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 95​% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi? 0.61  0.78  0.10  0.88  1.27  0.56  0.91 What is the confidence interval estimate of the population mean mu​?

Answer 1

Solution:

x	x2
0.61	0.3721
0.78	0.6084
0.1	0.01
0.88	0.7744
1.27	1.6129
0.56	0.3136
0.91	0.8281
x = 5.11	x2 = 4.5195

The sample mean is

Mean   = (x / n) )

= (0.61+0.78+0.10+0.88+1.27+0.56+0.91/ 7 )

= 5.11 / 4

= 15.5

Mean   = 0.73

c ) The sample standard is S

  S = ( x² ) - (( x)² / n ) n -1

= (4.5195 ( (5.11 )² / 7 ) 6

   = ( 4.5195 - 3.7303 / 6)

= (0.7892 / 6 )

= 0.1315

= 0.3627

The sample standard is = 0.36

Degrees of freedom = df = n - 1 = 7 - 1 = 6

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,6 =2.447

Margin of error = E = t_/2,df * (s /n)

= 2.447 * (0.36 / 7)

= 0.33

Margin of error = 0.33

The 95% confidence interval estimate of the population mean is,

- E < < + E

0.73 - 0.50 < < 0.73 + 0.33

0.30 < < 1.06

(0.30, 1.06 )