A company that produces ice cream, wants to make a plan of when should they calibrate their machines. Each machine is scheduled to fill in
a bucket of ice cream with 1 kg of ice cream. The company wants to identify when should they calibrate their machines, based on how much ice cream they drop.
If they know that the standard deviation of ice cream dropped deviation is 27 grams. They run 23 tests on the machine. Approximate the Confidence interval that shows that the machine operates without issues
with a 95% confidence interval.
What is the standard error?
=standard deviation / sqrt(n)
= 23/ sqrt(27)
= 5.62988918
What is the margin of error?
11.6756755
Find t-score first: =t.inv(0.975,22) = 2.07387307
Multiply t-score by standard error to find margin of error = = 2.07387307 * 5.62988918
= 11.6756755
What is the Confidence Interval, that is acceptable and shows that the machine doesn't need calibration?
There is no sample mean so in this case it’s:
Mu- 11.6756755 = lower bound of the confidence interval Mu+ 11.6756755 = Upper bound of the confidence interval
