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 is μ ± 1.96 SD.
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To calculate the 95% confidence interval for the average torque being applied during the assembly process, we can use the formula:
Confidence Interval = x̄ ± (Z * σ / √n)
Where:
- x̄ is the sample mean of the torque (150 lb-ft in this case)
- Z is the Z-score representing the desired confidence level (for a 95% confidence level, Z is approximately 1.96)
- σ is the standard deviation of the torque (3 lb-ft in this case)
- n is the sample size (35 in this case)
Let's now plug in the values:
Confidence Interval = 150 ± (1.96 * 3 / √35)
To calculate the value inside the parentheses:
1.96 * 3 ≈ 5.88
To calculate √35 (square root of 35):
√35 ≈ 5.92
Now, substituting these values into the confidence interval formula:
Confidence Interval = 150 ± (5.88 / 5.92)
Simplifying:
Confidence Interval = 150 ± 0.993
This gives us the range for the average torque being applied during the assembly process.
Confidence Interval = (150 - 0.993, 150 + 0.993)
Therefore, the 95% confidence interval for the average torque being applied during the assembly process is approximately (149.01 lb-ft, 150.99 lb-ft).