In order to determine whether or not there is a significant difference between the mean hourly wages paid by two companies (of the same industry), the following data have been accumulated.

company a

company b

sample size

80

60

sample mean

$16.75

$16.25

population standard deviation

$1.00

$.95



the test statistic is

a.

1.645.

b.

3.01.

c.

2.75.

d.

.098.

Test statistic = (mean 1 - mean 2) / sqrt( dev1/n1 + dev2/n2)

Mean 1 = $16.75

mean2 = $16.25

dev1 = $1.00

dev2 = $0.90

n1 = 80

n2 = 60

Test stat = (16.75 - 16.25) / sqrt(1/80 + 0.95/60)

Test stat = 3.01

The answer is B. 3.01

answer: 40.95

step-by-step explanation:

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497