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:
![H_0: p = 0.6](https://tex.z-dn.net/?f=H_0%3A%20p%20%3D%200.6)
The alternate hypothesis is:
![H_1: p > 0.6](https://tex.z-dn.net/?f=H_1%3A%20p%20%3E%200.6)
The test statistic is:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
0.6 is tested at the null hypothesis:
This means that ![\mu = 0.6, \sigma = \sqrt{0.6*0.4}](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.6%2C%20%5Csigma%20%3D%20%5Csqrt%7B0.6%2A0.4%7D)
Of 100 randomly selected subjects who followed the diet, 70 noticed an increase in their energy level.
This means that ![n = 100, X = \frac{70}{100} = 0.7](https://tex.z-dn.net/?f=n%20%3D%20100%2C%20X%20%3D%20%5Cfrac%7B70%7D%7B100%7D%20%3D%200.7)
Test statistic:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z = \frac{0.7 - 0.6}{\frac{\\sqrt{0.6*0.4}}{\sqrt{100}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B0.7%20-%200.6%7D%7B%5Cfrac%7B%5C%5Csqrt%7B0.6%2A0.4%7D%7D%7B%5Csqrt%7B100%7D%7D%7D)
![z = 2.04](https://tex.z-dn.net/?f=z%20%3D%202.04)
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.
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