Question

The following eight observations were drawn from a normal population whose variance is 100:

12

8

22

15

30

6

29

58

Part A

What is the standard error of the sample mean, based on the known population variance? Give your answer to two decimal places in the form x.xx

Standard error:

Part B

Find the lower and upper limits of a 90% confidence interval for the population mean. Give your answer to two decimal places in the form xx.xx

Lower limit:

Upper limit:

Part C

True or false:

If the population was not normally distributed the confidence interval calculation above would not be valid. Answer by writing T or F in the space provided.

Answer:

Answer 1

a)

σ = √100 = 10

Standard Error , SE = σ/√n =   10 / √    8   =   3.54

b)

Sample Size ,   n =    8
Sample Mean,    x̅ = ΣX/n =    22.5000
Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.6449   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   10.000   / √   8   =   3.5355
margin of error, E=Z*SE =   1.6449   *   3.536   =   5.815
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    22.50   -   5.815   =   16.685
Interval Upper Limit = x̅ + E =    22.50   -   5.815   =   28.315
90%   confidence interval is (   16.68   < µ <   28.32   )

c)

True - T