In our industry on average a firm paid out record year end bonuses of $125,500 per employee for 2008. we would like to take a sample of employees at our firm to see whether the mean year end bonus is different from the reported mean of $125,500 for the population. we sample 40 of our employees and we had a mean of $118,000. assume a population standard deviation of $30,000. setup for the p-value and critical value approaches. we want to be 95% confident about our finding.

if the answer could be broken down on how they were solved, that would be great.

1. sample mean =
2. population mean = ?
3. population standard deviation = ?
4. sample size = ?
5. alpha error = ?
6. confidence coefficient = ?
7. calculate test statistic = ?
8. rejection region critical value = ?

Step-by-step explanation:

Given that in our industry on average a firm paid out record year end bonuses of $125,500 per employee for 2008. We would like to take a sample of employees at our firm to see whether the mean year end bonus is different from the reported mean of $125,500 for the population.

1. Sample Mean = 118000

2. Population Mean = 125500

3. Population Standard Deviation = 30000

4. Sample size = 40

5. Alpha Error = 5%

6. Confidence Coefficient = 95%

7. Calculate Test Statistic = Mean difference /s td error = -1.581

8. Rejection Region Critical Value = If Z statistic is below -1.96 or above 1.96

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18