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 1.05 fluid Ounces and a standard deviation of 0.15 fluid Ounces. Assume that the syrup content of a beverage is normally distributed. Then, to check the claim that the syrup content per beverage is about one ounce, we use the following test He: - 1 against H: Hyl. Based on the samples, the p-value for the test is less than 0.05. True False

Answer 1

Given :

n = 25

= 1.05

T

s = 0.15

Test statistic t=ă-l sium = 1:05-1 0.15/525 = 1.6667 = 1.667 Degenee of freedom, df - nal 25-1 = 24 P-value P-value with test statistic t=1.667 and elf=24 is p-value = 2 x Plt? 1.667) 0.1085 since p-value is greter than significance level, 2=0.05 Therefore correct option b) false,