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. These data are obtained: Location 1 Victoria ni = 16 l = 19.82 pounds per year 81 = 5 pounds per year Location 2 Edmonton n2 = 16 15 = 25.06 pounds peż year 82 = 10 pounds per year (a) [2 marks) Construct a 90% confidence interval for M1 – 12, the true average difference in hazardous waste between Victoria and Edmonton.

Answer 1

3.

a.
TRADITIONAL METHOD
given that,
mean(x)=19.82
standard deviation , s.d1=5
number(n1)=16
y(mean)=25.06
standard deviation, s.d2 =10
number(n2)=16
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((25/16)+(100/16))
= 2.795
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 15 d.f is 1.753
margin of error = 1.753 * 2.795
= 4.9
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (19.82-25.06) ± 4.9 ]
= [-10.14 , -0.34]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=19.82
standard deviation , s.d1=5
sample size, n1=16
y(mean)=25.06
standard deviation, s.d2 =10
sample size,n2 =16
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 19.82-25.06) ± t a/2 * sqrt((25/16)+(100/16)]
= [ (-5.24) ± t a/2 * 2.795]
= [-10.14 , -0.34]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [-10.14 , -0.34] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
b.
Given that,
mean(x)=19.82
standard deviation , s.d1=5
number(n1)=16
y(mean)=25.06
standard deviation, s.d2 =10
number(n2)=16
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.753
since our test is two-tailed
reject Ho, if to < -1.753 OR if to > 1.753
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =19.82-25.06/sqrt((25/16)+(100/16))
to =-1.875
| to | =1.875
critical value
the value of |t α| with min (n1-1, n2-1) i.e 15 d.f is 1.753
we got |to| = 1.87472 & | t α | = 1.753
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.8747 ) = 0.08
hence value of p0.1 > 0.08,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.875
critical value: -1.753 , 1.753
decision: reject Ho
p-value: 0.08
we have enough evidence to support the claim that average level of hazardous waste is the same for the Victoria and Edmonton.
c.
Given data,
margin of error =2.5
standard deviation for =10
confidence level is 95%
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 10
ME =2.5
n = ( 1.96*10/2.5) ^2
= (19.6/2.5 ) ^2
= 61.47 ~ 62