Please only the formula and steps to do this problem. Thank you
An assembly process includes a torque wrench device that automatically tightens compressor housing bolts; the device has a known process standard deviation of ó = 3 lb-ft in the torque applied. A simple random sample of 35 nuts is selected, and the average torque to which they have been tightened is 150 lb-ft. What is the 95% confidence interval for the average torque being applied during the assembly process
.95 confidence interval for mean =
Mean ± .196 Standard Error
For calculation of SE, see previous post.
I hope this helps. Thanks for asking.
To find the 95% confidence interval for the average torque being applied during the assembly process, you can use the following formula:
Confidence Interval = X̄ ± (Z * σ / √n)
Where:
- X̄ is the sample mean (average torque to which the nuts have been tightened)
- Z is the z-score corresponding to the desired confidence level (95% confidence corresponds to a z-value of approximately 1.96)
- σ is the process standard deviation (3 lb-ft)
- n is the sample size (35 nuts)
Now let's calculate the confidence interval step by step:
Step 1: Calculate the margin of error
Margin of Error = Z * σ / √n
In this case, Z = 1.96, σ = 3, and n = 35, so the margin of error becomes:
Margin of Error = 1.96 * 3 / √35
Step 2: Calculate the confidence interval
Confidence Interval = X̄ ± Margin of Error
Given that X̄ = 150, the confidence interval becomes:
Confidence Interval = 150 ± (1.96 * 3 / √35)
Now you can evaluate this expression to find the confidence interval for the average torque being applied during the assembly process.