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.

0.53  

0.72  

0.11 

0.97 

1.32  

0.54 

0.90

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

ppmless than<muμless than<ppm

Answer 1

Solution =

We are given a data of sample size n = 7

0.53,0.72,0.11,0.97,1.32,0.54,0.90

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

=   

= (0.53 + 0.72.......+ 0.90)/7

= 0.727

Now ,

s=   

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

s = 0.386

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

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, d.f = n - 1 = 6

    =    =  0.025,6 = 2.447

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

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.447 * ( 0.386/ 7 )

= 0.357

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

( - E ) <   <  ( + E)

( 0.727 - 0.357 )   <   <  ( 0.727 + 0.357 )

0.370 ppm <   < 1.084 ppm

Required 95% confidence interval is ( 0.370 , 1.084 )