Math Problem Statement

A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 95​% confidence interval estimate of the mean amount of mercury in the population. 0.59  0.75  0.10  0.98  1.25  0.51  0.85

Solution

To construct a 95% confidence interval for the mean amount of mercury in the tuna sushi samples, we can follow these steps:

Step 1: Calculate the Sample Mean (xˉ\bar{x})

The sample data is:

  • 0.59
  • 0.75
  • 0.10
  • 0.98
  • 1.25
  • 0.51
  • 0.85

Calculating the mean: xˉ=0.59+0.75+0.10+0.98+1.25+0.51+0.857=4.9370.7043\bar{x} = \frac{0.59 + 0.75 + 0.10 + 0.98 + 1.25 + 0.51 + 0.85}{7} = \frac{4.93}{7} \approx 0.7043

Step 2: Calculate the Sample Standard Deviation (ss)

To find the sample standard deviation: s=(xixˉ)2n1s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} Where n=7n = 7.

Calculating each squared deviation:

  • (0.590.7043)20.0131(0.59 - 0.7043)^2 \approx 0.0131
  • (0.750.7043)20.0021(0.75 - 0.7043)^2 \approx 0.0021
  • (0.100.7043)20.3643(0.10 - 0.7043)^2 \approx 0.3643
  • (0.980.7043)20.0767(0.98 - 0.7043)^2 \approx 0.0767
  • (1.250.7043)20.2970(1.25 - 0.7043)^2 \approx 0.2970
  • (0.510.7043)20.0370(0.51 - 0.7043)^2 \approx 0.0370
  • (0.850.7043)20.0215(0.85 - 0.7043)^2 \approx 0.0215

Summing these squared deviations: (xixˉ)20.0131+0.0021+0.3643+0.0767+0.2970+0.0370+0.02150.8117\sum (x_i - \bar{x})^2 \approx 0.0131 + 0.0021 + 0.3643 + 0.0767 + 0.2970 + 0.0370 + 0.0215 \approx 0.8117

Now calculate ss: s=0.811760.13530.3686s = \sqrt{\frac{0.8117}{6}} \approx \sqrt{0.1353} \approx 0.3686

Step 3: Determine the Critical Value for 95% Confidence

For a 95% confidence level and n1=6n-1 = 6 degrees of freedom, we can find the critical t-value from a t-distribution table: t0.025,62.447t_{0.025, 6} \approx 2.447

Step 4: Calculate the Margin of Error (ME)

ME=tsn=2.4470.368672.4470.13950.3422ME = t^* \cdot \frac{s}{\sqrt{n}} = 2.447 \cdot \frac{0.3686}{\sqrt{7}} \approx 2.447 \cdot 0.1395 \approx 0.3422

Step 5: Construct the Confidence Interval

Lower limit=xˉME0.70430.34220.3621\text{Lower limit} = \bar{x} - ME \approx 0.7043 - 0.3422 \approx 0.3621 Upper limit=xˉ+ME0.7043+0.34221.0465\text{Upper limit} = \bar{x} + ME \approx 0.7043 + 0.3422 \approx 1.0465

Final Confidence Interval

Thus, the 95% confidence interval for the mean amount of mercury in the tuna sushi is approximately: (0.3621,1.0465)(0.3621, 1.0465)

Summary

The mean amount of mercury in the sampled tuna sushi is estimated to be between 0.3621 ppm and 1.0465 ppm with 95% confidence.

Would you like more details on any part of this process? Here are some related questions to consider:

  1. How do different sample sizes affect the width of a confidence interval?
  2. What assumptions are made when calculating a confidence interval for the mean?
  3. How would the confidence interval change if the sample contained more extreme values?
  4. Why is it important to monitor mercury levels in fish?
  5. What are some other common food safety guidelines regarding mercury levels?

Tip: Always check if your sample size is sufficient to make reliable inferences about the population.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Descriptive Statistics

Formulas

Sample Mean: \( \bar{x} = \frac{\sum x_i}{n} \)
Sample Standard Deviation: \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \)
Margin of Error: \( ME = t^* \cdot \frac{s}{\sqrt{n}} \)
Confidence Interval: \( (\bar{x} - ME, \bar{x} + ME) \)

Theorems

Central Limit Theorem
t-Distribution for Confidence Intervals

Suitable Grade Level

Grades 11-12