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Sagot :
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:
[tex]H_0: \mu = 0[/tex]
At the alternative hypothesis, we test if there was an increase, that is, the mean is greater than 0, so:
[tex]H_1: \mu > 0[/tex]
The test statistic is:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] 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 [tex]\mu = 0[/tex]
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 [tex]n = 104, X = 6.9, s = 55[/tex]
Value of the test-statistic:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{6.9 - 0}{\frac{55}{\sqrt{104}}}[/tex]
[tex]t = 1.28[/tex]
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.
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