The director of a state agency believes that the average starting salary for clerical employees in the state is less than ​$35 comma 000 per year. To test her​ hypothesis, she has collected a simple random sample of 100 starting clerical salaries from across the state and found that the sample mean is ​$34 comma 700.a.   State the appropriate null and alternative hypothesis.b. Assuming the population standard deviation is known to be ​$3 comma 500 and the significance level for the test is to be 0.10​, what is the critical value​ (stated in​ dollars)

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Answer:

There is not enough evidence to conclude that  average starting salary for clerical employees in the state is less than ​$35,000 per year.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 35,000 dollars per year

Sample mean, [tex]\bar{x}[/tex] = 34,700 dollars per year

Sample size, n = 100

Alpha, α = 0.10

Population standard deviation, σ = 3,500 dollars per year

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 35,000\text{ dollars per year}\\H_A: \mu < 35,000\text{ dollars per year}[/tex]

We use one-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{34700 - 35000}{\frac{3500}{\sqrt{100}} } = -0.8571[/tex]

Now, [tex]z_{critical} \text{ at 0.10 level of significance } = -1.28[/tex]

Since,  

[tex]z_{stat} > z_{critical}[/tex]

We fail to reject the null hypothesis and accept the null hypothesis.

Thus, there is not enough evidence to conclude that  average starting salary for clerical employees in the state is less than ​$35,000 per year.

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