Business Statistics Class PROBLEM 1 Who misses work more oft

Business Statistics Class

PROBLEM 1

Who misses work more often at the ABC Company: Women or Men? Are they different? Test at .06 significance level. Women: Average number of days absent = 18; standard deviation = 2.0; n = 40 Non-Smokers: Average number of days absent = 19; standard deviation = 4.0; n = 50;

Solution

Set Up Hypothesis
Null, There Is No-Significance between them Ho: u1=> u2
Alternate,Men miss more often than Women- H1: u1 < u2
Test Statistic
X(Mean)=18
Standard Deviation(s.d1)=2 ; Number(n1)=40
Y(Mean)=19
Standard Deviation(s.d2)=4; Number(n2)=50
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =18-19/Sqrt((4/40)+(16/50))
to =-1.54
| to | =1.54
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 39 d.f is 1.59
We got |to| = 1.54303 & | t | = 1.59
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Left Tail - Ha : ( P < -1.543 ) = 0.06545
Hence Value of P0.06 < 0.06545,Here We Do not Reject Ho

We have evidence to indicate the women misses more ofeten OR
There is no significanc between Men and women