Question

A machine produces metal pieces that are cylindrical in shape. A sample of these pieces is taken and the diameters are found to be 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, and 1.03 centimeters. Use these data to calculate three interval types and draw interpretations that illustrate the distinction between them in the context of the system. For all computations, assume an approximately normal distribution. The sample mean and standard deviation for the given data are 1.0056 and 0.0246, respectively. (Express answers in 4 decimal places.)

a. Find a 99% confidence interval on the mean diameter.
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μ Answer
b. Compute a 99% prediction interval on a measured diameter of a single metal piece taken from the machine.

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c. Find the 99% tolerance limits that will contain 95% of the metal pieces produced by this machine. Answer
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Answer 1

The 99% confidence interval for the mean diameter is (0.9845, 1.0267), the prediction interval for a single measurement is (0.9839, 1.0273), and the tolerance limits for 95% of pieces are (0.9574, 1.0538).

a. 99% Confidence Interval on the Mean Diameter:

The formula for a confidence interval is given by:

`\[ \text{Confidence Interval} = \bar{x} \pm Z \left( (s)/(√(n)) \right) \]`

For a 99% confidence interval, the Z-score is approximately 2.576 (assuming a normal distribution). Therefore,

`\[ \text{Confidence Interval} = 1.0056 \pm 2.576 \left( (0.0246)/(√(9)) \right) \]\[ \text{Confidence Interval} = 1.0056 \pm 2.576 \left( (0.0246)/(3) \right) \]\[ \text{Confidence Interval} = 1.0056 \pm 2.576 * 0.0082 \]\[ \text{Confidence Interval} = (1.0056 - 0.0211, 1.0056 + 0.0211) \]\[ \text{Confidence Interval} = (0.9845, 1.0267) \]`

b. 99% Prediction Interval on a Measured Diameter:

The formula for a prediction interval is given by:

`\[ \text{Prediction Interval} = \bar{x} \pm Z \left( (s)/(√(n)) \right) * \sqrt{1 + (1)/(n)} \]\[ \text{Prediction Interval} = 1.0056 \pm 2.576 \left( (0.0246)/(3) \right) * \sqrt{1 + (1)/(9)} \]`

`\[ \text{Prediction Interval} = 1.0056 \pm 2.576 * 0.0082 * \sqrt{(10)/(9)} \]\[ \text{Prediction Interval} = 1.0056 \pm 2.576 * 0.0082 * 1.0541 \]\[ \text{Prediction Interval} = (1.0056 - 0.0217, 1.0056 + 0.0217) \]\[ \text{Prediction Interval} = (0.9839, 1.0273) \]`

c. 99% Tolerance Limits for 95% of Metal Pieces:

For 95% of the metal pieces, the factor k is approximately 1.959 (assuming a normal distribution). Therefore,

`\[ \text{Tolerance Limits} = 1.0056 \pm 1.959 * 0.0246 \]\[ \text{Tolerance Limits} = 1.0056 \pm 1.959 * 0.0246 \]\[ \text{Tolerance Limits} = (1.0056 - 0.0482, 1.0056 + 0.0482) \]\[ \text{Tolerance Limits} = (0.9574, 1.0538) \]`

Therefore:

a. The 99% confidence interval on the mean diameter is (0.9845, 1.0267).

b. The 99% prediction interval on a measured diameter is (0.9839, 1.0273).

c. The 99% tolerance limits for 95% of metal pieces are (0.9574, 1.0538).