Question

An investigator compares the durability of two different compounds used in the manufacture of a certain automobile brake lining. A sample of 243 brakes using Compound 1 yields an average brake life of 37,866 miles. A sample of 268 brakes using Compound 2 yields an average brake life of 45,789 miles. Assume that the population standard deviation for Compound 1 is 4414 miles, while the population standard deviation for Compound 2 is 2368 miles. Determine the 95% confidence interval for the true difference between average lifetimes for brakes using Compound 1 and brakes using Compound 2. Step 2 of 3 : Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places.

Answer 1

The margin of error is of 623.2016.

The 95% confidence interval for the true difference between average lifetimes for brakes using Compound 1 and brakes using Compound 2 is (-8546.2016, -7299.7984).

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
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

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

Subtraction between 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 243 brakes using Compound 1 yields an average brake life of 37,866 miles. The population standard deviation for Compound 1 is 4414 miles.

This means that
`\mu_1 = 37866, s_1 = (4414)/(√(243)) = 283.16`

A sample of 268 brakes using Compound 2 yields an average brake life of 45,789 miles. The population standard deviation for Compound 2 is 2368 miles.

This means that
`\mu_2 = 45789, s_2 = (2368)/(√(268)) = 144.65`

Distribution of the difference between average lifetimes for brakes using Compound 1 and brakes using Compound 2.

The mean is:

`\mu = \mu_1 - \mu_2 = 37866 - 45789 = -7923`

The standard deviation is:

`s = √(s_1^2 + s_2^2) = √(283.16^2 + 144.64^2) = 317.96`

Confidence interval:

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

`\alpha = (1 - 0.95)/(2) = 0.025`

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

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = zs`

`M = 1.96*317.96 = 623.2016`

The margin of error is of 623.2016.

The lower end of the interval is the sample mean subtracted by M. So it is -7923 - 623.2016 = -8546.2016

The upper end of the interval is the sample mean added to M. So it is -7923 + 623.2016 = -7299.7984

The 95% confidence interval for the true difference between average lifetimes for brakes using Compound 1 and brakes using Compound 2 is (-8546.2016, -7299.7984).