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

Answer:

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).

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 $\mu$ 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 $s = \frac{\sigma}{\sqrt{n}}$.

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 $\mu_1 = 37866, s_1 = \frac{4414}{\sqrt{243}} = 283.16$

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 $\mu_2 = 45789, s_2 = \frac{2368}{\sqrt{268}} = 144.65$

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:

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{283.16^2 + 144.64^2} = 317.96$

Confidence interval:

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

$\alpha = \frac{1 - 0.95}{2} = 0.025$

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).