SOLUTION: A food snack manufacturer samples 7 bags of pretzels off the assembly line and weighs their contents. If the sample mean is 15.2 oz and the sample standard deviation is 0.70 oz, fi

Question 1200658: A food snack manufacturer samples 7 bags of pretzels off the assembly line and weighs their contents. If the sample mean is 15.2 oz and the sample standard deviation is 0.70 oz, find the 95% confidence interval of the true mean.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
sample size is 7.
sampole mean is 15.2.
sample standard deviation is .7
critical t-score at 6 degrees of freedom at 95% confidence interval is equal to plus or minus 2.446911839.
t-score formula is t = (x-m)/s
t is the t-score
x is the confidence limits of the true mean.
m is the sample mean
s is the standard error.
standard error = .7/sqrt(7) = .2645751311.
on the low side, you get:
minus 2.446911839 = (x-15.2)/.2645751322.
solve for x to get x = 2.446911839 * .2645751322 + 15.2 = 14.5
on the high side, you get:
2.446911839 = (x - 15.2)/.2645751322.
solve for x to get x = 2.446911839 * .2645751322 + 15.2 = 15.9
the 95% confidence interval is from 14.5 to 15.9.
the true population mean will be within those limits 95% of the time.

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