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In a particular city, the average salary of secretaries is $29,000 per year. Secretaries at Company A claim that they are paid less than the city average. In a sample of 45 secretaries, their average salary was $25,500 per year with a standard deviation of $5,000.
What is the test value for this hypothesis?

Respuesta :

Answer:

Value of z will be -4.695

Step-by-step explanation:

We have given [tex]\bar{x}=$25500[/tex]

Mean [tex]\mu =$29000[/tex]

Standard deviation [tex]\sigma=$5000[/tex]

Sample size n = 45

We have to find the z for the hypothesis

We know that z is given by

[tex]z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}=\frac{25500-29000 }{\frac{5000 }{\sqrt{45}}}=-4.695[/tex]

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