A national consumer agency selected independent random samples of 45 owners of newer cars (less than five years old) and 40 owners of older

Question

A national consumer agency selected independent random samples of 45 owners of newer cars (less than five years old) and 40 owners of older cars (more than five years old) to estimate the difference in mean dollar cost of yearly routine maintenance, such as oil changes, tire rotations, filters, and wiper blades. The agency found the mean dollar cost per year for newer cars was $195 with a standard deviation of $46. For older cars, the mean was $286 with a standard deviation of $58.

Required:
What represents the 95 percent confidence interval to estimate the difference (newer minus older) in the mean dollar cost of routine maintenance between newer and older cars?

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Thu Cúc 6 months 2021-08-21T06:38:27+00:00 1 Answers 6 views 0

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    2021-08-21T06:39:33+00:00

    Answer:

    The 95 percent confidence interval to estimate the difference (newer minus older) in the mean dollar cost of routine maintenance between newer and older cars is (-$113.44, -$68.56)

    Step-by-step explanation:

    Subtraction of normal variables:

    When we subtract the normal distributions, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

    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.

    45 owners of newer cars (less than five years old). The agency found the mean dollar cost per year for newer cars was $195 with a standard deviation of $46

    This means that:

    \mu_N = 195

    s_N = \frac{46}{\sqrt{45}} = 6.8573

    40 owners of older cars (more than five years old). For older cars, the mean was $286 with a standard deviation of $58.

    This means that:

    \mu_O = 286

    s_O = \frac{58}{\sqrt{40}} = 9.17

    Subtaction:

    Sample mean:

    \mu = \mu_N - \mu_O = 195 - 286 = -91

    Standard error:

    s = \sqrt{s_N^2 + s_O^2} = \sqrt{6.8573^2+9.17^2} = 11.45

    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

    In which s is the standard error.

    So

    M = zs = 1.96*11.45 = 22.44

    The lower end of the interval is the sample mean subtracted by M. So it is -91 – 22.44 = -$113.44

    The upper end of the interval is the sample mean added to M. So it is -91 + 22.44 = -$68.56

    The 95 percent confidence interval to estimate the difference (newer minus older) in the mean dollar cost of routine maintenance between newer and older cars is (-$113.44, -$68.56)

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