Question

In her book Red Ink Behaviors, Jean Hollands reports on the assessment of leading Silicon Valley companies regarding a manager's lost time due to inappropriate behavior of employees. Consider the following independent random variables. The first variable x₁ measures manager's hours per week lost due to hot tempers, flaming e-mails, and general unproductive tensions.

x1:

1

3

8

2

2

4

10

The variable x₂ measures manager's hours per week lost due to disputes regarding technical workers' superior attitudes that their colleagues are "dumb and dispensable".

x2:

10

3

6

5

9

4

10

3

(i) Use a calculator with sample mean and sample standard deviation keys to calculate x₁, s₁, x₂, and s₂. (Round your answers to two decimal places.)

x1	=
s1	=
x2	=
s2	=

(ii) Does the information indicate that the population mean time lost due to hot tempers is different (either way) from population mean time lost due to disputes arising from technical workers' superior attitudes? Use α = 0.05. Assume that the two lost-time population distributions are mound-shaped and symmetric.
(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: μ₁ = μ₂; H₁: μ₁ ≠ μ₂H₀: μ₁ = μ₂; H₁: μ₁ > μ₂    H₀: μ₁ ≠ μ₂; H₁: μ₁ = μ₂H₀: μ₁ = μ₂; H₁: μ₁ < μ₂

(b) What sampling distribution will you use? What assumptions are you making?

The standard normal. We assume that both population distributions are approximately normal with known standard deviations.The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.    The Student's t. We assume that both population distributions are approximately normal with known standard deviations.The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.

What is the value of the sample test statistic? (Test the difference μ₁ − μ₂. Do not use rounded values. Round your final answer to three decimal places.)


(c) Find (or estimate) the P-value.

P-value > 0.5000.250 < P-value < 0.500    0.100 < P-value < 0.2500.050 < P-value < 0.1000.010 < P-value < 0.050P-value < 0.010

Answer 1

i) = 4.29

   s1 = 3.40

   = 6.25

   s2 = 3.01

ii) a) The level of significance is 0.05.

b) The student's t . We assume that both population distributions are approximately normal with unknown standard deviations.

The test statistic is

c)

     

    

P-value = 2 * P(T < -1.175)

            = 2 * 0.1314

           = 0.2628

0.250 < P-value < 0.500