JordanJones04402
19.03.2021 •
Mathematics
The mean breaking strength of yarn used in manufacturing drapery material is required to be more than 100 psi. Past experience has indicated that the standard deviation of breaking strength is 2.4 psi. A random sample of 9 specimens is tested, and the average breaking strength is found to be 100.6 psi.
Required:
a. Should the fiber be judged acceptable? Use the P-value approach.
b. What is the probability of not rejecting the null hypothesis at α=0.05 if the fiber has a true mean breaking strength of 102 psi?
c. Find a 95% one-sided lower CI on the true mean breaking strength.
d. Use the CI found in part (d) to test the hypothesis.
e. What sample size is required to detect a true mean breaking strength of 101 with probability 0.95?
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Ответ:
H0: μ = 100 ; H1: μ > 100 ;
Test statistic = 0.75 ;
We fail to reject null ;
Kindly check explanation for the rest
Step-by-step explanation:
Given that :
H0: μ = 100
H1: μ > 100
Sample mean, x = 100.6 ; n = 9 ; sample Standard deviation, s = 2.4
Test statistic :
(x - μ) ÷ s/sqrt(n)
(100.6 - 100) ÷ 2.4 / sqrt(9)
0.6 ÷ 0.8 = 0.75
Probability of not rejecting the null :
P = P(Z > 0.75) = 0.22663
α = 0.05
Since, P > α ; we fail to reject the Null ; there is no sufficient evidence to accept the claim that mean strength is > 100 psi
μ + Zcritical*s/sqrt(n)
Zcritical at (1 - α) = 1.645
100 + 1.645*(2.4/3)
100 + 1.316 = 101.316
(x - μ) ÷ s/sqrt(n)
(101.316 - 102) ÷ 2.4 / sqrt(9)
-0.684 ÷ 0.8 = 0.75
= - 0.855
P(Z < - 0.855) = 0.1963 (Z probability calculator).
95% lower confidence interval :
x - error margin, E
E = Zcritical * s/sqrt(n)
E = 1.645 * 0.8 = 1.316
100.6 - 1.316 = 99.284
From part c :
Interval becomes ;
(99.284, 101.316) ; this interval contains the hypothesized mean value of 100. Hence we fail to reject the null.
Ответ:
is the correct answer to the given question.
Step-by-step explanation:
Missing information
In this question options are missing.Following are the option
(a)
(b)
(c)
(d)
As mention in the question On the highway, car is said to get 25 miles per gallon
We assume
also that the value of is fluctuate by 6 miles per gallon
This can be written as
Therefore
is the answer