Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. What is the planning value for the population standard deviation (to the nearest whole number)?

How large a sample should be taken if the desired margin of error is as shown below (to the nearest whole number)?

a. $500?
b. $200?
c. $100?

a) 217

b) 1351

c) 5403

Step-by-step explanation:

Given that:

confidence interval (c) = 0.95

The Z score of is from the z table is given as:

Range = $45000 - $30000 = $15000

The standard deviation (σ) is given as:

Sample size (n) is given as:

a) E = $500

≈ 217

b) ≈ 1351

c) ≈ 5403

this doesn't make sense lol

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068