To compare the salaries of two firms A and B two independent
To compare the salaries of two firms, A and B, two independent random samples of employees were taken. The following represents the findings (in thousands).
Company
1
2
3
4
5
6
7
8
9
10
A
41.5
39
47
102
89
62.5
55
50
48
62
B
38
90
80
62
55
60
61
71
At the .01 significance level, determine whether there is a mean difference between the salaries A and B. (Assume that the two population variances are equal.)
| Company | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| A | 41.5 | 39 | 47 | 102 | 89 | 62.5 | 55 | 50 | 48 | 62 |
| B | 38 | 90 | 80 | 62 | 55 | 60 | 61 | 71 |
Solution
Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=59.6
Standard Deviation(s.d1)=20.642 ; Number(n1)=10
Y(Mean)=64.625
Standard Deviation(s.d2)=15.873; Number(n2)=8
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =59.6-64.625/Sqrt((426.09216/10)+(251.95213/8))
to =-0.58
| to | =0.58
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 7 d.f is 3.499
We got |to| = 0.58374 & | t | = 3.499
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -0.5837 ) = 0.578
Hence Value of P0.01 < 0.578,Here We Do not Reject Ho

