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 ?

the confidence interval is 1.2282 and 0.269

Step-by-step 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

=

it's both, now can you answer my most recent question?

need more detail

step-by-step explanation:

lele