Question

Studies of hazardous wastes are conducted at two locations. Wastes of interest are those generated by households and small businesses. Examples of such wastes are used oil from automobiles, batteries, antifreeze, paint and paint thinner, and solvents. If needed, you may assume the amount of yearly hazardous waste generated follows a normal distribution. These data are obtained:

ni = - S1 = Location 1 Location 2 16 16 X1 22 19.82 pounds per year 25.06 pounds per year 5 pounds per year 10 pounds per year n2 = S2 =

(a) Construct a 90% confidence interval for µ1 − µ2, the true average difference in hazardous waste between Location 1 and Location 2.

(b) Using your confidence interval from part (a), is it reasonable to say the average level of hazardous waste is the same for Location 1 and Location 2 ? Explain.

(c) Using the data from the setup as a pilot study, find the common sample size needed (that is, n1 = n2 = n) to estimate the true average difference in hazardous waste between Location 1 and Location 2 to within 2.5 pounds per year with 95% confidence.

Answer 1

a)

Sample #1   ---->   1                  
mean of sample 1,    x̅1=   19.820                  
standard deviation of sample 1,   s1 =    5.000                  
size of sample 1,    n1=   16                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   25.060                  
standard deviation of sample 2,   s2 =    10.000                  
size of sample 2,    n2=   16                  
                          
difference in sample means =    x̅1-x̅2 =    19.8200   -   25.1   =   -5.24  

Degree of freedom, DF=   n1+n2-2 =    30              
t-critical value =    t α/2 =    1.6973   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    7.9057              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    2.7951              
margin of error, E = t*SE =    1.6973   *   2.80   =   4.74  
                      
difference of means =    x̅1-x̅2 =    19.8200   -   25.060   =   -5.2400
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -5.2400   -   4.7440   =   -9.9840
Interval Upper Limit=   (x̅1-x̅2) + E =    -5.2400   +   4.7440   =   -0.4960

..........

B)

0 does not lie in the interval, so do not reject Ho

so, it is not reasonable to say the average level of hazardous waste is the same for Location 1 and Location 2

........

Please let me know in case of any doubt.

Thanks in advance!

Please upvote!