Question

A food safety guideline is that the mercury in fish should be below 1 part per million. Listed below are the amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 98% 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  0.68  0.10  0.98  1.37  0.55  0.97

Required:
What is the confidence interval estimate of the population mean μ ?

Answer 1

the confidence interval is 1.2282 and 0.269

Explanation:

n = 7

the mean of the sample =

0.59+0.68+0.10+0.98+1.37+0.55+0.97/7

= 5.25/7

= 0.74857= 0.7486

the standard dev!aton of the sample

= √(0.59-0.7486)²+(0.68-0.7486)²+(0.1-0.7486)²+(0.98-0.7486)²+(1.37-0.7486)²+(0.55-0.7486)²+(0.97-0.7486)²/n-1

= √0.02515+0.0047+0.4206+0.0535+0.3861+0.03944+0.0490/6

= √0.9786/6

= √0.163083

= 0.4038

degree of freedom = 7-1= 6

crtcal value at 2% significance and df of 6 = 3.143

confidence interval =

0.7486±3.143(0.4038/√7)

= 0.7486±3.143(0.1526)

0.7486+0.4796 = 1.2282

0.7486-0.4795 = 0.269

=