Respuesta :
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:
[tex]H_0: p \leq 0.245[/tex]
At the alternative hypothesis, we test if this proportion is greater than 24.5%, that is:
[tex]H_1: p > 0.245[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.245 is tested at the null hypothesis:
This means that [tex]\mu = 0.245, \sigma = \sqrt{0.245*0.755}[/tex]
Survey 86,664 randomly sampled U.S. adults. Of the adults surveyed, 23,053 said that they were obese.
This means that [tex]n = 86664, X = \frac{23053}{86664} = 0.266[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.266 - 0.245}{\frac{\sqrt{0.245*0.755}}{\sqrt{86664}}}[/tex]
[tex]z = 14.37[/tex]
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
