To construct a 95% confidence interval estimate of the mean amount of mercury in tuna sushi, we'll follow these steps:

1. Step 1: Gather the data. The mercury amounts in ppm are: $0.55, 0.77, 0.10, 0.91, 1.30, 0.56, 0.87$

2. Step 2: Calculate the sample mean ($\bar{x}$) and the sample standard deviation ($s$).

   • Sample mean ($\bar{x}$): $\bar{x} = \frac{0.55 + 0.77 + 0.10 + 0.91 + 1.30 + 0.56 + 0.87}{7} = \frac{5.06}{7} = 0.7229 \, \text{ppm}$

   • Sample standard deviation ($s$): First, calculate the squared differences from the mean for each data point: $(0.55 - 0.7229)^2 = 0.0299, \quad (0.77 - 0.7229)^2 = 0.0022, \quad (0.10 - 0.7229)^2 = 0.3876$ $(0.91 - 0.7229)^2 = 0.0351, \quad (1.30 - 0.7229)^2 = 0.3335, \quad (0.56 - 0.7229)^2 = 0.0267, \quad (0.87 - 0.7229)^2 = 0.0217$ Summing these squared differences: $\sum = 0.0299 + 0.0022 + 0.3876 + 0.0351 + 0.3335 + 0.0267 + 0.0217 = 0.8367$ Now, calculate the variance (since $n = 7$): $s^2 = \frac{0.8367}{7 - 1} = \frac{0.8367}{6} = 0.1394$ Finally, take the square root to get the standard deviation: $s = \sqrt{0.1394} \approx 0.3734 \, \text{ppm}$

3. Step 3: Determine the critical value for a 95% confidence interval. Since the sample size is $n = 7$, we use the t-distribution with $n - 1 = 6$ degrees of freedom. The critical value $t_{\alpha/2}$ for a 95% confidence level and 6 degrees of freedom is approximately 2.447 (from t-distribution tables or a calculator).

4. Step 4: Calculate the margin of error (E). $E = t_{\alpha/2} \times \frac{s}{\sqrt{n}} = 2.447 \times \frac{0.3734}{\sqrt{7}} \approx 2.447 \times 0.1411 \approx 0.3455 \, \text{ppm}$

5. Step 5: Construct the confidence interval. The confidence interval is given by: $\bar{x} - E \leq \mu \leq \bar{x} + E$ Substituting the values: $0.7229 - 0.3455 \leq \mu \leq 0.7229 + 0.3455$ $0.3774 \leq \mu \leq 1.0684$

6. Conclusion: The 95% confidence interval for the mean mercury content in tuna sushi is approximately $[0.3774, 1.0684] \, \text{ppm}$.

Since the upper bound of the confidence interval (1.0684 ppm) is slightly above the safety guideline of 1 ppm, there is a possibility that the mean mercury content in tuna sushi exceeds the recommended limit. However, since the interval includes values both below and above 1 ppm, it doesn't definitively show that the mean mercury content is too high.

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Would you like further details on any part of this process or have any questions?

Here are 5 questions that could extend this topic:

1. How does the confidence level affect the width of the confidence interval?
2. What happens to the confidence interval if the sample size is increased?
3. How would using a different level of significance (e.g., 99%) change the results?
4. What assumptions do we make when using a t-distribution for confidence intervals?
5. Could the presence of outliers significantly affect the confidence interval?

Tip: Always ensure the sample data is representative of the population when constructing confidence intervals to avoid misleading conclusions.