Question

#6. The level of lead in the blood was determined for a sample of 152 male and 86 female hazardous- waste workers resulting in mean t standard error of 5.5 t 0.3 for the men and 3.8 ± 0.2 for the women. (a) Provide a 92% CI on the difference between the true average blood levels for male and female workers (b)Is there sufficient evidence that the true average blood level for male workers exceeds that of female workers by more than 2 units Give a relevant P-value

Answer 1

a)

Sample #1   ---->   male           
mean of sample 1,    x̅1=   5.500000          
size of sample 1,    n1=   152          
                  
Sample #2   ---->   female          
mean of sample 2,    x̅2=   3.800          
size of sample 2,    n2=   86          
                  
difference in sample means = x̅1-x̅2 =    5.500   -   3.8000   =   1.7000
                  
std error , SE =    √(s1²/n1+s2²/n2) =√( 0.3²+0.2² ) = 0.3606          

Level of Significance ,    α =    0.08
DF = min(n1-1 , n2-1 )=   85          
t-critical value , t* =    1.7719   (excel formula =t.inv(α/2,df)      
              
margin of error, E = t*SE =    1.7719   *   0.361   =   0.6389
                  
difference of means = x̅1-x̅2 =    5.5000   -   3.800   =   1.7000
confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    1.7000   -   0.6389   =   1.0611
Interval Upper Limit = (x̅1-x̅2) + E =    1.7000   -   0.6389   =   2.3389

b)

Ho :   µ1 - µ2 =   2
Ha :   µ1-µ2 >   2

t-statistic = ((x̅1-x̅2)-µd)/SE = ( (1.7000 -2) /   0.3606   ) =   -0.8321

DF = min(n1-1 , n2-1 )=   85  
      
      
p-value =    0.7961    [excel function: =T.DIST.RT(t stat,df) ]
Decision:   p value >α , Do not reject Ho