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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 fluid ounces and the sample standard deviation is fluid ounces. Find a 95% two-sided confidence interval on the mean volume of syrup dispensed. Assume the population is approximately normally distributed.

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funkeyusuff52

Answer:

You can be 95% confident that the population mean (μ) falls between 0.5837 and 1.3363.

Step-by-step explanation:

Calculation

M = 0.96

Z = 1.96

sM = √(0.96)^2/25) = 0.19

μ = M ± Z(sM)

μ = 0.96 ± 1.96*0.19

μ = 0.96 ± 0.3763

Result

M = 0.96, 95% CI [0.5837, 1.3363].

You can be 95% confident that the population mean (μ) falls between 0.5837 and 1.3363.

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