Question

In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer. Are America's top chief executive officers (CEOs) really worth all that money? One way to answer this question is to look at row B, the annual company percentage increase in revenue, versus row A, the CEO's annual percentage salary increase in that same company. Suppose a random sample of companies yielded the following data: B: Percent increase for company 16 24 12 18 6 4 21 37 A: Percent increase for CEO 15 25 19 14 -4 19 15 30 Do these data indicate that the population mean percentage increase in corporate revenue (row B) is different from the population mean percentage increase in CEO salary? Use a 5% level of significance. Solve the problem using the critical region method of testing. (Let d = B − A. Round your answers to three decimal places.) test statistic = critical value = ±

Answer 1

B	A	Difference
16	15	1
24	25	-1
12	19	-7
18	14	4
6	-4	10
4	19	-15
21	15	6
37	30	7

∑d = 5

∑d² = 477

n = 8

Mean , x̅d = Ʃd/n = 5/8 = 0.6250

Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] = √[(477-(5)²/8)/(8-1)] = 8.2278

Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd ≠ 0

Test statistic:

t = (x̅d)/(sd/√n) = (0.625)/(8.2278/√8) = 0.2149

df = n-1 = 7

Critical value :

Two tailed critical value, t-crit = T.INV.2T(0.05, 7) = ± 2.365

Decision:

Do not reject the null hypothesis

Conclusion:

There is not enough evidence to conclude that the population mean percentage increase in corporate revenue (row B) is different from the population mean percentage increase in CEO salary