Question 1202553: A beer producer wants to find out whether its existing bottling equipment is working to standards. 151 bottles were selected for testing: the average beer fill level is x̄ = 50.1mm and the sample variance s^2 = 4mm^2.
Determine the 99% confidence interval for the mean value.
Determine the 90% confidence interval for the variance.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample size is 151.
sample mean is 50.1 millimeters.
sample variance is 4 millimeters.
sample standard deviation is square root of 4 = 2 millimeters.
standard error is 2/sqrt(151) = 1.628 rounded to 4 decimal places.
critical t-score at 99% two-tailed confidence inteval with 150 degrees of freedom is plus or minus 2.609 rounded to 3 decimal places.
critical t-score at 90% two-tailed confidence interval with 150 degrees of freedom is plus or minus 1.656 rounded to 3 decimal places.
mean of the sample is assumed to be 50.1.
mean of the variance is assumed to be 4.
use the t-score formula to find the raw score of the confidence interval.
formula is t = (x-m)/s
t is the t-score.
x is the critical raw score.
m is the mean of the sample.
s is the standard error.
for the mean, the formula becomes:
2.609 = (x - 50.1) / 1.628 on the high side.
solve for x to get x = 2.609 * 1.628 + 50.1 = 54.35 rounded to 2 decimal places.
-2.609 = (x - 50.1) / 1.628 on the low side.
solve for x to get x = -2.609 * 1.628 + 50.1 = 45.85 rounded to 2 decimal places.
for the variance, the formula becomes:
1.656 = (x - 4) / 1.628 on the high side.
solve for x to get x = 1.656 * 1.628 + 4 = 6.7 rounded to 2 decimal places.
-1.656 = (x - 4) / 1.628 on the low side.
solve for x to get x = -1.656 * 1.628 + 4 = 1.30 rounded to 2 decimal places.
your 99% confidence interval for the mean is 45.85 to 54.35.
your 99% confidence interval for the variance is 1.30 to 6.70.
i'm pretty sure about the mean of the sample.
i think that's the way to analyze the variance of the sample, but i'm not surre.
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