Answer:
The p-value of the test is 0.1017, which is greater than the standard significance level of 0.05, which means that this is not good evidence that the mean real compensation μ of all CEOs increased that year.
Step-by-step explanation:
At the null hypothesis, we test if there was no increase, that is, the mean is 0, so:
![H_0: \mu = 0](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%20%3D%200)
At the alternative hypothesis, we test if there was an increase, that is, the mean is greater than 0, so:
![H_1: \mu > 0](https://tex.z-dn.net/?f=H_1%3A%20%5Cmu%20%3E%200)
The test statistic is:
![t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
In which X is the sample mean,
is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.
0 is tested at the null hypothesis:
This means that ![\mu = 0](https://tex.z-dn.net/?f=%5Cmu%20%3D%200)
104 companies, adjusted for inflation, in a recent year. The mean increase in real compensation was x¯=6.9%, and the standard deviation of the increases was s=55%.
This means that ![n = 104, X = 6.9, s = 55](https://tex.z-dn.net/?f=n%20%3D%20104%2C%20X%20%3D%206.9%2C%20s%20%3D%2055)
Value of the test-statistic:
![t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![t = \frac{6.9 - 0}{\frac{55}{\sqrt{104}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B6.9%20-%200%7D%7B%5Cfrac%7B55%7D%7B%5Csqrt%7B104%7D%7D%7D)
![t = 1.28](https://tex.z-dn.net/?f=t%20%3D%201.28)
P-value of the test:
The p-value of the test is a right-tailed test(test if the mean is greater than a value), with 104 - 1 = 103 df and t = 1.28.
Using a t-distribution calculator, this p-value is of 0.1017.
The p-value of the test is 0.1017, which is greater than the standard significance level of 0.05, which means that this is not good evidence that the mean real compensation μ of all CEOs increased that year.