Question

A student researcher compares the ages of cars owned by students and cars owned by faculty at a local state college. A sample of 224 cars owned by students had an average age of 5.06 years. A sample of 233 cars owned by faculty had an average age of 7.19 years. Assume the standard deviation is known to be 3.42 years for age of cars owned by students and 2.81 years for age of cars owned by faculty. Determine the 95% confidence interval for the difference between the true mean ages for cars owned by students and faculty. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval.

Answer 1

The formula for estimation is:

μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)}

where:

M₁ & M₂ = sample means
t = t statistic or critical value determined by confidence level =2.13
s_{(M1 - M2)} = standard error = √((s²_p/n₁) + (s²_p/n₂))

So Calculation

Pooled Variance
s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 4440.19 / 455 = 9.76

Standard Error
s_{(M1 - M2)} = √((s²_p/n₁) + (s²_p/n₂)) = √((9.76/224) + (9.76/233)) = 0.29

Confidence Interval
μ₁ - μ₂ = (M₁ - M₂) ± ts_{(M1 - M2)} = 2.13 ± (1.97 * 0.29) = 2.13 ± 0.5745

95% CI [1.5555, 2.7045].