Math Problem Statement

A company produces regulation ultimate discs. The discs have a mean mass of 175.0 g, with a standard deviation of 0.9 g. To ensure that few discs are rejected, the quality control manager must ensure that the mean mass of the discs lies in the acceptable range of 174.8 g to 175.2 g. During each shift, a random sample of discs is selected, and the mass of each disc in the sample is measured. The table shown helps guide the sampling process. a) What is the confidence interval and margin of error this company is using for its quality control tests? b) Approximately how many discs should be measured to ensure the mean mass is within ±0.2 g, 99% of the time?

Solution

Let's analyze and solve the given problem. Here’s the breakdown:


Given Data:

  • Mean mass of the discs (μ\mu) = 175.0 g
  • Standard deviation (σ\sigma) = 0.9 g
  • Acceptable range: 174.8gxˉ175.2g174.8 \, \text{g} \leq \bar{x} \leq 175.2 \, \text{g}
  • Confidence level = 99% (assumed from part b).

Part (a): Confidence interval and margin of error

The margin of error (ME) for a confidence interval is determined using the formula: ME=zσn\text{ME} = z \cdot \frac{\sigma}{\sqrt{n}} Where:

  • zz: critical value for the desired confidence level (e.g., z=2.576z = 2.576 for 99% confidence).
  • σ\sigma: population standard deviation.
  • nn: sample size.

We would solve this if the sample size is provided or inferred.


Part (b): Sample size for a specific margin of error

The formula for determining sample size is: n=(zσME)2n = \left( \frac{z \cdot \sigma}{\text{ME}} \right)^2 Here:

  • Desired margin of error (ME\text{ME}) = 0.2 g
  • z=2.576z = 2.576 (critical value for 99% confidence)
  • σ=0.9g\sigma = 0.9 \, \text{g}

Substitute values: n=(2.5760.90.2)2n = \left( \frac{2.576 \cdot 0.9}{0.2} \right)^2

First, calculate the numerator: 2.5760.9=2.31842.576 \cdot 0.9 = 2.3184

Now divide by the margin of error: 2.31840.2=11.592\frac{2.3184}{0.2} = 11.592

Square this value: n=(11.592)2=134.38n = (11.592)^2 = 134.38

Since sample size must be a whole number, round up: n=135n = 135

Answer to part (b): Approximately 135 discs must be measured.


Conclusion:

(a) Requires sample size data to calculate the confidence interval.

(b) Approximately 135 discs should be measured to ensure the mean mass is within ±0.2 g with 99% confidence.


Would you like me to calculate part (a) further (if sample size is known), or explain anything else? Let me know!


Related Questions:

  1. What is the formula to calculate zz-scores for a confidence interval?
  2. How is the standard error of the mean calculated in quality control?
  3. How does increasing the sample size affect the margin of error?
  4. What assumptions are made when calculating confidence intervals for normal distributions?
  5. Can we use these formulas if the data is not normally distributed?

Tip: Always round up sample size calculations to ensure the desired confidence level is maintained.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Margin of Error
Sample Size Calculation

Formulas

Margin of Error (ME) = z * (σ / √n)
Sample Size (n) = (z * σ / ME)^2

Theorems

Central Limit Theorem

Suitable Grade Level

Grades 10-12