Question

A random sample of 23 items is drawn from a population whose standard deviation is unknown. The sample mean is x⎯⎯x¯ = 820 and the sample standard deviation is s = 25. Use Excel to find your answers.

(a)	Construct an interval estimate of μ with 99% confidence. (Round your answers to 3 decimal places.)

The 99% confidence interval is from  to

(b)	Construct an interval estimate of μ with 99% confidence, assuming that s = 50. (Round your answers to 3 decimal places.)

The 99% confidence interval is from  to

(c)	Construct an interval estimate of μ with 99% confidence, assuming that s = 100. (Round your answers to 3 decimal places.)

The 99% confidence interval is from  to

(d)	Describe how the confidence interval changes as s increases.
	The interval stays the same as s increases. The interval gets wider as s increases. The interval gets narrower as s increases.

Answer 1

a)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   22          
't value='   tα/2=   2.8188   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   25.0000   / √   23   =   5.2129
margin of error , E=t*SE =   2.8188   *   5.2129   =   14.6938
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    820.00   -   14.693782   =   805.3062
Interval Upper Limit = x̅ + E =    820.00   -   14.693782   =   834.6938
99%   confidence interval is (   805.306   < µ <   834.694   )

b)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   22          
't value='   tα/2=   2.8188   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   50.0000   / √   23   =   10.4257
margin of error , E=t*SE =   2.8188   *   10.4257   =   29.3876
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    820.00   -   29.387563   =   790.6124
Interval Upper Limit = x̅ + E =    820.00   -   29.387563   =   849.3876
99%   confidence interval is (   790.612   < µ <   849.388   )

c)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   22          
't value='   tα/2=   2.8188   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   100.0000   / √   23   =   20.8514
margin of error , E=t*SE =   2.8188   *   20.8514   =   58.7751
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    820.00   -   58.775127   =   761.2249
Interval Upper Limit = x̅ + E =    820.00   -   58.775127   =   878.7751
99%   confidence interval is (   761.225   < µ <   878.775   )

d)

The interval gets wider as s increases.