certain application, along with 50 specimens of chips with theold design. The average speed, in MHz, for the new chips was495.6, and the standard deviation was 19.4. The average speedfor the old chips was 481.2, and the standard deviation was14.3.
a) Can you conclude a significant difference inthe mean speed between the new and old chips?
b) Verify your result in part a) using theappropriate confidence interval.
c) A sample of 60 even older chips had anaverage speed of 391.2 MHz with a standard deviation of 17.2MHz. Someone claims that the new chips average more than 100MHz faster than these very old ones. Do the data provideconvincing evidence for this claim?
Any helpplease!
Explanation / Answer
a) Can you conclude a significant difference inthe mean speed between the new and old chips?
Ho:(new) = (old)
Ha: (new) not equal (old)
The test statistic is Z=(xbar(new) -xbar(old))/(s^2(new)/n + s^2(old)/n)
=(495.6-481.2)/sqrt(19.4^2/50 + 14.3^2/50)
=4.22
If =0.05, Z(0.025) =1.96 (check normaltable), we reject Ho since Z=4.22 > 1.96. Thus, we conclude asignificant difference in the mean speed between the new and oldchips.
b) Verify your result in part a) using theappropriate confidence interval.
(xbar(new) -xbar(old))±Z*(s^2(new)/n + s^2(old)/n)
--> (495.6-481.2)±1.96*sqrt(19.4^2/50 +14.3^2/50)
--> (7.7196 , 21.0804)
c) A sample of 60 even older chips had anaverage speed of 391.2 MHz with a standard deviation of 17.2MHz. Someone claims that the new chips average more than 100MHz faster than these very old ones. Do the data provideconvincing evidence for this claim?
Z=[(xbar(new) -xbar(old))-100]/(s^2(new)/n + s^2(old)/n)
=(495.6-391.2-100)/sqrt(19.4^2/50 +17.2^2/60)
=1.25
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