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?

0.56, 0.71, 0.11, 0.96, 1.26, 0.50, 0.86

What is the confidence interval estimate of the population mean mu

μ​?

Answer 1

	x	(x-xbar)^2
	0.56	0.00018	Mean(x)=xbar=sum(x)/n	0.546571
	0.71	0.190595	standard deviation(s)=sum(x-xbar)^2/n-1	0.33309
	0.11	0.170923	n	7
	0.96	0.17688	for 90 % confidence level with degree of freedom (n-1)=13
	0.126	0.002169	=1-c%=1-0.99=0.01	0.01
	0.5	0.098237	degrres of freedom	6
	0.86	10.75465	t=critical value find using t-table with corresponding df=(n-1)	3.707428
sum	3.826	11.39364	Margin of error =t*s/sqrt(n)	0.466751
			LCL=xbar-ME	0.07982
			UCL=xbar+ME	1.013322

#

the confidence interval estimate of the population mean μ​ is

(xbar-ME<μ<xbar+ME)

(0.07982<μ<1.0133)