Question

A study of the pay of corporate chief executive officers (CEOs) examined the increase in cash compensation of the CEOs of 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%. Is this good evidence that the mean real compensation μ of all CEOs increased that year? The hypotheses are

Answer 1

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:

$H_0: \mu = 0$

At the alternative hypothesis, we test if there was an increase, that is, the mean is greater than 0, so:

$H_1: \mu > 0$

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ 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 $\mu = 0$

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 $n = 104, X = 6.9, s = 55$

Value of the test-statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{6.9 - 0}{\frac{55}{\sqrt{104}}}$

$t = 1.28$

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.