A food safety guideline is that the mercury in fish should be below 1 part per m

Question

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. Does it appear that there is too much mercury in tuna sushi? 6. 0.51 0.71 0.09 0.93 1.34 0.49 0.88 What is the confidence interval estimate of the population mean ? ppm (Round to three decimal places as needed.) Does it appear that there is too much mercury in tuna sushi? O A. No, because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe. O B. Yes, because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values exceeds 1 ppm, so at least some of the fish have too much mercury. C. Yes, because it is possible that the mean is greater than 1 ppm. Also, at least one of the sample values exceeds 1 ppm, so at least some of the fish have too much mercury O D. No, because it is not possible that the mean is greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe

Explanation / Answer

Solution:- Given that values: 0.51,0.71,0.09,0.93,1.34,0.49,0.88

mean = sum of terms/number of terms = 4.95/7 = 0.7071

sd = = 0.3973

n =7 , df = n-1 = 7-1 = 6

t0.025 = 2.447

95% confidence interval of the population mean :

=> X +/- t*s/sqrt(n)

=  0.7071 +/- 2.447*(0.3973/sqrt(7))

=  0.340 , 1.075 (rounded)

=> option B.

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