Question Help In a study of cell phone usage and brain hemisphen dominance, an Internet survey was e-mailed to 6957 subjects randomly

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In a study of cell phone usage and brain hemisphen dominance, an Internet survey was e-mailed to 6957 subjects randomly selected from an online group involved with ears. There were 1319 surveys returned Use a 0.01
significance level to test the claim that the return rate is less than 20% Use the P-value method and use the normal distnbution as an approximation to the binomial distrbution
Identify the null hypothesis and alternative hypothesis

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Diễm Kiều 5 months 2021-09-05T09:10:37+00:00 1 Answers 5 views 0

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    2021-09-05T09:12:25+00:00

    Answer:

    The null hypothesis is H_0: p \geq 0.2

    The alternative hypothesis is H_1: p < 0.2

    The p-value of the test is of 0.0154 > 0.01, which means that at the 0.01 significance level, there is not significant evidence that the return rate is less than 20%.

    Step-by-step explanation:

    Test the claim that the return rate is less than 20%.

    At the null hypothesis, we test if the return rate is of at least 20%, that is:

    H_0: p \geq 0.2

    At the alternative hypothesis, we test if the return rate is less than 20%, that is:

    H_1: p < 0.2

    Normal distribution as an approximation to the binomial distribution to find the Z-score.

    Binomial probability distribution

    Probability of exactly x successes on n repeated trials, with p probability.

    Can be approximated to a normal distribution, using the expected value and the standard deviation.

    The expected value of the binomial distribution is:

    E(X) = np

    The standard deviation of the binomial distribution is:

    \sqrt{V(X)} = \sqrt{np(1-p)}

    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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

    When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

    0.2 is tested at the null hypothesis. Sample of 6957.

    This means that p = 0.2, n = 6957. So

    \mu = E(X) = np = 6957*0.2 = 1391.4

    \sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{6957*0.2*0.8} = 33.36

    1319 surveys returned

    This means that, due to continuity correction, X = 1319 + 0.5 = 1319.5. So

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

    Z = \frac{1319.5 - 1391.4}{33.36}

    Z = -2.16

    P-value of the test:

    The p-value of the test is the probability of 1319 or less people returning the survey, which is the p-value of Z = -2.16.

    Looking at the z-table, Z = -2.16 has a p-value of 0.0154.

    The p-value of the test is of 0.0154 > 0.01, which means that at the 0.01 significance level, there is not significant evidence that the return rate is less than 20%.

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