Question

Records of 40 used passenger cars and 40 used pickup trucks (none used commercially) were randomly selected to investigate whether there was any difference in the mean time in years that they were kept by the original owner before being sold. For cars the mean was 5.3 years with standard deviation 2.2 years. For pickup trucks the mean was 7.1 years with standard deviation 3.0 years. Construct the 95% confidence interval for the difference in the means based on these data

Answer 1

The 95% confidence interval for the difference in means = (-2.953, -0.647)

Explanation:

The confidence interval for difference in means formula =

μ1 - μ2 ± z × √(σ²1/n1 + σ²2/n2)

Where

σ = standard deviation

n = number of samples

μ = mean

Passenger cars = 1

Pick up trucks = 2

μ1 =5.3 , σ1 =2.2 , n2 = 40

μ1 =7.1, σ1 =3.0 , n2 = 40

z = 95 % confidence interval = 1.96

μ1 - μ2 ± z × √(σ²1/n1 + σ²2/n2)

= 5.3 - 7.1 + 1.96 × √2.2²/40 + 3.0²/40

= -1.8 ± 1.96× √0.121 + 0.225

= -1.8 ± 1.96 ×√0.346

= -1.8 ± 1.96 × 0.5882176468

Confidence interval = -1.8 ± 1.1529065877

= -1.8 - 1.1529065877

= -2.9529065877

≈ -2.953

= -1.8 + 1.1529065877

= -0.6470934123

≈ -0.647

Therefore, the 95% confidence interval for the difference in means = (-2.953, -0.647)