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 ()
The sample data is:
- 0.59
- 0.75
- 0.10
- 0.98
- 1.25
- 0.51
- 0.85
Calculating the mean:
Step 2: Calculate the Sample Standard Deviation ()
To find the sample standard deviation: Where .
Calculating each squared deviation:
Summing these squared deviations:
Now calculate :
Step 3: Determine the Critical Value for 95% Confidence
For a 95% confidence level and degrees of freedom, we can find the critical t-value from a t-distribution table:
Step 4: Calculate the Margin of Error (ME)
Step 5: Construct the Confidence Interval
Final Confidence Interval
Thus, the 95% confidence interval for the mean amount of mercury in the tuna sushi is approximately:
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:
- How do different sample sizes affect the width of a confidence interval?
- What assumptions are made when calculating a confidence interval for the mean?
- How would the confidence interval change if the sample contained more extreme values?
- Why is it important to monitor mercury levels in fish?
- 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
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