A politician claims that the mean salary for managers in his state is more than national mean $82,000. The salaries (in dollars) for a random sample of 30 managers in the state are listed. At α = 0.07, is there enough evidence to support the claim?
86,809
|
72290
|
81862
|
79764
|
79511
|
75716
|
72039
|
83358
|
90425
|
95658
|
85221
|
72387
|
85468
|
76680
|
76529
|
98304
|
97118
|
74618
|
90788
|
83777
|
99303
|
92608
|
79489
|
72045
|
79296
|
90872
|
72865
|
82948
|
84042
|
93539
|
Solution:
a) Identification of Null and alternative hypotheses:
Null hypothesis:
H0 The mean salary for managers in the state is not more than $82000
H0 µ ≤ 82000
alternative hypothesis:
H0 The mean salary for managers in the state is more than $82000
H0 µ > 82000
Level of significance (α) = 0.07 One tailed test
Assumptions:
Population is normally distributed.
Data Information: In the above data we find
n (Sample size) = 30
σ (Population standard deviation) = 8315.90
X‾ (Sample mean) = 83410.97
b) Calculation of Test Statistic:
![403_distributed value.png](https://www.expertsmind.com/CMSImages/403_distributed%20value.png)
c) Finding P value
![179_finding p value.png](https://www.expertsmind.com/CMSImages/179_finding%20p%20value.png)
P value = P (Z > Z observed)
= P (Z > 0.93)
= 1 - P(Z < 0.93)
= 1- 0.8238
= 0.1762
d) Decision Rule:
Reject H0 if P value is less than the level of significance.
Decision:
Since P value (observed level of significance) = 0.1762 is greater than α (level of significance) = 0.07, we fail to reject H0.