Question

Using techniques from an earlier section, we can find a confidence interval for μ_d. Consider a random sample of n matched data pairs A, B. Let d = B − A be a random variable representing the difference between the values in a matched data pair. Compute the sample mean d of the differences and the sample standard deviation s_d. If d has a normal distribution or is mound-shaped, or if n ≥ 30, then a confidence interval for μ_d is as follows. d − E < μ_d < d + E , where

E = t_c

s_d

n

c = confidence level (0 < c < 1)

t_c = critical value for confidence level c and d.f. = n − 1

B: Percent increase for company	16	16	30	18	6	4	21	37
A: Percent increase for CEO	22	27	19	14	−4	19	15	30

(a) Using the data above, find a 95% confidence interval for the mean difference between percentage increase in company revenue and percentage increase in CEO salary. (Round your answers to two decimal places.)

lower limit	=
upper limit	=

Answer 1

sample mean, xbar = 0.75
sample standard deviation, s = 9.9964
sample size, n = 8
degrees of freedom, df = n - 1 = 7

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.365

ME = tc * s/sqrt(n)
ME = 2.365 * 9.9964/sqrt(8)
ME = 8.359

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (0.75 - 2.365 * 9.9964/sqrt(8) , 0.75 + 2.365 * 9.9964/sqrt(8))
CI = (-7.61 , 9.11)

lower limit     = -7.61
upper limit     = 9.11