This Question: 1 pt 47 of 17 (6 complete)> This Quiz: 17 pts possible A food saf
ID: 2949611 • Letter: T
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This Question: 1 pt 47 of 17 (6 complete)> This Quiz: 17 pts possible A food safety guideline is that the mercury in fish should be below 1 part per milion (ppm). Listed below are the amounts of meury (pp) found in tuna sushi sampled at different stores in a major city. Construct a 98% confidence interval estimate of the mean amount ofrmercury in the population. Does it appear that there is too much mercury in tuna sushi? 0.50 0.80 0.09 0.95 1.34 0.51 0.83 What is the confidence interval estimate of the population mean ?? ppm ppm Round to three decimal places as needed.) Does it appear that there is too much mercury in tuna sushi? OA. 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. 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 ? ? Yes, because it is possible that the mean is greater than 1 ppm. Also at least one of the sa mple values exceeds 1 ppm. so at least some of he fish have too much mercury. D. 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. Click to select your answer(s).Explanation / Answer
solution :
Given that 0.50,0.80,0.09,0.95,1.34,0.51,0.83
=> Mean x = 0.717
=> Mean = sum of terms/number of terms
= ( 0.50 + 0.80 + 0.09 + 0.95 + 1.34 + 0.51 + 0.83 )/7
= 0.717
=> Standard deviation s = 0.397
=> n = 7
=> df = n-1 = 6
=> For 98% confidence interval , t = 3.143
=> the 98% confidence interval of the population mean is
=> x +/- t*s/sqrt(n)
=> 0.717 +/- 3.143*0.397/sqrt(7)
=> 0.245 < ? < 1.189
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