Two random samples one of 15 Los Angeles area commuters and

Two random samples, one of 15 Los Angeles area commuters and another of 15 San Francisco Bay area commuters were independently chosen. The following table shows the sample average and sample standard deviation of daily commuting miles for each sample group.

Los Angeles commuters x1 = 57.4 s1 = 12.4

San Francisco commuters x2 = 52.8 s2 = 13.8

Assume that the populations of commuting miles driven by the commuters in these two cities are normal populations with equal variances ² . Find both a 90% and a 95% confidence interval for the difference in mean commuting miles between the two populations. What conclusion do you draw?

Solution

X (Mean)=57.4; Standard Deviation (s.d1)=12.4
Number(n1)=15
Y(Mean)= 52.8; Standard Deviation(s.d2)=13.8
Number(n2)=15

WHEN SD ARE EQUAL, AT 0.1 LOS              
              
CI = (x1 - x2) ± t a/2 * S^2 * Sqrt ( 1 / n1 + 1 / n2 )              
Where,               
x1 = Mean of Sample 1, x2 = Mean of sample2              
sd1 = SD of Sample 1, sd2 = SD of sample2              
a = 1 - (Confidence Level/150)              
ta/2 = t-table value              
CI = Confidence Interval               
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (14*153.76 + 14*190.44) / (30- 2 )
S^2 = 172.1
S = Sqrt(172.1)   = 13.119
t with (n1+n2-2) i.e 28 d.f is 1.701          
CI = [ ( 57.4- 52.8) ±t a/2 * S * Sqrt( 1/15 + 1/15)]              
= [ ( 57.4- 52.8) ± t a/2 * 2.898 * Sqrt( 1/15 + 1/15 ) ]              
= [ 4.6 ± (1.701 * 13.119 * Sqrt( 1/15 + 1/15 )) ]              
= [-3.548 , 12.748]              
              
WHEN SD ARE EQUAL, AT 0.05 LOS  
t with (n1+n2-2) i.e 28 d.f is 2.048
CI = [ ( 57.4- 52.8) ±t a/2 * S * Sqrt( 1/15 + 1/15)]              
= [ ( 57.4- 52.8) ± t a/2 * 2.898 * Sqrt( 1/15 + 1/15 ) ]              
= [ 4.6 ± (2.048 * 13.119 * Sqrt( 1/15 + 1/15 )) ]              
= [-5.2107 , 14.41]