\"Durable press\" cotton fabrics are treated to improve their recovery from wrin
ID: 3206269 • Letter: #
Question
"Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. Unfortunately, the treatment also reduces the strength of the fabric. The breaking strength of untreated fabric is normally distributed with mean 51.9 pounds and standard deviation 2.7 pounds. The same type of fabric after treatment has normally distributed breaking strength with mean 25.2pounds and standard deviation 1.9 pounds. A clothing manufacturer tests 5 specimens of each fabric. All 10 strength measurements are independent. (Round your answers to four decimal places.)
(a) What is the probability that the mean breaking strength of the 5 untreated specimens exceeds 50 pounds?
(b) What is the probability that the mean breaking strength of the 5 untreated specimens is at least 25 pounds greater than the mean strength of the 5 treated specimens?
Explanation / Answer
a. From given information, mu=51.9, sigma=2.7, n=5.
Applying CLT, mu=Xbar, sigmaxbar=sigma/sqrt n=2.7/sqrt 5.
Substitute the values in Z score formula, Z=(X-Xbar)/sigmaxbar
P(X>50)=P[Z>(50-51.9)/(2.7/sqrt 5)]=P[Z>-1.57]=1-0.0582=0.9418.
b. Assume the hypotheses as follows:
H0:mu1-mu2<25 (mean breaking strength of untreated specimen is 25 pounds less than treated specimen)
H1:mu1-mu2>=25 (mean breaking strength of untreated specimen is atleast 25 pounds greater than that of treated specimen)
Assume the randomization condition, independence assumption, and independent group assumptions are reasonably met.
Compute the 2-sample t test statistic (equal variance).
For two equal sized samples, the ratio of standard deviation1/standard deviation 2 is less than 2.
The pooled standard deviation, sp=sqrt [{(n1-1)s1^2+(n2-1)s2^2}/(n1+n2-2)}], where, s denotes sample proportion, n is sample size.
=sqrt[{(5-1)2.7^2+(5-1)1.9^2}/(5+5-2)}
=2.3345
t={(x1bar-x2bar)-25}/sp sqrt{1/n1+1/n2}, where, xbar is sample mean.
={(51.9-25.2)-25}/{2.3345 sqrt(1/5+1/5)}
=1.15
Compute subsequent p value.
p value: 0.141 (ans)
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