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 90​% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi? 0.54  0.82  0.09  0.96  1.28  0.54  0.96

What is the confidence interval estimate of the population mean?

Use this information to draw an appropriate conclusion about whether there could be too much mercury in tuna sushi

Answer 1

Solution

Solution:

We are given a data of sample size n = 7

0.54  0.82  0.09  0.96  1.28  0.54  0.96

Using this, first we find sample mean() and sample standard deviation(s).

=   

= (0.54 + 0.82 + 0.09 + 0.96 + 1.28 + 0.54 + 0.96)/7

= 0.7414

Now ,

s=   

Using given data, find Xi - for each term.Take square for each.Then we can easily find s.

s= 0.38671387

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

Our aim is to construct 90% confidence interval.   

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05

Also, d.f = n - 1 = 6  

    =    =  _0.05,6 = 1.943

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

The margin of error is given by

E =  /2,d.f. * ( / n)

= 1.943 * ( 0.38671387 / 7 )

=  0.2840

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

( - E ) <   <  ( + E)

(0.7414 - 0.2840)   <   <  (0.7414 + 0.2840)

0.4574  <   <  1.0254

Required interval is (0.4574 , 1.0254)

Now , it is given that "a food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm)""

The value ' 1 ' lie in the interval . Upper limit of the interval is greater than 1 . So we can say that  there could be too much mercury in tuna sushi"