Question

Part C only:

Two catalysts are being considered for a chemical process. Sixteen batches were completed using Catalyst 1, providing an average yield of x1=84 and a standard deviation of s1=1.87. Thirteen batches were run with Catalyst 2, and a mean yield of x2=89 was found, along with s2=1.48. Assume the underlying yield data are Normally distributed.

1. Is it believable that the variances of the two populations differ? Use the appropriate hypothesis test with α = 0.05.
2. Based on your answer to part (a), calculate the appropriate 99% CI to test if the mean yields for the two catalysts differ. Give your conclusions with context.
3. If 4 batches made with Catalyst 1 were found to be below the yield specification, and 3 batches made using Catalyst 2 were below specification, calculate the 95% CI to determine if the proportion of batches that fall under specification is the same for the two Catalysts.

Answer 1

a.
Given that,
sample 1
s1^2=3.4969, n1 =16
sample 2
s2^2 =2.1904, n2 =13
null, Ho: σ^2 = σ^2
alternate, H1: σ^2 != σ^2
level of significance, α = 0.05
from standard normal table, two tailed f α/2 =3.177
since our test is two-tailed
reject Ho, if F o < -3.177 OR if F o > 3.177
we use test statistic fo = s1^1/ s2^2 =3.4969/2.1904 = 1.596
| fo | =1.596
critical value
the value of |f α| at los 0.05 with d.f f(n1-1,n2-1)=f(15,12) is 3.177
we got |fo| =1.596 & | f α | =3.177
make decision
hence value of |fo | < | f α | and here we do not reject Ho
ANSWERS
---------------
null, Ho: σ^2 = σ^2
alternate, H1: σ^2 != σ^2
test statistic: 1.596
critical value: -3.177 , 3.177
decision: do not reject Ho
we do not have enough evidence to support the claim that it is believable that the variances of the two populations differ
b.
TRADITIONAL METHOD
given that,
mean(x)=84
standard deviation , s.d1=1.87
number(n1)=16
y(mean)=89
standard deviation, s.d2 =1.48
number(n2)=13
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((3.497/16)+(2.19/13))
= 0.622
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 12 d.f is 2.179
margin of error = 2.179 * 0.622
= 1.356
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (84-89) ± 1.356 ]
= [-6.356 , -3.644]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=84
standard deviation , s.d1=1.87
sample size, n1=16
y(mean)=89
standard deviation, s.d2 =1.48
sample size,n2 =13
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 = [( 84-89) ± t a/2 * sqrt((3.497/16)+(2.19/13)]
= [ (-5) ± t a/2 * 0.622]
= [-6.356 , -3.644]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-6.356 , -3.644] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
c.
TRADITIONAL METHOD
given that,
sample one, x1 =4, n1 =16, p1= x1/n1=0.25
sample two, x2 =3, n2 =13, p2= x2/n2=0.231
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.25*0.75/16) +(0.231 * 0.769/13))
=0.159
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.159
=0.312
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.25-0.231) ±0.312]
= [ -0.293 , 0.331]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =4, n1 =16, p1= x1/n1=0.25
sample two, x2 =3, n2 =13, p2= x2/n2=0.231
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.25-0.231) ± 1.96 * 0.159]
= [ -0.293 , 0.331 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ -0.293 , 0.331] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean P1-P2