Question

A drapery store manager was interested in determining whether a new employee can install vertical blinds faster than an employee who has been with the company for two years. The manager takes independent samples of 10 records of installations times of vertical blinds of each of the two employees and computes the following information. Test whether the new employee installs vertical blinds faster, on the average, than the veteran employee, at the level of significance 0.05.
New Employee Veteran Employee
Sample Size 10 10
Sample Mean 22.2 min 24.8 min
Standard Deviation 0.90 min 0.75 min

Required:
What is the appropriate conclusion for this test?

Answer 1

Answer:

we arrive at the conclusion that the new employee installs the vertical blinds faster on average.

Step-by-step explanation:

null hypothesis; m₁-m₂ =0

alternative hypothesis; m₁-m₂<0

Given the information in this question, we first have to solve for the t statistic

$t=\frac{22-24.8}{\sqrt{\frac{0.90^{2} }{10}+\frac{0.75^{2} }{10} } }$

$t=\frac{-2.6}{\sqrt{o.081+0.05625} }$

$t=\frac{-2.6}{0.37047}$

t = -7.018

the degree of freedom = n₁+n₂-2

= 10+10-2

= 18

Alpha α = 0.05

from these results the t critical value = -1.734

because the test statistic -7.018 < -1.734,

at 0.05 level of testing, we arrive at the conclusion that the new employee installs the vertical blinds faster on average than the veteran.