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 99% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi? 0.59 0.75 011 098 1.31 0.57 094 What is the confidence interval estimate of the population mean ? (Round to three decimal places as needed) Enter your answer in the edit fields and then click Check Answer part remaining

Explanation / Answer

Solution:

Given data 0.59, 0.75, 0.11, 0.98, 1.31, 0.57, 0.94
Sample Mean = 0.75
SD = 0.381
Sample Size (n) = 7
Standard Error (SE) = SD/sqrt(n) = 0.1440
alpha () = 1-0.99 = 0.01
we use t-distribution as population standard deviation is unknown
t(/2, n-1 ) = 3.7074
Margin of Error (ME) = t(/2,n-1)* SE = 0.5339

90% confidence interval is given by:
CI = Sample Mean ± (Margin of Error)
= 0.75 ± 0.5339
= (0.2161, 1.2839)

$0.2161 ppm < < $1.2839 ppm

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