Question

3. 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.

3. 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 = X1 = Location 1 Location 2 Victoria Edmonton 16 16 19.82 pounds per year 25.06 pounds per year $1 = 5 pounds per year S2 10 pounds per year n2 = X2 (a) [2 marks] Construct a 90% confidence interval for Mi hazardous waste between Victoria and Edmonton. M2, the true average difference in

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 =    19.82-25.06=   -5.2400

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 =    1.697*2.7951=   4.7440
      
difference in sample means = x̅1 - x̅2 =    19.82-25.06=   -5.2400
confidence interval is       
Interval Lower Limit= (x̅1-x̅2) - E =    -5.24-4.744=   -9.9840
Interval Upper Limit= (x̅1-x̅2) + E =    -5.24+4.744=   -0.4960
CI ( -9.984 , -0.496)

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B)

0 DO NOT LIE WITHIN THE INTERVAL , SO REJECT Ho

it is not reasnalble to say that means are same

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c)

true mean= µ =   2.5  
hypothesized mean=µo =    0  
α=   0.05  
std dev,σ=   2.7951  
power= 1- ß =   0.95  
ß =    0.05  
δ=µ - µo =    2.5  
Zα/2=   1.9600  
Z (ß ) =    1.6449  
n = ( ( Z(ß)+Z(α) )*σ / δ )² =    ((1.645+1.96)*2.7951/2.5)^2=   16.24
so, sample size=   17  

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