Question

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant at around 2.1 years. A survey of 44 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.2 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level

e) What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round your answers to three decimal places.)

i) Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to two decimal places.)

Answer 1

e)

Ho :   µ =   19  
Ha :   µ <   19   (Left tail test)
          
Level of Significance ,    α =    0.050  
population std dev ,    σ =    2.1000  
Sample Size ,   n =    44  
Sample Mean,    x̅ =   18.2000  
          
'   '   '  
          
Standard Error , SE = σ/√n =   2.1/√44=   0.3166  
Z-test statistic= (x̅ - µ )/SE =    (18.2-19)/0.3166=   -2.53
          
critical z value, z* =       -1.6449   [Excel formula =NORMSINV(α/no. of tails) ]
          
p-Value   =   0.0058   [ Excel formula =NORMSDIST(z) ]
Decision:   p-value<α, Reject null hypothesis       

i)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.960   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   2.1/√44=   0.3166          
margin of error, E=Z*SE =   1.9600   *   0.3166   =   0.6205
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    18.20   -   0.620499   =   17.5795
Interval Upper Limit = x̅ + E =    18.20   -   0.620499   =   18.8205
95%   confidence interval is (   17.58   < µ <   18.82   )