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?

0.51  

0.78  

0.10 

0.90  

1.35  

0.59  

0.96

What is the confidence interval estimate of the population mean muμ​?

ppmless than<muμless than<nothing ppm

​(Round to three decimal places as​ needed.)

Answer 1

99% Confidence interval estimate for the mean amount of Hg in the population:

From the data: , $\bar{x}$ = 0.7414, s = 0.3943, n = 7

Since population standard deviation is unknown, the tcritical (2 tail) for $\alpha$ = 0.01, for df = n -1 = 6, is 3.7074

The Confidence Interval is given by $\bar{x}$ $\pm$ ME, where

0.3430.5525 ME = tcritical *-= 3.7074 * = 0.5525 Vn

The Lower Limit = 0.7414 - 0.5525 = 0.1889

The Upper Limit = 0.7414 + 0.5525 = 1.294

The 99% Confidence interval (0.1889 < $\mu$ < 1.294)

Since the confidence interval contains the value less than 1 and values greater than 1, then the null hypothesis cannot be rejected. Therefore there is insufficient evidence that the mercury levels is more than 1 part per million.

______________________________________________________________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)².

#	X	Mean	(x - mean)2
1	0.51	0.7414	0.05354596
2	0.78	0.7414	0.00148996
3	0.1	0.7414	0.41139396
4	0.9	0.7414	0.02515396
5	1.35	0.7414	0.37039396
6	0.59	0.7414	0.02292196
7	0.96	0.7414	0.04778596
Total	5.19		0.9326857

n	7
Sum	5.19
Average	0.7414
SS(Sum of squares)	0.93268572
Variance = SS/n-1	0.155
Std Dev=Sqrt(Variance)	0.3943