A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line.The machine that dispenses dressing is working properly when 8 ounces are dispensed.The standard deviation of the process is 0.15 ounces.A sample of 48 bottles is selected periodically,and the filling line is stopped if there is evidence that the mean amount dispensed is different from 8 ounces.Suppose that the mean amount dispensed in a particular sample of 48 bottles is 7.983 ounces.Calculate a confidence interval to test the hypotheses at $\alpha$ = .05.
A)[7.947 8.019]
B)[7.941 8.025]
C)[7.933 8.033]
D)[7.955 8.011]
E)[7.916 8.050]

Question

Quizplus · Accepted Answer

To calculate the confidence interval, we use the formula:

CI = x̄ ± zα/2 * σ/√n

where x̄ = 7.983, σ = 0.15, n = 48, and zα/2 is the critical value of the standard normal distribution for a level of significance α/2.

From the table of the standard normal distribution, we find that zα/2 = 1.96 for α = 0.05.

Plugging in the values, we get:

CI = 7.983 ± 1.96 * 0.15/√48
   = [7.941, 8.025]

Therefore, the best choice is B, which matches the calculated confidence interval.

A Manufacturer of Salad Dressings Uses Machines to Dispense Liquid α\alphaα

A Manufacturer of Salad Dressings Uses Machines to Dispense Liquid $\alpha$