Respuesta :
Answer:
The p-value of the test is 0.0207.
Step-by-step explanation:
Test of the claim that more than 60% of the people following a particular diet will experience increased energy (H1: p > 0.6).
The null hypothesis is:
[tex]H_0: p = 0.6[/tex]
The alternate hypothesis is:
[tex]H_1: p > 0.6[/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.
0.6 is tested at the null hypothesis:
This means that [tex]\mu = 0.6, \sigma = \sqrt{0.6*0.4}[/tex]
Of 100 randomly selected subjects who followed the diet, 70 noticed an increase in their energy level.
This means that [tex]n = 100, X = \frac{70}{100} = 0.7[/tex]
Test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.7 - 0.6}{\frac{\\sqrt{0.6*0.4}}{\sqrt{100}}}[/tex]
[tex]z = 2.04[/tex]
P-value of the test:
The p-value of the test is the probability of finding a sample proportion of at least 0.7, which is 1 subtracted by the p-value of z = 2.04.
Looking at the z-table, z = 2.04 has a p-value of 0.9793
1 - 0.9793 = 0.0207
The p-value of the test is 0.0207.
