Math Problem Statement

Canadians are becoming more conscious of the importance of good nutrition, and some researchers believe that we may be altering our diets to include less red meat and more fruits and vegetables. To test this theory, a researcher decides to select hospital nutritional records for subjects surveyed 10 years ago and to compare the average amount of beef consumed per year to the amounts consumed by an equal number of subjects she will interview this year. She knows that the amount of beef consumed annually by Canadians ranges from 0 to approximately 47 kg. How many subjects should the researcher select for each group if she wishes to estimate the difference in the average annual per-capita beef consumption correct to within 2.5 kg with 99% confidence? (Round your answer up to the nearest whole number.)

Solution

To solve this problem, we can use the formula for determining the sample size for estimating the difference in means with a specified confidence level. The formula for the sample size nn for each group is given by:

n=2(zα/2σ)2E2n = \frac{2 \left( z_{\alpha/2} \cdot \sigma \right)^2}{E^2}

Where:

  • zα/2z_{\alpha/2} is the Z-score corresponding to the desired confidence level.
  • σ\sigma is the standard deviation of the population (estimated range of beef consumption).
  • EE is the margin of error (the maximum allowable difference between the sample estimate and the true population parameter).

Step 1: Identify the values needed for the calculation

  • Confidence level: 99%, so zα/2=2.576z_{\alpha/2} = 2.576 (from Z-tables).
  • Range of beef consumption: The amount of beef consumed annually ranges from 0 to 47 kg, so the standard deviation σ\sigma can be approximated using half the range: σ474=11.75kg\sigma \approx \frac{47}{4} = 11.75 \, \text{kg} (This is based on the assumption that the data is uniformly distributed).
  • Margin of error E=2.5E = 2.5 kg.

Step 2: Plug values into the formula

Now, substitute the values into the formula for nn:

n=2(2.57611.75)22.52n = \frac{2 \left( 2.576 \cdot 11.75 \right)^2}{2.5^2}

n=2(30.2)26.25n = \frac{2 \left( 30.2 \right)^2}{6.25}

n=2912.046.25n = \frac{2 \cdot 912.04}{6.25}

n=1824.086.25=291.85n = \frac{1824.08}{6.25} = 291.85

Step 3: Round up to the nearest whole number

Since we can't have a fraction of a subject, we round up to the nearest whole number:

n292n \approx 292

Conclusion

The researcher should select 292 subjects for each group to estimate the difference in average beef consumption with 99% confidence, correct to within 2.5 kg.


Would you like further explanation or clarification on any step? Here are some related questions you may find interesting:

  1. How do you calculate sample size when the population standard deviation is unknown?
  2. What is the meaning of a Z-score and how is it used in confidence intervals?
  3. How does the margin of error impact the required sample size?
  4. Why do we round the sample size up in these kinds of calculations?
  5. What would happen if we used a different confidence level, like 95% instead of 99%?

Tip: When estimating sample size, always consider using the largest possible estimate for the population standard deviation to ensure the sample size is large enough for the desired accuracy.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Sample Size Estimation

Formulas

n = (2 * (z_α/2 * σ)^2) / E^2

Theorems

Central Limit Theorem
Normal Distribution Properties

Suitable Grade Level

Undergraduate Statistics / Advanced High School