Download Posting: 6998

From the table we can find that of the Employee Pre-Test the mean, median and
mode are almost the same, so we can say that all of them can be used to describe the
data set. But usually we use Mean=69.04.
However, for the Employee Post-Test, the mean, median and mode are quite
different form each other, since mean and mode are greatly affected by some unusual
extreme figures, we have to apply Median=74 to describe the data set.
And of the Strategy-Formulation Group, mean, median and mode are exactly
the same, so we can say that all of them can be used to describe the data set. Still we
use Mean=88.
6) Use a hypothesis test to determine whether statistically significant improvement
has occurred between the two tests.
We are to test the difference between means. And we use 95% level of confidence.
Let the difference between means be Md=M2-M1, then:
Ho: M1 = M2 or Md=M2-M1=0
H1: M1 <> M2 or Md=M2-M1<>0
for Employee Pre-Test: M1=69.04, n1=200, Standard Deviation is S1=3.310
for Employee Post-Test: M2=76.11, n2=200, Standard Deviation is S2=7.704
then Md=76.11-69.04=7.07
And the standard error of difference between mean is
Smd=SQRT((S1^2+S2^2)/n)= SQRT((3.310^2+7.704^2)/200)=0.593
Then t^=(Md-0)/Smd=7.07/0.593=11.92
This is a two-tailed test, so we should find the 97.5% confidence level t value(with
2.5% level of significance on either side of the t distribution). And the degree of
freedom is df=n1+n2-2=200+200-2=398
From the t-table, we can find the t*(97.5%, 398)=1.960
Since t^ > t*, we have to reject Ho that there’s no difference between means at
95% level of confidence.
Therefore, we should say that statistically significant improvement has occurred
between the two tests at 95% level of confidence.
7) The CEO would like the mean score for all employees to be no more than ten
percentage points lower than the mean score of the strategy-formulation group.
Given the CEO’s goals, should we conduct further training?
Again, this is a test of the difference between means. And we still use 95% level
of confidence.
First we should find the mean score for all employees “Me”, which can be
calculated by:
Me=(M1*n1+M2*n2)/(n1+n2)=(69.04*200+76.11*200)/(200+200)=72.58
And the standard deviation: Se=6.898 (calculation can be referred to the attached
EXCEL file.)
While the mean score of the strategy-formulation group is Ms=88
Again, let the difference between means be Md=Ms-Me, then we are to test
whether the difference between means is greater than 10 percentage points:
Ho: Md=Ms-Me=10
H1: Md=Ms-Me>10
Here, for all employees, Me=72.58, Ne=400, Se=6.898
for strategy-formulation group: Ms=88, n2=25, Ss=2.041
then Md=88-72.58=15.42
And the estimated standard error,
, when Ne<>Ns, the two values of variance
were simply averaged as they are for equal sample sizes, then the estimate based on
the smaller sample size would count as much as the estimate based on the larger
sample size. The formula for MSE is:
MSE = SSE/df
where df is the degrees of freedom (n1 - 1 + n2 - 1):
df=400-1+25-1=423
and SSE is: SSE = SSEe + SSEs
SSEe is actually the variation of Xe.
So SSEe = Se2(Ne-1)=6.898^2*(400-1)=18985.38
Similarly, SSEs= Ss2(Ns-1)= 2.041^2*(25-1)= 99.98
So SSE= SSEe + SSEs=18985.38+99.98=19085.36
Then MSE = SSE/df=19085.36/423=45.12
We should use the harmonic mean of the two sample sizes. The harmonic mean (nh)
of n1 and n2 is:
=2/(1/400+1/25)=47.06
The formula for the estimated standard error of the difference between means
becomes:
Smd=SQRT(2*45.12/47.06)=1.385
Then t^=(Md-10)/Smd=(15.42-10)/1.385=3.913
This is a one-tailed test, so we should find the 95% confidence level t value. And the
degree of freedom is df=423
From the t-table, we can find the t*(95%, 423)=1.645
Since t^ < t*, we fail to reject Ho that the difference between means is just 10
percentage points at 95% level of confidence.
Therefore, we should say that the mean score for all employees is no more than ten
percentage points lower than the mean score of the strategy-formulation group at 95%
level of confidence. So we needn’t conduct further training.