Question

Jose wants to advertise how many chocolate chips are in each Big Chip cookie at his bakery. he randomly selects a sample of 67 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 19.5 and a standard deviation of 2.6. What is the 90% 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

Given a sample of n = 67 cookies and the number of chocolate chips per cookie in the sample has a mean of M = 19.5 and a standard deviation of s = 2.6, assuming the data is from a normally distributed population the confidence interval is calculated as:

μ = M ± t(s_M)

where:

M = sample mean

df = degree of freedom = n-1
t = t statistic determined by confidence level and the degree of freedom.
s_M = standard error = √(s²/n)

Since the population standard deviation is unknown hence t-distribution is applicable.

Now the calculation as;

M = 19.5

df = 67-1=66
t = 1.668 calculated using the excel formula for t-distribution which is =T.INV.2T(0.10, 66)
s_M = √(2.6²/67) = 0.3176

μ = M ± t(s_M)
μ = 19.5 ± 1.668*0.3176
μ = 19.5 ± 0.5298

Thus the confidence interval at 90% confidence level is computed as:

90% CI [18.970, 20.030].