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)

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.