Question 1205251: A quality controller of a company plans to inspect the average diameter of small bolts made. A random sample of 6 bolts was selected. The sample is computed to be 2.0016mm and the sample standard deviation 0.0012mm. Construct the 99% confidence interval for all bolts made.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample size is 6.
sample mean is 2.0016 millimeters.
sample standard deviation is .0012 millimeters.
since the sample mean and sample standard deviation are used, then the 99% confidence interval will be based on the t-score, rather than the z-score.
99% two tail confidence interval for t-score with 5 degrees of freedom (degrees of freedom equal sample size minus 1) is equal to t = plus or minus 4.032142983.
since you're looking at a distribution of sample means for samples of 6 elements each, the standard error is used.
standard error = standard deviation / sqrt(sample size) = .0012 / sqrt(6) = .0004898979486.
t-score formula is t = (x-m)/s
t is the critical t-score
x is the critical sample mean.
m is the sample mean.
s is the standard error.
on the low side of the confidence interval, the t-score formula becomes:
-4.032142983 = (x - 2.0016) / .0004898979486.
solve for x to get x = 1.999624661.
on the high side of the confidence interval, the t-score formula becomes:
4.032142983 = (x - 2.0016) / .0004898979486.
solve for x to get x = 2.003575339.
your two tail 99% confidence interval is from 1.999624661 to 2.003575339.
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