An investigator compares the durability of two different compounds used in the manufacture of a certain automobile brake lining. A sample of

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.

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Trúc Chi 6 months 2021-08-28T11:03:49+00:00 1 Answers 1 views 0

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    2021-08-28T11:05:15+00:00

    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).

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