It is believed that the average amount of money spent per U.S. household per week on food is about $99, with standard deviation $13. A rando

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It is believed that the average amount of money spent per U.S. household per week on food is about $99, with standard deviation $13. A random sample of 16 households in a certain affluent community yields a mean weekly food budget of $100. We want to test the hypothesis that the mean weekly food budget for all households in this community is higher than the national average. State the null and alternative hypotheses for this test, the test statistic and determine if the results significant at the 5% level.

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RI SƠ 5 years 2021-07-20T01:12:41+00:00 1 Answers 25 views 0

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  1. Maris
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    2021-07-20T01:14:00+00:00
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    Answer:

    The null hypothesis is H_0: \mu = 99

    The alternate hypothesis is H_a: \mu > 99

    The test statistic is z = 0.31

    The pvalue of the test is 0.3783 > 0.05, which meas that the results are not significant at the 5% level.

    Step-by-step explanation:

    It is believed that the average amount of money spent per U.S. household per week on food is about $99. We want to test the hypothesis that the mean weekly food budget for all households in this community is higher than the national average.

    At the null hypothesis, we test if the mean is the national average of 99, that is:

    H_0: \mu = 99

    At the alternate hypothesis, we test if the mean is greater than the national average of 99, so:

    H_a: \mu > 99

    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.

    99 is tested at the null hypothesis:

    This means that \mu = 99

    A random sample of 16 households in a certain affluent community yields a mean weekly food budget of $100. Standard deviation of $13.

    This means that n = 16, X = 100, \sigma = 13

    Value of the test statistic:

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

    z = \frac{100 - 99}{\frac{13}{\sqrt{16}}}

    z = 0.31

    The test statistic is z = 0.31.

    Determine if the results significant at the 5% level.

    The pvalue of the test is the probability of finding a sample mean above 100, which is 1 subtracted by the pvalue of z = 0.31.

    Looking at the ztable, z = 0.31 has a pvalue of 0.6217

    1 – 0.6217 = 0.3783

    The pvalue of the test is 0.3783 > 0.05, which meas that the results are not significant at the 5% level.

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