Researchers believe that 24.5% of the adults in the United States are obese. A Gallup poll examined the rate of obesity in America with a su

Question

Researchers believe that 24.5% of the adults in the United States are obese. A Gallup poll examined the rate of obesity in America with a survey 86,664 randomly sampled U.S. adults. Of the adults surveyed, 23,053 said that they were obese. Is this enough evidence to state that more than 24.5% of all adults in the U.S. are obese

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Sigridomena 4 years 2021-08-25T10:22:54+00:00 1 Answers 17 views 0

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  1. Latifah
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    2021-08-25T10:24:36+00:00
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    Answer:

    The p-value of the test is 0 < 0.05(standard significance level), which means that this is enough evidence to state that more than 24.5% of all adults in the U.S. are obese

    Step-by-step explanation:

    Researchers believe that 24.5% of the adults in the United States are obese. Test if there is enough evidence to state that more than 24.5% of all adults in the U.S. are obese.

    At the null hypothesis, we test if the proportion is of 24.5% or less, that is:

    H_0: p \leq 0.245

    At the alternative hypothesis, we test if this proportion is greater than 24.5%, that is:

    H_1: p > 0.245

    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.

    0.245 is tested at the null hypothesis:

    This means that \mu = 0.245, \sigma = \sqrt{0.245*0.755}

    Survey 86,664 randomly sampled U.S. adults. Of the adults surveyed, 23,053 said that they were obese.

    This means that n = 86664, X = \frac{23053}{86664} = 0.266

    Value of the test statistic:

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

    z = \frac{0.266 - 0.245}{\frac{\sqrt{0.245*0.755}}{\sqrt{86664}}}

    z = 14.37

    P-value of the test and decision:

    The p-value of the test is the probability of finding a sample proportion of 0.266 or higher, which is 1 subtracted by the p-value of z = 14.37.

    The p-value of z = 14.37 is 1.

    1 – 1 = 0

    The p-value of the test is 0 < 0.05(standard significance level), which means that this is enough evidence to state that more than 24.5% of all adults in the U.S. are obese

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