A manufacturer claims that the average tensile strength of thread A exceeds the

Question

A manufacturer claims that the average tensile strength of thread A exceeds the average tensile strength of thread B by more than 7 kilograms. To test his claim, 50 pieces of each type of thread are tested under similar conditions. Type A thread had an average tensile strength of 867 kilograms with known standard deviation of A-628 kilograms, while type B thread had an average tensile strength of 77.8 kilograms with known standard deviation of 5.61 kilograms. Test the manufacturer's claim at -0.05. Assuming equal samples for the two population, how large should the samples be if the power of our test is to be 0.95 when the true difference between thread types A and B is 11 kilograms (A-11B-11 Created by Paint X

Explanation / Answer

Given that,
mean(x)=86.7
standard deviation , 1 =6.28
number(n1)=50
y(mean)=77.8
standard deviation, 2 =5.61
number(n2)=50
null, Ho: u1 = u2
alternate, H1: 1 > u2
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=86.7-77.8/sqrt((39.4384/50)+(31.4721/50))
zo =7.4734
| zo | =7.4734
critical value
the value of |z | at los 0.05% is 1.645
we got |zo | =7.473 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail -Ha : ( p > 7.4734 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 > u2
test statistic: 7.4734
critical value: 1.645
decision: reject Ho
p-value: 0

ii)

Given that,
mean(x)=86.7
standard deviation , 1 =6.28
number(n1)=50
y(mean)=77.8
standard deviation, 2 =5.61
number(n2)=50
null, Ho: u1 = u2
alternate, H1: 1 != u2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=86.7-77.8/sqrt((39.4384/50)+(31.4721/50))
zo =7.4734
| zo | =7.4734
critical value
the value of |z | at los 0.05% is 1.96
we got |zo | =7.473 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 7.4734 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 != u2
test statistic: 7.4734
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0

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