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 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? 0.50 0.80 0.09 0.90 1.24 0.50 0.80 What is the confidence interval estimate of the population mean? ppm- ppm (Round to three decimal places as needed) Does it appear that there is too much mercury in una sushi? O A No, because it is not possible that the mean is greater than pom. Also, at least one of the sample vist than mo least some of the fish are so OB Yes, tecause it is possible that the mean is greater than om Alsoas one of the sample second most some of the fish have too much mercury OC No, because it is possible that the mean is not greater than 1 pom. Also, at least one of the sample vistes han pomo at least some of the fish are so OD. Yos, because it is possible that the mean is not greater than 1 ppm Als, at least one of the sample values crom, so last some of the fish have too much mercury

Answer 1

Solution:

Sample size = n = 7

0.59,0.80,0.09,0.90,1.24,0.50,0.89

Using calculator ,

sample mean = = 0.71571428571429

T

sample standard deviation = s = 0.36518749261428

Note that, Population standard deviation($\sigma$) is unknown. So we use t distribution.

Our aim is to construct 95% confidence interval.

$\therefore$ c = 0.95

$\therefore$$\alpha$ = 1- c = 1- 0.95 = 0.05

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

Also, d.f = n - 1 = 7 - 1 = 6

$\therefore$  
ta/2.d.f-
  =  
ta/2.n-1
  =  $url=https://latex.codecogs.com/gif.latex?t&t=1596154678&ymreqid=f4764063-0d40-e2b0-1cde-f00003012f00&sig=nsh.iU8gBt1m17juFlQCGQ--~D$_0.025,6 = 2.365

( use t table or t calculator to find this value..)

The margin of error is given by

E =  $url=https://latex.codecogs.com/gif.latex?t&t=1596154678&ymreqid=f4764063-0d40-e2b0-1cde-f00003012f00&sig=nsh.iU8gBt1m17juFlQCGQ--~D$$\alpha$/2,d.f. * ($url=https://latex.codecogs.com/gif.latex?s&t=1596154678&ymreqid=f4764063-0d40-e2b0-1cde-f00003012f00&sig=VkBTm5BZsBYP690PYvFc4g--~D$ / $\sqrt{}$ n )

=  2.365 * (0.36518749261428 / $\sqrt{}$ 7)

= 0.32643597923

Now , confidence interval for mean($\mu$) is given by:

($url=http://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1596154678&ymreqid=f4764063-0d40-e2b0-1cde-f00003012f00&sig=SDOdAXSmU0epkCUE0SD6Cw--~D$ - E ) <  $\mu$ <  ($url=http://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1596154678&ymreqid=f4764063-0d40-e2b0-1cde-f00003012f00&sig=SDOdAXSmU0epkCUE0SD6Cw--~D$ + E)

(0.71571428571429 - 0.32643597923)   <  $\mu$ <  (0.71571428571429 + 0.32643597923)

0.38927830648 <  $\mu$ <  1.04215026495

0.389 <  $\mu$ <  1.042

Answer :

0.389 ppm <  $\mu$ <  1.042 ppm

Does it appear that there is too much mercury ?

Option A. is correct.