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 sa
mple 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 = ?
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
Suppose x-6 is listed as a possible answer. That is zero if x = 6. So put x=6 into the original x^2 + 4x - 60 to get 6^2 + 4*6 - 60 = 36 + 24 -60 = 0. hence x-6 is a factor.
Answer (x-6)
on this hand :x-5 is not a factor, because plugging x=5 into does not work.
The coordinates are plotted and connected from one point to the other to see the triangle form. At first, it looks like an isosceles triangle, but after getting the distances, it is a SCALENE triangle.
