An investigator compares the durability of two different compounds used in the manufacture of a certain automobile brake lining. A sample of 212 brakes using Compound 1 yields an average brake life of 47,895 miles. A sample of 180 brakes using Compound 2 yields an average brake life of 49,767 miles. Assume that the population standard deviation for Compound 1 is 1590 miles, while the population standard deviation for Compound 2 is 4152 miles.
1. Determine the 90% confidence interval for the true difference between average lifetimes for brakes using Compound 1 and brakes using Compound 2.
2. Find the point estimate for the true difference between the population means.

Answer:

1. The 90% confidence interval for the true difference between average lifetimes for brakes using Compound 1 and brakes using Compound 2 is (-2411.84, -1332.16).

2. The point estimate for the true difference between the population means is of -1872.

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction of Normal Variables:

When two normal variables are subtracted, the mean is the subtraction of the means while the standard deviation is the square root of the sum of the variances.

A sample of 212 brakes using Compound 1 yields an average brake life of 47,895 miles. The population standard deviation for Compound 1 is 1590 miles.

This means that [tex]\mu_1 = 47895, \sigma_1 = 1590, n = 212, s_1 = \frac{1590}{\sqrt{212}} = 109.2[/tex]

A sample of 180 brakes using Compound 2 yields an average brake life of 49,767 miles. The population standard deviation for Compound 2 is 4152 miles.

This means that [tex]\mu_2 = 49767, \sigma_2 = 4152, n = 180, s_2 = \frac{4152}{\sqrt{180}} = 309.47[/tex]

True difference between average lifetimes for brakes using Compound 1 and brakes using Compound 2.

This is the distribution 1 - 2. So

[tex]\mu = \mu_1 - \mu_2 = 47895 - 49767 = -1872[/tex]

This is also the point estimate for the true difference between the population means, which is question 2.

[tex]s = \sqrt{s_1^2+s_2^2} = \sqrt{109.2^2+309.47^2} = 328.17[/tex]

90% confidence interval

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.9}{2} = 0.05[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.05 = 0.95[/tex], so Z = 1.645.

Now, find the margin of error M as such

[tex]M = zs = 1.645*328.17 = 539.84[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is -1872 - 539.84 = -2411.84

The upper end of the interval is the sample mean added to M. So it is -1872 + 539.84 = -1332.16.

The 90% confidence interval for the true difference between average lifetimes for brakes using Compound 1 and brakes using Compound 2 is (-2411.84, -1332.16).

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