Question

A postmix beverage machine is adjusted to release a certain amount of syrup into a chamber where it is mixed with carbonated water. A random sample of 25 beverages was found to have a mean syrup content of t= 1.17 fluid Ounces and the sample standard deviation is s = 0.015 fluid Ounces. Find a 95% two-sided confidence interval on the mean volume of syrup dispensed. Assume population is approximately normally distributed. beverages was found to have mangrup content 863 = 1.17 rue ounces and Round your answers to 3 decimal places. sus

Answer 1

Degree of freedom=n-1=25-1=24

Critical value at 95% confidence interval(t_a/2)=2.06

Confidence interval is given by

i Ita/2 * vn 0.015 = 1.17 + 2.06 ** V25 = 1.17 +0.00618 = (1.1638, 1.1762) hence Ans: 1.164 <u<1.176