2. [30 pts] In a sample of 100 steel wires, the average breaking strength is 50

Question

2. [30 pts] In a sample of 100 steel wires, the average breaking strength is 50 kN, with a standard deviation of 2 kN. a) Find a 9S% confidence interval for the mean breaking strength of this type of wire. b) Find a 99% confidence interval for the mean breaking strength ofthis type of wire. c) An engineer claims that the mean breaking strength is between 49.7 kN and 50.3 kN. With what level of confidence can this statement be made? mean breaking strength to within ±0.3 kN? mean breaking strength to within ±0.3 kN? d) How many wires must be sampled so that a 95% confidence interval specifies the e) How many wires must be sampled so that a 99% confidence interval specifies the

Explanation / Answer

Solution:

Given that x = 50, s = 2, n = 100
= 0.05

a) 95% Confidence interval for the mean breaking strength of this type of wire.
Z/2 = Z0.025 = 1.96
CI = x ± Z/2 *s/n = 50 ± 1.96 *2/100
= 50 ± 1.96 *2/100
= 50 ± 0.392
= (49.6080, 50.3920)
b) 99% Confidence interval for the mean breaking strength of this type of wire.
Z/2 = Z0.005 = 2.58
CI = x ± Z/2 *s/n = 50 ± 2.58 *2/100
= 50 ± 2.58 *2/100
= 50 ± 0.516
= (49.484, 50.516)
c) An engineer claims that the mean breaking strength is between 49.7k and 50.3kN. With what level of confidence can this statement be made.
P(49.7 < X < 50.3) = P(49.750/(2/100)< z <50.350/(2/100))
= P(-1.5 < z < 1.5)
= 0.8664
d) = 0.05,
Z/2 = Z0.025 = 1.96
The necessary sample size is found by solving the equation:
Z/2 *s/n = 0.3
(1.96) *2/n = 0.3
n = (1.96*2/0.3)^2
= 170.74 171

e) = 0.01,
Z/2 = Z0.005 = 2.58
The necessary sample size is found by solving the equation:
Z/2 *s/n = 0.3
(2.58) *2/n = 0.3
n = (2.58 *2/0.3)^2
= 295.84 296

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