A quality control engineer is interested in the mean length of sheet insulation being cut automatically by machine. The desired mean length of the insulation is 15 feet. It is known that the standard deviation in the cutting length is 0.2 feet. A sample of 60 cut sheets yields a mean length of 15.14 feet. This sample will be used to obtain a 99% confidence interval for the mean length cut by machine. The confidence interval goes from ________ to ________.

1 answer

To calculate the 99% confidence interval for the mean length of sheet insulation cut by the machine, we can use the formula for the confidence interval of the mean:

\[ \text{CI} = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right) \]

Where:

  • \(\bar{x}\) is the sample mean
  • \(z\) is the z-value for the desired confidence level
  • \(\sigma\) is the population standard deviation
  • \(n\) is the sample size

Given:

  • \(\bar{x} = 15.14\) feet
  • \(\sigma = 0.2\) feet
  • \(n = 60\)
  • For a 99% confidence interval, the z-value is approximately 2.576 (you can find this value in z-tables or using normal distribution calculators).

Now, let's calculate the margin of error (ME):

\[ \text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{0.2}{\sqrt{60}} \approx \frac{0.2}{7.746} \approx 0.0258 \]

Now we calculate the margin of error:

\[ \text{Margin of Error (ME)} = z \times SE = 2.576 \times 0.0258 \approx 0.0665 \]

Now we can calculate the confidence interval:

\[ \text{Lower Limit} = \bar{x} - ME \approx 15.14 - 0.0665 \approx 15.0735 \] \[ \text{Upper Limit} = \bar{x} + ME \approx 15.14 + 0.0665 \approx 15.2065 \]

Thus, the 99% confidence interval for the mean length cut by the machine goes from approximately 15.0735 feet to 15.2065 feet.