Question

The mean salary of federal government employees on the General Schedule is $59,593. The average salary of 30 state employees who do similar work is $58,800 with \sigmaσσ= $1500. At the 0.01 level of significance, can it be concluded that state employees earn on average less than federal employees?

Answer 1

`z=(58800-59593)/((1500)/(√(30)))=-2.896`

`p_v =P(Z<-2.896)=0.00189`

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average for the employes earn is less than 59593$ at 1% of significance.

Step-by-step explanation:

1) Data given and notation

`\bar X=58800` represent the sample mean

`\sigma=1500` represent the population standard deviation

`n=30` sample size

`\mu_o =59593` represent the value that we want to test

`\alpha=0.01` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 59593, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 59593`

Alternative hypothesis:
`\mu < 59593`

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

`z=(58800-59593)/((1500)/(√(30)))=-2.896`

4)P-value

Since is a left tailed test the p value would be:

`p_v =P(Z<-2.896)=0.00189`

5) Conclusion

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average for the employes earn is less than 59593$ at 1% of significance.