Question

5) A nutritionist wants to compare the mean protein content of grilled chicken sandwiches from Arby's and McDonald's. You randomly select several chicken sandwiches from both restaurants and find the following: Arby's finds from a sample of 15 sandwiches, a mean protein content of 29g with a sample standard deviation of 1.5g. McDonald's finds from a sample of 20 sandwiches, a mean protein content of 25g with a sample standard deviation of 2g. Find a 95% confidence interval for the true difference in mean protein contents.

Answer 1

formula for estimation is:

μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)}

where:

M₁ & M₂ = sample means
t = t statistic determined by confidence level
s_{(M1 - M2)} = standard error = √((s²_p/n₁) + (s²_p/n₂))

Calculation

Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 107.5 / 33 = 3.26

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((3.26/15) + (3.26/20)) = 0.62

Confidence Interval
μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)} = 4 ± (2.03 * 0.62) = 4 ± 1.25

The 95% confidence interval is (2.75, 5.25)