Question

Question 7 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 2.9 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 a = 0.05, should the fiber be judged acceptable? the tolerance is +/-2% (b) What is the probability of not rejecting the null hypothesis at Q = 0.05 if the 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% (c) Find a 95% one-sided lower confidence interval bound on the true mean breaking strength. Round your answer to three decimal place (e.g. 98.765). the tolerance is +/-2% (d) Use the confidence interval found in part (c) to test the hypothesis c ho. (e) What sample size is required to detect a true mean breaking strength of 101 with probability 0.95?

Answer 1

gives The O that, Hypothesis dae, Ho l= 100 psi Hi: 47100 psi. vs O = 2.9 psi A ŽE 100.6 psi To find, calculate the The test Z . X-4 de p. value Statistic = 100.6 - 100 2. 99 0.6 0.9667 CUCI 0-6206 Vodoro = 0.621 z p value - przy 0.6217 = 0.26463 I p. value = 0.268 Ans - pivalue = 0.268 IF the fiber be L=0.05, should judged acceptable ?

Here, I b.value = 0.2687 cue accept Ho at 5 % Signis conce. 250.05 lever of Ans These the - Accept Ho Ans tho is not sufficient evidence against sul hipothesis Ho. 6 what is the probability of not 2 rejecting the neell hypothesis at. G 2-0.05 if the Fibers has a toue mean breaking Strength of 102 psi . Here, -0.05 = 1.645 The corresponding Sample nean is the popwation mean increased by the pooduct of zt score and SD. X = ut Za 6 = 100+ 1.645. (2.9) = 100 + 1.590167 = 101.5902 * -The L. z value, = x - 4 = 101.5902-102 2.9 15g

= -0.4098 0.9667 -0.42391 -0.4239 3 of then, Find the probability the hull bypothesis Osing poobability table not Rejecting the normal - PC Z <-0-4239) = P (27 0°4239) - P (27 0.42) 0.33724 Poobability 10.3372 Ans - 0.337- o find a 95.). One-side loves confidence Interval bound on the true mean bolaking Stoength > Here 20.05 = 1.645 . Find M. E ME = S .6) - ).645. (2.9) Jg = 1.5902 ME 4 x = 1006

c. I is The loves boundary of X-ME 10.6 - 1.5902 eg.oog8 Aos a lower boundoroy & ggr gg.010 @ Ose the confidence toterval found in part 6 to test the hypothesis. ve note that C.I in port © contain the hypothesized mean 100 psi, which indicates that it is likely that the population mean Is 100 psi and thus e can not Reject Ho (Accept Ho) - There is not sufficient evidence to Reject to the claim that the meon. breaking strength as more than 100 psi & thus fiber should not be Gudged Acceptable... Ans - Accept Ho conat sample size is required to derect a toue mean balaking . Strength lol with probability 0.95 ? given pover 20.95 U= 101

Here - pover - 0.095 1-0.95 E 0.05 = TB E = 104-1000 E = 1 Marogin of erosor find, sample size; Pra nel KatzB) 6 ) ? Cohdence level o L = 0.95 Za = Zoos = 1.64 = zo.05 = 1.64 For - 7 (ZaZB) 2 (1.64 +1.64) x 2.g. 071 g0.478

& Hose The Hypothesis arol, vs Ho! U-15 mm He: 47175 mm n4 ) X = 181 = 20 To bodo P. value = ? the test statistica Z = X-es 181 - 175 20 154 = 6 z = 0.6 p. value = p z 70.6] = 0. 274253 p.value = 0. 2743 Ans