\"Durable press\" cotton fabrics are treated to improve their recovery from wrin
ID: 3363554 • Letter: #
Question
"Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. "Wrinkle recovery angle" measures how well a fabric recovers from wrinkles. Higher is better. Here are data on the wrinkle recovery angle (in degrees) for the same fabric specimens discussed in the previous exercise:
A manufacturer wants to know how large is the difference in mean wrinkle recovery angle.
The mean and the standard deviation for the Permafresh are:
¯x=
Round to 3 decimal places.
s=
Round to 3 decimal places.
The mean and the standard deviation for the Hylite are:
¯x=
Round to 3 decimal places.
s=
Round to 3 decimal places.
The standard error is:
SE = Round to 3 decimal places.
The critical value from the distribution for a confidence interval of 98% is:
t*= Use degrees of freedom of 3. Use Technology Round to 3 decimal places.
Using the the answer from above (all rounded to 3 decimal places), calculate by hand a 98% confidence interval for the difference in mean wrinkle recovery angle:
to . Hint Round to 3 decimal places.
Explanation / Answer
TRADITIONAL METHOD
given that,
mean(x)=121.6
standard deviation , s.d1=16.1028 = 16.103
number(n1)=5
y(mean)=173.25
standard deviation, s.d2 =22.1265 = 22.123
number(n2)=4
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((259.3/5)+(489.582/4))
= 13.201
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table,right tailedand
value of |t | with min (n1-1, n2-1) i.e 3 d.f is 3.482
margin of error = 3.48 * 13.201
= 45.938
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (121.6-173.25) ± 45.938 ]
= [-97.588 , -5.712]
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DIRECT METHOD
given that,
mean(x)=121.6
standard deviation , s.d1=16.1028
sample size, n1=5
y(mean)=173.25
standard deviation, s.d2 =22.1265
sample size,n2 =4
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 = [( 121.6-173.25) ± t a/2 * sqrt((259.3/5)+(489.582/4)]
= [ (-51.65) ± t a/2 * 13.201]
= [-97.588 , -5.712]
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interpretations:
1. we are 98% sure that the interval [-97.588 , -5.712] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion
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