Question

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?

Answer 1

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.