In: Statistics and Probability
Associate’s Degree |
Bachelor’s Degree |
$10.75 |
$10.75 |
$11.25 |
$12.25 |
$11.05 |
$11.60 |
Associate’s Degree |
Bachelor’s Degree |
$10.00 |
$9.80 |
$9.70 |
$9.40 |
$11.00 |
$12.60 |
$10.10 |
$11.80 |
$9.40 |
$10.40 |
$9.20 |
$11.90 |
$11.10 |
$12.70 |
A social researcher is interested in understanding the effect of college education on wages. Listed below are the hourly wages for two groups of employees all working in the same position at a particular technology firm. The workers in one group have earned an associate’s degree while members of the other group hold at least a bachelor’s degree. Test the null hypothesis that there are no differences in mean wages by degrees held by employees. What assumptions about the data must be made to use this test? |
Ho : µ1 - µ2 = 0
Ha : µ1-µ2 ╪ 0
Level of Significance , α =
0.05
Sample #1 ----> 1
mean of sample 1, x̅1= 10.355
standard deviation of sample 1, s1 =
0.7657
size of sample 1, n1= 10
Sample #2 ----> 2
mean of sample 2, x̅2= 11.320
standard deviation of sample 2, s2 =
1.1653
size of sample 2, n2= 10
difference in sample means = x̅1-x̅2 =
10.3550 - 11.3 =
-0.97
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 0.9860
std error , SE = Sp*√(1/n1+1/n2) =
0.4409
t-statistic = ((x̅1-x̅2)-µd)/SE = ( -0.9650
- 0 ) / 0.44
= -2.189
Degree of freedom, DF= n1+n2-2 =
18
p-value =
0.0421 (excel function: =T.DIST.2T(t stat,df)
)
Conclusion: p-value <α , Reject null
hypothesis
There is enough evidence to conclude that there is a differences in
mean wages by degrees held by employees.
Assumption:
population of two samples have equal variance
two population are normally distributed