Question 298823: 1. By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the standard deviation is 5.1 seconds.
Answer each of the following (show all work):
(A) How many measurements should be made in order to be 95% certain that the maximum error of estimation will not exceed 1.5 seconds?
(B) What sample size is required for a maximum error of 2.5 seconds?
2. Determine the critical region and critical values for z that would be used to test the null hypothesis at the given level of significance, as described in each of the following:
(A) Ho: ≤ 20 and Ha: > 20, α = 0.05
(B) Ho: = 25and Ha: ≠ 25, α = 0.01
(C) Ho: ≥ 18and Ha: < 18, α = 0.01
3. A 99% confidence interval estimate for a population mean was computed to be (42.0 to 50.0). Determine the mean of the sample, which was used to determine the interval estimate (show all work).
Answer by alanc(27)   (Show Source):
You can put this solution on YOUR website! A.) Ok, we know our confidence interval, it is a 95% interval.
Our variable of interest X : the amount of time a component moves from one station to another.
s = 5.1 sec (Standard deviation)
We want to find n: the number of measurements to make such that our error doesn't exceed 1.5
I will assume this is a Normal distribution
Error is X - u. The Z variable is z = (x-u)/[s/sqrt(n)]
We want the Probability(X-u <= 1.5 ) >= 0.95
So: P((x-u)/[s/sqrt(n)] <= 1.5/[5.1*sqrt(n)] ) >= 0.95
But since the normal distribution is symmetric about mean, I will look for a z-value corresponding to 0.95/2 = 0.475, which is z= 1.96
from the z-score chart this corresponds to : 0 < z <= 1,96
Z = (x-u)/[s/sqrt(n)] <= 1.5/[5.1*sqrt(n)]
therefore sqrt(n)* 1.5/5.1 = 1.96
n = (5.1*1.96/1.5)^2
n = 44.408
so I would say 44 or more tests.
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The estimator for standard deviation is s/sqrt(n) not s/sqrt(n -1)
that is why I have Z = (x-u)/[s/sqrt(n)], if I used s/sqrt(n-1) as my estimator, then Z = (x-u)/[s/sqrt(n - 1)] would be my Z statistic and then n would be 45 or more tests.
I assumed we used a normal distribution because the number of measurements was large. If the number of samples is small, we would be forced to use the t-distribution.
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