A manufacturer of nails claims that only 4% of its nails are defective. A random sample of 20 nails is selected, and it is found that two of

Question

A manufacturer of nails claims that only 4% of its nails are defective. A random sample of 20 nails is selected, and it is found that two of them, 10%, are defective. Is it fair to reject the manufacturer’s claim based on this observation?

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Helga 3 years 2021-08-11T17:51:46+00:00 1 Answers 14 views 0

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    2021-08-11T17:53:20+00:00

    Answer:

    The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer’s claim based on this observation.

    Step-by-step explanation:

    A manufacturer of nails claims that only 4% of its nails are defective.

    At the null hypothesis, we test if the proportion is of 4%, that is:

    H_0: p = 0.04

    At the alternative hypothesis, we test if the proportion is more than 4%, that is:

    H_a: p > 0.04

    The test statistic is:

    z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

    In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

    4% is tested at the null hypothesis

    This means that \mu = 0.04, \sigma = \sqrt{0.04*0.96}

    A random sample of 20 nails is selected, and it is found that two of them, 10%, are defective.

    This means that n = 20, X = 0.1

    Value of the test statistic:

    z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

    z = \frac{0.1 - 0.04}{\frac{\sqrt{0.04*0.96}}{\sqrt{20}}}

    z = 1.37

    P-value of the test and decision:

    Considering an standard significance level of 0.05.

    The p-value of the test is the probability of finding a sample proportion above 0.1, which is 1 subtracted by the p-value of z = 1.37.

    Looking at the z-table, z = 1.37 has a p-value of 0.9147

    1 – 0.9147 = 0.0853

    The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer’s claim based on this observation.

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