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

Answer 1

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