A manufacturer receives parts from two suppliers. A simple random sample of 400 parts from supplier 1 finds 20 defective. A simple random sa

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A manufacturer receives parts from two suppliers. A simple random sample of 400 parts from supplier 1 finds 20 defective. A simple random sample of 200 parts from supplier 2 finds 20 defective. Let p1 and p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. Test whether the defective rates of the parts from two suppliers are significant different at the 1% significance level. Conduct a hypothesis testing. Answer the next three questions. 12. Test statistic

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Thông Đạt 1 week 2021-07-21T21:47:49+00:00 1 Answers 3 views 0

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    2021-07-21T21:48:57+00:00

    Answer:

    The test statistic is z = -2.1

    The p-value of the test is 0.0358 > 0.01, which means that the defective rates of the two suppliers are not significant different at the 1% significance level.

    Step-by-step explanation:

    Before testing the hypothesis, we need to understand the central limit theorem and subtraction of 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.

    For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

    Subtraction between normal variables:

    When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

    A simple random sample of 400 parts from supplier 1 finds 20 defective.

    This means that:

    p_1 = \frac{20}{400} = 0.05, s_1 = \sqrt{\frac{0.05*0.95}{400}} = 0.0109

    A simple random sample of 200 parts from supplier 2 finds 20 defective.

    This means that:

    p_2 = \frac{20}{200} = 0.1, s_2 = \sqrt{\frac{0.1*0.9}{200}} = 0.0212

    Let p1 and p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. Test whether the defective rates of the parts from two suppliers are significant different at the 1% significance level.

    At the null hypothesis, we test if they are equal, that is, if the subtraction of the proportions is 0. So

    H_0: p_1 - p_2 = 0

    At the alternate of the null hypothesis, we test if they are different, that is, if the subtraction of the proportions is different of 0. So

    H_a: p_1 - p_2 \neq 0

    The test statistic is:

    z = \frac{X - \mu}{s}

    In which X is the sample mean, \mu is the value tested at the null hypothesis and s is the standard error.

    0 is tested at the null hypothesis:

    This means that \mu = 0

    From the sample proportions:

    X = p_1 - p_2 = 0.05 - 0.1 = -0.05

    s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0109^2 + 0.0212^2} = 0.0238

    Value of the test statistic:

    z = \frac{X - \mu}{s}

    z = \frac{-0.05 - 0}{0.0238}

    z = -2.1

    P-value of the test and decision:

    The p-value of the test is the probability that the sample proportion differs from 0 by at least 0.05, that is, P(|z| < 2.1), which is 2 multiplied by the p-value of Z = -2.1.

    Z = -2.1 has a p-value of 0.0179

    2*0.0179 = 0.0358.

    The p-value of the test is 0.0358 > 0.01, which means that the defective rates of the two suppliers are not significant different at the 1% significance level.

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