Question

1.14. A random sample of 22 shearing pins is taken in a study of the Rockwell hardness of the pin head. Measurements on the Rockwell hardness are made for each of the 22, yielding an average value of 58.50 with a sample standard deviation of 15.5. Assuming the measurements are normally distributed.

Construct a 95% two-sided confidence interval for the mean Rockwell hardness. (2 Points)

If true standard deviation is equal to 12, how large a sample size is necessary if the error at 99% confidence level is equal to 3? (2 Points)

Answer 1

Given n= 22 , xbar = 58.50 , s = 15.5

The 95% confidence interval for population mean is

xbar - E < < xbar + E

Where E = t​​​​​​a/2*(s/√n)

For a = 0.05 , d.f = n -1 = 21

t​​​​​​0.025,21 = 2.08

E = 2.08*(15.5/√22) = 6.87

58.50 - 6.87 < < 58.50 + 6.87

51.63 < < 65.37

b)

If population standard deviation = 12

Confidence level c = 0.99

Margin of error E = 3

The required sample size n

n = ( Z​​​​​​a/2* /E)2

For a = 0.01 , Z​​​​​​0.005 = 2.58

n = ( 2.58*12/3)2

n = 107