Question

Noah wants to advertise how many chocolate chips are in each Big Chip cookie at his bakery. He randomly selects a sample of 41 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 8.1 and a standard deviation of 1.8. What is the 80% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Assume the data is from a normally distributed population. Round answers to 3 decimal places where possible.

Answer 1

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 8.1

S = 1.8

n = 41

df = n – 1 = 40

Confidence level = 80%

Critical t value = 1.3031

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 8.1 ± 1.3031*1.8/sqrt(41)

Confidence interval = 8.1 ± 0.3663

Lower limit = 8.1 - 0.3663 = 7.734

Upper limit = 8.1 + 0.3663 = 8.466

Confidence interval = (7.73, 8.466)