In: Statistics and Probability
A Human Resources Director has created two different online training tutorials for newly hired employees. The employees must pass a test before they can work with clients. The director wants to know if one tutorial is more effective than the other. The director collects the following sample data for the test scores:
Sample Mean |
Sample Standard Deviation |
Sample Size |
|
Tutorial 1 |
80.7 |
7.8 |
41 |
Tutorial 2 |
75.3 |
8.1 |
41 |
Test at the 5% level (with 80 degrees of freedom) if tutorial 1 leads to higher mean scores than tutorial 2.
Given that,
mean(x)=80.7
standard deviation , s.d1=7.8
number(n1)=41
y(mean)=75.3
standard deviation, s.d2 =8.1
number(n2)=41
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.684
since our test is right-tailed
reject Ho, if to > 1.684
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =80.7-75.3/sqrt((60.84/41)+(65.61/41))
to =3.0749
| to | =3.0749
critical value
the value of |t α| with min (n1-1, n2-1) i.e 40 d.f is 1.684
we got |to| = 3.07487 & | t α | = 1.684
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:right tail - Ha : ( p > 3.0749 ) = 0.00189
hence value of p0.05 > 0.00189,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 3.0749
critical value: 1.684
decision: reject Ho
p-value: 0.00189
we have enough evidence to support the claim that if tutorial 1
leads to higher mean scores than tutorial 2.