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:

 

c) Finding P value

 

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.

Posted Date: 3/16/2013 3:09:55 AM | Location : United States

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