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? 0.57  0.81  0.11  0.96  1.24  0.58  0.80 What is the confidence interval estimate of the population mean mu​?

Answer 1

Solution:

Solution:

x	x2
0.57	0.3249
0.81	0.6561
0.11	0.0121
0.96	0.9216
1.24	1.5376
0.58	0.3364
0.8	0.64
$\sum$x = 5.07	$\sum$x2 = 4.4287

Mean $\bar x$    = ($\sum$x / n) )The sample mean is $\bar x$

= (0.57+0.81+0.11+0.96+1.24+0.58+0.80 / 7 )

= 5.07 / 7

= 0.7243

Mean $\bar x$    = 0.72

c ) The sample standard is S

  S = $\sqrt{}$ ( $\sum$ x² ) - (( $\sum$ x)² / n ) n -1

= $\sqrt{}$ (4.4287 ( (5.07 )² / 7 ) 6

   = $\sqrt{}$ (4.4287 - 3.6721 / 6)

=$\sqrt{}$ (0.7566 / 6 )

= $\sqrt{}$ 0.1261

= 0.3551

The sample standard is = 0.35

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

At 95% confidence level the t is ,

$\alpha$  = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,6 =2.447

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.447 * (0.35 / $\sqrt$ 7)

= 0.32

Margin of error = 0.32

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

$\bar x$ - E < $\mu$ < $\bar x$ + E

0.73 - 0.32 < $\mu$ < 0.72 + 0.32

0.41 < $\mu$ < 1.04

( 0.41. 104 )