In: Statistics and Probability
4. The following are 20 observations of response times (in seconds) from a random sample of participants on a cognitive psychology task:
1.7 0.8 4.3 2.9 2.3 1.1 2.2 1.8 2.0 1.2 4.4 1.6 3.8 1.5 2.8
3.3 1.8 2.5 2.7 1.6
(a) Calculate the mean and unbiased standard deviation of the response times.
(b) Construct and interpret a 95% confidence interval for μ, which is the true mean response time. Assume that σ = 1.5.
(c) Construct and interpret a 90% confidence interval for μ, which is the true mean response time. Assume that σ is unknown.
(d) What would happen to the width of your confidence interval if our sample size increased? Explain.
a)
Sample Mean,    x̅ = ΣX/n =    2.3150
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) )
=   1.0246
b)
Level of Significance ,    α =   
0.05          
'   '   '      
   
z value=   z α/2=   1.960   [Excel
formula =NORMSINV(α/2) ]      
          
       
Standard Error , SE = σ/√n =   1.5000   /
√   20   =   0.3354
margin of error, E=Z*SE =   1.9600  
*   0.3354   =   0.6574
          
       
confidence interval is       
           
Interval Lower Limit = x̅ - E =    2.32  
-   0.657392   =   1.6576
Interval Upper Limit = x̅ + E =    2.32  
-   0.657392   =   2.9724
95%   confidence interval is (  
1.6576   < µ <   2.9724  
)
c)
Level of Significance ,    α =   
0.1          
degree of freedom=   DF=n-1=   19  
       
't value='   tα/2=   1.729   [Excel
formula =t.inv(α/2,df) ]      
          
       
Standard Error , SE = s/√n =   1.0246   /
√   20   =   0.2291
margin of error , E=t*SE =   1.7291  
*   0.2291   =   0.3961
          
       
confidence interval is       
           
Interval Lower Limit = x̅ - E =    2.32  
-   0.396149   =   1.9189
Interval Upper Limit = x̅ + E =    2.32  
-   0.396149   =   2.7111
90%   confidence interval is (  
1.9189   < µ <   2.7111  
)
d)
width of your confidence interval will get decreased if our sample size increased