The mean breaking strength of yarn used in manufacturing drapery material is req

Question

The mean breaking strength of yarn used in manufacturing drapery material is required to be at least 100 psi. Past experience has indicated that the standard deviation of breaking strength is 3.1 psi. A random sample of 9 specimens is tested, and the average breaking strength is found to be 100.6 psi Statistical Tables and Charts (a) Calculate the P-value. Round your answer to 3 decimal places (e.g. 98.765) If -0.05 , should the fiber be judged acceptable? No. Yes. is +/-2% (b) what is the probability of not rejecting the null hypothesis at = 0.05 if th e fiber has a true mean breaking strength of 102 psi? Round your answer to 3 decimal places (e.g. 98.765) the tolerance is +/-2%

Explanation / Answer

PART A.
Given that,
population mean(u)=100
standard deviation, =3.1
sample mean, x =100.6
number (n)=9
null, Ho: =100
alternate, H1: >100
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-u/(s.d/sqrt(n))
zo = 100.6-100/(3.1/sqrt(9)
zo = 0.58065
| zo | = 0.58065
critical value
the value of |z | at los 5% is 1.645
we got |zo| =0.58065 & | z | = 1.645
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : right tail - ha : ( p > 0.58065 ) = 0.28074
hence value of p0.05 < 0.28074, here we do not reject Ho
ANSWERS
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p-value: 0.28074

it is not acceptable

PART B.
Given that,
Standard deviation, =3.1
Sample Mean, X =100.6
Null, H0: <100
Alternate, H1: >100
Level of significance, = 0.05
From Standard normal table, Z /2 =1.6449
Since our test is right-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-100)/3.1/(n) < -1.6449 OR if (x-100)/3.1/(n) > 1.6449
Reject Ho if x < 100-5.09919/(n) OR if x > 100-5.09919/(n)
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Suppose the size of the sample is n = 9 then the critical region
becomes,
Reject Ho if x < 100-5.09919/(9) OR if x > 100+5.09919/(9)
Reject Ho if x < 98.30027 OR if x > 101.69973
Implies, don't reject Ho if 98.30027 x 101.69973
Suppose the true mean is 102
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(98.30027 x 101.69973 | 1 = 102)
= P(98.30027-102/3.1/(9) x - / /n 101.69973-102/3.1/(9)
= P(-3.58038387 Z -0.29058387 )
= P( Z -0.29058387) - P( Z -3.58038387)
= 0.3857 - 0.0002 [ Using Z Table ]
= 0.3855
For n =9 the probability of Type II error is 0.3855

PART C.
given that,
standard deviation, =3.1
sample mean, x =100.6
population size (n)=9
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 100.6 ± Z a/2 ( 3.1/ Sqrt ( 9) ) ]
= [ 100.6 - 1.645 * (1.033) , 100.6 + 1.645 * (1.033) ]
= [ 98.9,102.3 ]

PART D.
Fail to Reject Ho

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