Question

A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the cable can hold. The weight limit will be reported on cable packaging. The engineers take a random sample of 45 cables and apply weights to each of them until they break. The mean breaking weight for the 45 cables is 768.2 lb. The standard deviation of the breaking weight for the sample is 15.1 lb. Using the sample information as given, construct a confidence interval for the mean breaking strength of the new steel cable.

if a 97% level of confidence is used instead, the new interval estimate is:

a. (762.402, 773.998)

b. (764.259, 772.141)

c. (763.315, 773.085)

d. (738.2, 798.2)

If the engineers would like to adjust the sample size so that the width of the interval estimate will not be more than 5 lbs. with 95% confidence. What should the appropriate new sample size be? [Note: In the calculation, use 15.1 lbs. as the standard deviation.]

a. 140

b. 141

c. 225

d. 226

Answer 1

До D n = 45 x = 768.2 SoD = Isol. confidence interval at 97"). 9. AL 97.1 - 2-value 2017oog so confidence interval x + 2x 6 sla 768.2 + (2017009) /isol प 768.2 - (2.17009) (15.1) 768.27 (2.17009) (15.1)? + 6.7082 6.7082 [768.2-4.8848 768.2 +9.8848] 5 option (c) is correct answer. 763.315 773.085 9 So,

QS 2 ps nol- be Interval width สา0 M95) S , 95% confidence interval SoD = 15.1 At 95% confidence interval z value = 1.96 lale know laridth 2{margin of error) 5 = 2 (margin of error ME ㄷ 2.5 ME on 2=1.96 at 95%. Col. = (1-96) (15.1) 1.96) (15.1) 2.5 110 8384 -) n = (11.8384) 2 n i.l 1 yo. 1477 n = 140 is correct answer. so oprion (A)