Question

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with σ = 100. The composition of the bar has been slightly modified, but the modification is not believed to have affected either the normality or the value of σ.

(a) Assuming this to be the case, if a sample of 64 modified bars resulted in a sample average yield point of 8439 lb, compute a 90% CI for the true average yield point of the modified bar. (Round your answers to one decimal place.)

(_,_) lb

(b) How would you modify the interval in part (a) to obtain a confidence level of 96%? (Round your answer to two decimal places.)

---Select--- The standard deviation The sample size The value of z The sample mean should be changed to ___.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 8439

Population standard deviation = = 100

Sample size = n = 64

a)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645* (100 / 64)

= 20.6

At 90% confidence interval estimate of the population mean is,

- E < < + E

8439 - 20.6 < < 8439 + 20.6

8418.4 < < 8459.6

(8418.4 , 8459.6 )

b)

At 96% confidence level the z is ,

= 1 - 96% = 1 - 0.96 = 0.04

/ 2 = 0.04 / 2 = 0.02

Z_/2 = Z_0.02 = 2.054

Margin of error = E = Z_/2* ( /n)

= 2.054* (100 / 64)

= 25.68

At 96% confidence interval estimate of the population mean is,

- E < < + E

8439 - 25.68 < < 8439 + 25.68

8413.32 < < 8464.68

(8413.32 , 8464.68 )

The value of z should be changed to 2.054