Question 1029050: A group of students from a college are expressing some concern with the amount of coffee that is being dispensed from a certain machine. The standard deviation for the population of fills is .24 oz.
1. If a margin of error of .08 is wanted for a 95% confidence interval, what sample size should be used?
2. If each of 100 students take a sample of 100 fills from the machine, and calculates the 95% confidence interval for the population mean fill, how many intervals are expected to have the value of the mean?
A) 38 B) 2 C) 5 D) 76 E) 95
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! This has to be +/-0.04
use t(0.975 with df=unknown)
t*0.24/sqrt(n)=0.04
square everything
t^2*0.0576/n=0.0016
t^2*0.0576/0.0016=n; I have to approximate t and then use that to get n.
t^2*36=n
will use 1.96 to approximate t. That squared is 3.84
t=3.84*36=138.24 or 139. Use calculator for that
t(0.975, df=139) is +/-0.0403
for n=140, the margin of error is +/-0.04
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95 are expected. This is another definition of a confidence interval. One does not know which 95 will contain it, only that 95 will be expected to and 95% with an infinite number.
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