Question

A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 31.52 years and the standard deviation is 10.24 years. ​a) Construct a 95​% confidence interval for the mean age of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be 11.0 ​years? 49 46 11 30 33 40 49 40 30 35 23 34 26 33 48 34 31 22 32 20 21 23 18 22 38

Answer 1

a)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   10.24
Sample Size ,   n =    25
Sample Mean,    x̅ = ΣX/n =    31.52

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   24          
't value='   tα/2=   2.0639   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   10.24 / √   25   =   2.0479
margin of error , E=t*SE =   2.0639   *   2.048   =   4.227
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    31.52   -   4.227   =   27.2934
Interval Upper Limit = x̅ + E =    31.52   -   4.227   =   35.7466
95%   confidence interval is (   27.29   < µ <   35.75   )

b)

margin of error , E=t*SE =   2.0639   *   2.048   =   4.227

c)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   11.000   / √   25   =   2.2000
margin of error, E=Z*SE =   1.9600   *   2.200   =   4.312
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    31.52   -   4.312   =   27.2081
Interval Upper Limit = x̅ + E =    31.52   -   4.312   =   35.8319
95%   confidence interval is (   27.21   < µ <   35.83   )