SOLUTION: A machine that stuffs a cheese filled snack product can be adjusted for the amount of cheese injected into each unit. A simple random sample of 30 units is selected, and the averag
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Question 1178027: A machine that stuffs a cheese filled snack product can be adjusted for the amount of cheese injected into each unit. A simple random sample of 30 units is selected, and the average amount of cheese injected is found to be X= 3.5 grams. If the process standard deviation is known to be o =0.25grams, construct a 99% confidence interval for the average amount of cheese being injected by machine.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
sample size = 30
population standard deviation = .25
sample mean = 3.5
standard error = population standard deviation / square root of sample size = .25/sqrt(30) = .0456435465.
z-score of two tail confidence interval of 90% = plus or minus 2.5758.
z-score formula = (x - m) / se
x is raw score
m is mean
se is standard error.
plus side raw score = .25/sqrt(30) * 2.5758 + 3.5 = 3.617568647
minus side raw score = .25/sqrt(30) * -2.5758 + 3.5 = 3.382431353
at 99% two tail confidence interval, the average amount of cheese injected will be between 3.6176 and 3.3824 grams.
visually, this looks like this:
when you're dealing with the mean of a sample, you have to use the standard error rather than the standard deviation.
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