Question

In: Statistics and Probability

1(a) Using a driving simulator, a human factors psychologist measured the time to press the brake...

1(a) Using a driving simulator, a human factors psychologist measured the time to press the brake pedal after a visual alert for younger and older drivers. The mean reaction time for a sample of 20 young drivers was 2.8 sec with a standard deviation of .3 sec, and 3.2 sec for a sample of 18 older drivers with a standard deviation of .4. Compute a 99% confidence interval for the difference between the population means for the two groups using the t-distribution. What is the lower limit?

(b) Using a driving simulator, a human factors psychologist measured the time to press the brake pedal after a visual alert for younger and older drivers. The mean reaction time for a sample of 20 younger drivers was 2.8 sec with a standard deviation of .3 sec, and 3.2 sec for a sample of 18 older drivers with a standard deviation of .4. Compute a 95% confidence interval for the difference between the population means for the two groups using the t-distribution. What is the lower limit?

Solutions

Expert Solution

a)

Level of Significance ,    α =    0.01
Degree of freedom, DF=   n1+n2-2 =    36              
t-critical value =    t α/2 =    2.7195   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    0.3508              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.1140              
margin of error, E = t*SE =    2.7195   *   0.11   =   0.31  
                      
difference of means =    x̅1-x̅2 =    2.8000   -   3.200   =   -0.4000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -0.4000   -   0.3099   =   -0.7099

b)

α=0.05

Degree of freedom, DF=   n1+n2-2 =    36              
t-critical value =    t α/2 =    2.0281   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    0.3508              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.1140              
margin of error, E = t*SE =    2.0281   *   0.11   =   0.23  
                      
difference of means =    x̅1-x̅2 =    2.8000   -   3.200   =   -0.4000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -0.4000   -   0.2311   =   -0.6311

please revert for doubt..


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