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 99​% 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  0.72  0.10  0.88  1.35  0.56  0.97 What is the confidence interval estimate of the population mean mu​? nothing ppmless thanmuless than nothing ppm ​(Round to three decimal places as​ needed.)

Answer 1

Solution :

We are given a data of sample size n = 7

0.54,0.72,0.10,0.88,1.35,0.56,0.97

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

=   

= (0.54 + 0.72 +.......+ 0.97) / 7

= 0.731

Now ,

s=   

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

s = 0.393

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

Our aim is to construct 99% confidence interval.

   c = 0.99

= 1 - c = 1 - 0.99 = 0.01

  /2 = 0.01 2 = 0.005

Also, d.f = n - 1 = 6

    =    =  0.005,6 = 3.707

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

The margin of error is given by

E =  /2,d.f. * ( / n )

= 3.707 * ( 0.393 / 7 )

= 0.551

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

( - E ) <   <  ( + E)

( 0.731 - 0.551 )   <   <  ( 0.731+ 0.551 )

0.180 <   < 1.282

Required 99% confidence interval is ( 0.180 , 1.282 )