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german
2 years ago
14

The mean breaking strength of yarn used in manufacturing drapery material is required to be more than 100 psi. Past experience h

as 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?
Mathematics
1 answer:
Ganezh [65]2 years ago
8 0

Answer:

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.

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