Question

Are America's lop chief executive officers (CEDs) really worth all that money? One way to answer this question is to look at row the annual company percentage increase in revenue, versus row A, the CEO's annual percentage salary increase in that same comparty Suppose a random sample of companies yielded me following data A: Percan increase for CEO B Percent increase for company 12 18 14 18 6 4 21 37 18.28 22 14 4 19 15 30 - 5 Do these data indicate that the population mean porcentage increase in corporale revenue (row B) is different Bom the population mean percentage increase in CEO salary? Use a 5% level of significance a Hypotheses Me - MÓTwo tail A 14 21 Sample test statistic 12 18 18 28 12-18--6 18-28--10 18 14 18-14-4 4 6.4=10 19 4-192-15 37 30 37-30-7 15 21-15-6 14-22-8 d d-d/n - 12/8 = -1.5(n = number of pairs = 8 and d-6 - 10 - 8+4 + 10 - 15+6+7--12) Standard deviation of d: 59.3197 (See chapter 3 notes. How to find sample mean and sample standard deviation of -8.-10.-8..,7) Sample test statistic - = -0.455 (STAT, TESTS, TTest Type x = d-1.5, 8 - 9.3197, n-8.4) Find the P-value: 0.6827 Since the P-value, we fail to reject H. Conclusion in the context of the application: Al the 5% level of significance the evidence is insufficient to claim a difference in population mean percentage increases for corporale revenue and CEO salary

Answer 1

Differei ce = Sample Sample Sample 2 1 1 - Sample 2 12 18 -6 18 28 -10 14 22 -8 18 14 4. 6 4. 2 4 19 -15 21 15 6 37 30 7 Average 16.25 18.75 -2.5 St. Dev. 10.264 8.259 8.281 n 8 8 8

From the sample data, it is found that the corresponding sample means are: X1 = 16.25 X2 = 18.75 Also, the provided sample standard deviations are: $1 = 10.264 S2 = 8.259 and the sample size is n = 8. For the score differences we have D = -2.5 SD = 8.281 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: up = 0 Ha: MD 0 This corresponds to a two-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 7. Hence, it is found that the critical value for this two-tailed test is tc = 2.365, for a = 0.05 and df = 7. The rejection region for this two-tailed test is R= {t: 1t| > 2.365}. (3) Test Statistics The t-statistic is computed as shown in the following formula: Õ t = -2.5 8.281/78 = -0.854 sp/vn (4) Decision about the null hypothesis = Since it is observed that t| 0.854 <tc 2.365, it is then concluded that the null hypothesis is not rejected. = Using the P-value approach: The p-value is p = 0.4214, and since p 0.4214 > 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean ui is different than um, at the 0.05 significance level. Confidence Interval The 95% confidence interval is -9.423 MD < 4.423. Graphically. p-value = Test 0.42141098676077 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Thus there is insufficient evidence to conclude that there is difference in population mean percent increase for corporate revenue and CEO salary.

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