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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 a \( 95 \% \) confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi? 0.590 .750 .090 .951 .350 .550 .91 What is the confidence interval estimate of the population mean \( \mu \) ? \( \square \mathrm{ppm}<\mu<\square \mathrm{ppm} \) (Round to three decimal places as needed)

Ask by Gough Dawson. in the United States
Mar 15,2025

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The 95% confidence interval for the mean mercury level is \(0.314\,\text{ppm} < \mu < 0.884\,\text{ppm}\). Since all values are below 1 ppm, it does not appear that there is too much mercury in tuna sushi.

Solución

1. Calculate the sample mean, \(\bar{x}\): \[ \bar{x} = \frac{0.590+0.750+0.090+0.951+0.350+0.550+0.910}{7}=\frac{4.191}{7}\approx0.599\text{ ppm} \] 2. Calculate the sample standard deviation, \(s\). First, compute the deviations and square them: \[ \begin{array}{rcl} (0.590-0.599)^2 &\approx& (-0.009)^2\approx 0.000081,\\[1mm] (0.750-0.599)^2 &\approx& (0.151)^2\approx 0.022801,\\[1mm] (0.090-0.599)^2 &\approx& (-0.509)^2\approx 0.259081,\\[1mm] (0.951-0.599)^2 &\approx& (0.352)^2\approx 0.123904,\\[1mm] (0.350-0.599)^2 &\approx& (-0.249)^2\approx 0.062001,\\[1mm] (0.550-0.599)^2 &\approx& (-0.049)^2\approx 0.002401,\\[1mm] (0.910-0.599)^2 &\approx& (0.311)^2\approx 0.096721. \end{array} \] Sum these values: \[ 0.000081+0.022801+0.259081+0.123904+0.062001+0.002401+0.096721\approx0.567 \] Then, the sample variance is: \[ s^2=\frac{0.567}{7-1}=\frac{0.567}{6}\approx0.0945 \] Taking the square root: \[ s\approx\sqrt{0.0945}\approx0.3075\text{ ppm} \] 3. Compute the \(95\%\) confidence interval for the population mean. The formula is: \[ \bar{x}\pm t_{\frac{\alpha}{2},\,n-1}\frac{s}{\sqrt{n}} \] With \(n=7\) and degrees of freedom \(n-1=6\), the critical \(t\)-value for a \(95\%\) confidence level (i.e. \(\alpha=0.05\)) is approximately: \[ t_{0.025,6}\approx 2.447 \] 4. Calculate the margin of error: \[ E = 2.447\cdot\frac{0.3075}{\sqrt{7}} \] Since \(\sqrt{7}\approx2.6458\), \[ E\approx2.447\cdot\frac{0.3075}{2.6458}\approx2.447\cdot0.1163\approx0.285\text{ ppm} \] 5. Determine the lower and upper bounds: \[ \text{Lower bound} = 0.599 - 0.285 \approx 0.314\text{ ppm}\\[1mm] \text{Upper bound} = 0.599 + 0.285 \approx 0.884\text{ ppm} \] Rounded to three decimals: \[ 0.314\,\text{ppm} < \mu < 0.884\,\text{ppm} \] Since the food safety guideline is that mercury should be below 1 ppm, and the entire confidence interval is below 1 ppm, it does not appear that there is too much mercury in tuna sushi.

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Bonus Knowledge

To determine the confidence interval estimate of the population mean \( \mu \) for the mercury levels found in tuna sushi, we will first calculate the sample mean and the standard deviation. Then, we will use the t-distribution to construct the confidence interval. ### Step 1: Calculate the Sample Mean (\( \bar{x} \)) Given the mercury amounts: 0.590, 0.750, 0.090, 0.951, 0.350, 0.550, 0.910 \[ \bar{x} = \frac{0.590 + 0.750 + 0.090 + 0.951 + 0.350 + 0.550 + 0.910}{7} = \frac{3.191}{7} \approx 0.456 \] ### Step 2: Calculate the Sample Standard Deviation (s) First, we find the variance: \[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1} \] \[ s^2 = \frac{(0.590 - 0.456)^2 + (0.750 - 0.456)^2 + (0.090 - 0.456)^2 + (0.951 - 0.456)^2 + (0.350 - 0.456)^2 + (0.550 - 0.456)^2 + (0.910 - 0.456)^2}{7 - 1} \] Calculating each term: - \( (0.590 - 0.456)^2 \approx 0.017856 \) - \( (0.750 - 0.456)^2 \approx 0.086016 \) - \( (0.090 - 0.456)^2 \approx 0.134956 \) - \( (0.951 - 0.456)^2 \approx 0.244036 \) - \( (0.350 - 0.456)^2 \approx 0.011556 \) - \( (0.550 - 0.456)^2 \approx 0.008884 \) - \( (0.910 - 0.456)^2 \approx 0.204196 \) Summing these: \[ \sum = 0.017856 + 0.086016 + 0.134956 + 0.244036 + 0.011556 + 0.008884 + 0.204196 \approx 0.607036 \] Now calculate the variance: \[ s^2 = \frac{0.607036}{6} \approx 0.1011727 \] \[ s = \sqrt{0.1011727} \approx 0.318 \] ### Step 3: Find the t-value for a \( 95 \% \) Confidence Level The degrees of freedom (df) is \( n - 1 = 7 - 1 = 6 \). For \( 95\% \) confidence and \( df = 6 \), the critical t-value (from t-table) is approximately \( 2.447 \). ### Step 4: Calculate the Margin of Error The margin of error (ME) is calculated as: \[ ME = t \cdot \frac{s}{\sqrt{n}} = 2.447 \cdot \frac{0.318}{\sqrt{7}} \approx 2.447 \cdot 0.120 \approx 0.29364 \] ### Step 5: Construct the Confidence Interval Finally, we calculate the confidence interval: \[ \bar{x} - ME < \mu < \bar{x} + ME \] \[ 0.456 - 0.29364 < \mu < 0.456 + 0.29364 \] \[ 0.16236 < \mu < 0.74964 \] ### Rounded to three decimal places: \[ \boxed{0.162 \, \mathrm{ppm} < \mu < 0.750 \, \mathrm{ppm}} \] ### Conclusion Since the upper limit of the confidence interval (0.750 ppm) is below the threshold of 1 ppm, it does not appear that there is too much mercury in tuna sushi based on this sample.

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