Identification of null and alternative hypotheses, Basic Statistics

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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

c) Finding P value

  179_finding p value.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.


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