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.