A consumer group has determined that the distribution of life spans for gas ovens has a mean of 15.0 years and a standard deviation of 4.2 y

Question

A consumer group has determined that the distribution of life spans for gas ovens has a mean of 15.0 years and a standard deviation of 4.2 years. The distribution of life spans for electric ovens has a mean of 13.4 years and a standard deviation of 3.7 years. Both distributions are moderately skewed to the right. Suppose we take a simple random sample of 35 gas ovens and a second simple random sample of 40 electric ovens. Suppose we take a simple random sample of 35 gas ovens and a second SRS of 40 electric ovens. Which of the following best describes the sampling distribution of barXG – bar XE, the difference in mean life span of gas and electric ovens?
A. Mean = 1.6 years, standard deviation = 7.9 years, shape: moderately right-skewed.
B. Mean = 1.6 years, standard deviation = 0.92 years, shape: approximately Normal.
C. Mean = 1.6 years, standard deviation = 0.92 years, shape: moderately right skewed.
D. Mean = 1.6 years, standard deviation = 0.40 years, shape: approximately Normal.
E. Mean = 1.6 years, standard deviation = 0.40 years, shape: moderately right skewed.

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Thu Giang 5 years 2021-08-19T19:40:32+00:00 1 Answers 126 views 0

Answers ( )

  1. Cherry
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    2021-08-19T19:42:18+00:00
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    Answer:

    B. Mean = 1.6 years, standard deviation = 0.92 years, shape: approximately Normal.

    Step-by-step explanation:

    To solve this question, we need to understand the normal probability distribution and the central limit theorem.

    Normal Probability Distribution:

    Problems of normal distributions can be solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

    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 of normal variables:

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

    35 gas ovens

    A consumer group has determined that the distribution of life spans for gas ovens has a mean of 15.0 years and a standard deviation of 4.2 years. This means that:

    \mu_G = 15, \sigma_G = 4.2, n = 35, s_G = \frac{4.2}{\sqrt{35}} = 0.71

    40 electric ovens.

    The distribution of life spans for electric ovens has a mean of 13.4 years and a standard deviation of 3.7 years.

    \mu_E = 13.4, \sigma_E = 3.7, n = 40, s_E = \frac{3.7}{\sqrt{40}} = 0.585

    Which of the following best describes the sampling distribution of barXG – bar XE, the difference in mean life span of gas and electric ovens?

    By the Central Limit Theorem, the shape is approximately normal.

    Mean: \mu = \mu_G - \mu_E = 15 - 13.4 = 1.6

    Standard deviation:

    s = \sqrt{s_G^2+s_E^2} = \sqrt{(0.71)^2+(0.585)^2} = 0.92

    So the correct answer is given by option b.

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