In semiconductor manufacturing, wet chemical etching is often used to remove sil

Question

In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the and known to follow a normal distribution. Two different etching solutions have been compared using two random samples of 10 wafers for each solution. The observed etch rates are as follows (in mils per minute: Solution 1 9.99.5 9.0 9.6 10.2 10.6 10.3 10 10.3 10.1 Solution 2 10.2 10 10.6 10.2 10.7 10.7 10.410.4 10.5 10.3 Test to see if the etch rates for the two solutions have similar variations. Answer the following questions about testing to see if there is a difference in the average etch rates for the two solutions. Why should you use a two sample method instead of paired difference? Should you use pooled or not pooled standard deviation? Explain Calculate and interpret a 95% confidence interval for the difference in average etch rates for the two solutions

Explanation / Answer

Given that,
mean(x)=9.95
standard deviation , s.d1=0.4696
number(n1)=10
y(mean)=10.4
standard deviation, s.d2 =0.2309
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =9.95-10.4/sqrt((0.22052/10)+(0.05331/10))
to =-2.719
| to | =2.719
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
we got |to| = 2.71935 & | t | = 2.262
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.7193 ) = 0.024
hence value of p0.05 > 0.024,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.719
critical value: -2.262 , 2.262
decision: reject Ho
p-value: 0.024
we have enough evidence to support the claim
b.
we use difference of sample means instead of paired difference because given two samples are different solutions and dependent each other
c.
we use pooled variance not pooled standard deviation

d.
TRADITIONAL METHOD
given that,
mean(x)=9.95
standard deviation , s.d1=0.4696
number(n1)=10
y(mean)=10.4
standard deviation, s.d2 =0.2309
number(n2)=10
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((0.221/10)+(0.053/10))
= 0.165
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
margin of error = 2.262 * 0.165
= 0.374
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (9.95-10.4) ± 0.374 ]
= [-0.824 , -0.076]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=9.95
standard deviation , s.d1=0.4696
sample size, n1=10
y(mean)=10.4
standard deviation, s.d2 =0.2309
sample size,n2 =10
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 = [( 9.95-10.4) ± t a/2 * sqrt((0.221/10)+(0.053/10)]
= [ (-0.45) ± t a/2 * 0.165]
= [-0.824 , -0.076]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-0.824 , -0.076] 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

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