In a sample of 100 steel wires the average breaking strength, x = 50 kN, with a

Question

In a sample of 100 steel wires the average breaking strength, x = 50 kN, with a standard deviation, sigma = 4 kN. a. Find a 90% confidence interval for the true mean breaking strength of this type of wire. b. Find a 95% confidence interval for the true mean breaking strength of this type of wire. c. An engineer claims that the true mean breaking strength is between 49.53 kN and 50.47 kN. With what level of confidence can this statement be made? d. How many wires must be sampled so that a 95% confidence interval specifies the true mean breaking strength to within plusminus .3 kN?

Explanation / Answer

a)for std error=std deviation/(n)1/2 =0.4

for 90% CI, z=1.645

hence confidence interval =sample mean -/+ z*std error =49.3421 ; 50.6579

b) for 95% CI, z=1.96

hence confidence interval =sample mean -/+ z*std error =49.2160 ; 50.7840

c)for margin of error =(50.47-49.53)/2 =0.47

hence z=margin of error/std error =0.47/0.4 =1.175

for above confidence inteval is 76%

d)for margin of error E=0.3

for 95% CI, z=1.96

hence sample size =(z*std deviation/E)2 =683

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