Question

Suppose μ₁ and μ₂ are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. The data follows: m = 10, x = 116.8, s₁ = 5.28, n = 7, _y = 129.6, and s₂ = 5.47. Calculate a 99% CI for the difference between true average stopping distances for cars equipped with system 1 and cars equipped with system 2. (Give answers accurate to 2 decimal places.)

Lower limit
Upper limit

Answer 1

Solution:

Confidence interval for difference between two population means is given as below:

Confidence interval = (X1bar – X2bar) ± t*sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given data, we have

X1bar = 116.8

X2bar = 129.6

S1 = 5.28

S2 = 5.47

n1 = 10

n2 = 7

df = n1 + n2 – 2 = 10 + 7 - 2 = 15

α = 0.01

Confidence level = 99%

Critical t value = 2.9467

(by using t-table)

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(10 – 1)* 5.28^2 + (7 – 1)* 5.47^2]/(10 + 7 – 2)

S_p² = 28.6954

(X1bar – X2bar) = 116.8 - 129.6 = -12.8

Confidence interval = (X1bar – X2bar) ± t*sqrt[S_p²*((1/n1)+(1/n2))]

Confidence interval = -12.8 ± 2.9467*sqrt[28.6954*((1/10)+(1/7))]

Confidence interval = -12.8 ± 2.9467*2.6399

Confidence interval = -12.8 ± 7.7789

Lower limit = -12.8 - 7.7789 = -20.58

Upper limit = -12.8 + 7.7789 = -5.02

Confidence interval = (-20.58, -5.02)

Lower limit = -20.58

Upper limit = -5.02