Math Problem Statement

A coin-operated coffee machine made by BIG Corporation was designed to discharge a mean of eight ounces of coffee per cup. If it dispenses more than that on average, the corporation may lose money, and if it dispenses less, the customers may complain. BIG Corporation would like to estimate the mean amount of coffee, μ, dispensed per cup by this machine. BIG will choose a random sample of cup amounts dispensed by this machine and use this sample to estimate μ. Assuming that the standard deviation of cup amounts dispensed by this machine is 0.43 ounces, what is the minimum sample size needed in order for BIG to be 99% confident that its estimate is within 0.07 ounces of μ?

Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole number that satisfies the requirements).

(If necessary, consult a list of formulas

Solution

To find the minimum sample size required, we need to use the formula for the sample size in a confidence interval for the mean when the population standard deviation is known. The formula is:

n=(Zα/2σE)2n = \left( \frac{Z_{\alpha/2} \cdot \sigma}{E} \right)^2

where:

  • nn is the sample size,
  • Zα/2Z_{\alpha/2} is the critical value from the standard normal distribution corresponding to the desired confidence level (99% in this case),
  • σ\sigma is the population standard deviation (0.43 ounces),
  • EE is the margin of error (0.07 ounces).

Step 1: Find the critical value Zα/2Z_{\alpha/2}

For a 99% confidence level, the critical value Zα/2Z_{\alpha/2} corresponds to the value where 99% of the distribution is within this range. In terms of the standard normal distribution, we look for the critical value such that 0.5% is in each tail (since 10.99=0.011 - 0.99 = 0.01, and half of that is 0.005). The critical value Zα/2Z_{\alpha/2} for 99% confidence is:

Zα/22.576Z_{\alpha/2} \approx 2.576

Step 2: Plug the values into the formula

We have:

  • Zα/2=2.576Z_{\alpha/2} = 2.576,
  • σ=0.43\sigma = 0.43,
  • E=0.07E = 0.07.

Now, substitute these into the sample size formula:

n=(2.5760.430.07)2n = \left( \frac{2.576 \cdot 0.43}{0.07} \right)^2

Step 3: Simplify the expression

First, calculate the value inside the parentheses:

2.5760.430.07=1.107680.0715.824\frac{2.576 \cdot 0.43}{0.07} = \frac{1.10768}{0.07} \approx 15.824

Now, square this result:

n=(15.824)2250.417n = (15.824)^2 \approx 250.417

Step 4: Round to the next whole number

Since the sample size must be a whole number and we need the minimum size, round 250.417 up to the next whole number:

n=251n = 251

Thus, the minimum sample size needed is 251.


Would you like further details or have any questions? Here are five related questions to deepen understanding:

  1. How does the confidence level affect the sample size?
  2. What is the effect of increasing the margin of error on the sample size?
  3. What if the population standard deviation was unknown? How would the approach change?
  4. Why do we square the result in the sample size formula?
  5. How is the critical value Zα/2Z_{\alpha/2} determined for different confidence levels?

Tip: The margin of error decreases as the sample size increases, making your estimate of the population mean more precise.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Interval
Sample Size Estimation

Formulas

n = (Z_{α/2} * σ / E)^2

Theorems

Central Limit Theorem
Standard Normal Distribution

Suitable Grade Level

College level, Statistics