A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles. The machine that dispenses dressings is working properly when the mean amount dispensed is 330 mL. The population standard deviation of the amount dispensed is 4 mL. A sample of 50 bottles is selected periodically, and the filling line is stopped if there is evidence that the mean amount dispensed is different from 330 mL.
Suppose that the mean amount dispensed in a particular sample of 50 bottles is 329.7 mL. Is there evidence that the population mean amount is different from 330 mL? (Use a 0.05 leve

Question

A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles. The machine that dispenses dressings is working properly when the mean amount dispensed is 330 mL. The population standard deviation of the amount dispensed is 4 mL. A sample of 50 bottles is selected periodically, and the filling line is stopped if there is evidence that the mean amount dispensed is different from 330 mL.
Suppose that the mean amount dispensed in a particular sample of 50 bottles is 329.7 mL. Is there evidence that the population mean amount is different from 330 mL? (Use a 0.05 leve

dumbcow · Answer

find t-value

t = (329.7 -330)/(4/sqrt50)
take absolute value since this is 2-tailed test

degrees of freedom = n-1 = 49

look up p-value in t-distribution table
if p < .05, then there is significant evidence the mean is different

kropot72 · Answer

The value of t is found from the following equation:

$$t=((m- \mu)\sqrt{n})\div \sigma$$
where m is the mean of a sample of n  values
mu is the mean of the parent population
sigma is the standard deviation 

The value of t = 0.53
Taking the degrees of freedom to be 49, reference to a t-distribution table gives a probability of 0.6 that t = 0.53. The probability of 0.6 is much greater than 0.05 therefore there is insignificant evidence that the population mean amount is different from 330 mL.