An assembly process includes a torque wrench that automatically tightens engine bolts. The wrench has a...

An assembly process includes a torque wrench that automatically tightens engine bolts. The wrench has a known process standard deviation of 4 foot pounds of torque. A sample of 25 bolts is selected and the average torque to which they have been tightened is 120 foot pounds. What is the 95% confidence interval for the average torque being applied by this wrench?

Solutions

Expert Solution

solution

Given that,

= 120

s =4

n = 25

Degrees of freedom = df = n - 1 =25 - 1 = 24

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,24 = 2.064 ( using student t table)

Margin of error = E = t/2,df * (s / $\sqrt$ n)

= 2.064 * (4 / $\sqrt$ 25) = 1.6512

The 95% confidence interval estimate of the population mean is,

- E < < + E

120-1.6512 < < 120+1.6512

118.3488 < 121.6512

( 118.3488,121.6512)

orchestra answered 1 year ago

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