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Dmitriy789 [7]
3 years ago
11

Question 6 (1.25 points)

Mathematics
1 answer:
Lisa [10]3 years ago
4 0

Answer:

Test statistic Z= 1.713

The calculated Z- value =  1.7130 < 2.576 at 0.01 level of significance

Null hypothesis is accepted

There is no difference between the  mean annual salary of all lawyers in a city is  different from $110,000

Step-by-step explanation:

<u><em>Step(i):-</em></u>

A researcher wants to test if the mean annual salary of all lawyers in a city is

different from $110,000

Mean of the Population  μ = $110,000

Sample size 'n' = 53

Mean of the sample x⁻ = $114,000.

standard deviation of the Population = $17,000,

Level of significance = 0.01

Null hypothesis :

There is no difference between the  mean annual salary of all lawyers in a city is  different from $110,000

H₀:  x⁻ =  μ

Alternative Hypothesis :  x⁻ ≠  μ

<u><em>Step(ii)</em></u>:-

Test statistic

                 Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }

                 Z = \frac{114000-110000}{\frac{17000}{\sqrt{53} } }

                Z =  1.7130

Tabulated value Z = 2.576 at 0.01 level of significance

The calculated Z- value =  1.7130 < 2.576 at 0.01 level of significance

Null hypothesis is accepted

There is no difference between the  mean annual salary of all lawyers in a city is  different from $110,000

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