Are America's top chief executive officers (CEOs) really worth all that money?

Let mu1 be the mean for corporation

mu2 be the mean for CEO



Ho: mu1=mu2

Ha: mu1>mu2



The test statistic is

t=(xbar1-xbar2)/sqrt(s1^2/n1+s2^2/n2)

=(10.75-8.875)/sqrt(8.328094^2/8+11.54417^2/8)

=0.37



The p-value= P(t>0.37) =0.3585



Since the p-value is larger than 0.05, we do not reject Ho.



So we can not conclude that the population mean percentage increase in corporate revenue is greater than the population mean percentage increase in CEO salary

percent change for corporation data :
sample size ( n ) = 8
sample mean (x1-bar) = 19.75
sample s.d. ( s1 ) = 9.9718

percent change for CEO data :
sample size ( n ) = 8
sample mean (x2-bar) = 17.5
sample s.d. ( s2 ) = 9.4472

Hypotheses of interest : Ho: μ1 = μ2 against Hr : μ1 ≥ μ2..
where μ1, μ2 are the population means percentage increase in corporate revenue and population mean percentage increase in CEO salary.

Here the test is right tailed t -test.

Test Statistics :
t = [ (x1-bar - x2-bar ) - ( μ1 - μ2 ) ] / √ [ ( s1^2 / n ) + ( s2^2 / n ) ]

Here,
t = [ (19.75 - 17.5 ) - 0 ] / √ [ ( 9.97182 / 8 ) + ( 9.44722 / 8 ) ] = 2.25 / 4.8565 = 0.4633

Here our test statistics is student's t statistic.

The degrees of freedom is the number of independent estimates of variance = (n - 1) + (n - 1) = 14

So, P-value = P ( t ≥ 0.4633 ) = 0.3251

Since the P-value is significantly greater than the level of significance = 5%, We can not reject the null Ho: μ1 = μ2.

Hence we may accept the null hypothesis and can conclude that these data do not indicate that the population mean percentage increase in corporate revenue is greater than the population mean percentage increase in CEO salary.