Suppose that the amount of coffee in each cup dispensed by a

Suppose that the amount of coffee in each cup dispensed by a vending machine follows a normal distribution. The operator of the vending machine claims that the amount of coffee in each cup has a mean of 12.5 ounces and a standard deviation of 1.8 ounces. The mean of the distribution can be set by adjusting the filling machinery, while the standard deviation reflects the precision of the filling machinery. Construct a 90% confidence interval estimate for the actual mean amount of coffee per cup if 40 randomly selected cups had a mean of 13.0 ounces. use 1.8 standard deviation. Does the interval created in the problem above contain 12.5 ounces (the mean content claimed by the operator)? What might this tell us about the operator\'s claim?

Solution

CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=13
Standard deviation( sd )=1.8
Sample Size(n)=40
Confidence Interval = [ 13 ± Z a/2 ( 1.8/ Sqrt ( 40) ) ]
= [ 13 - 1.64 * (0.285) , 13 + 1.64 * (0.285) ]
= [ 12.533,13.467 ]

The interval created above does n\'t contain 12.5 ounces