In: Statistics and Probability
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?
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..