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?

0.51  0.70  0.10  0.97  1.37  0.55  0.87   

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

Answer 1

Solution:

We are given a data of sample size n = 7

0.51 , 0.70 , 0.10 , 0.97 , 1.37 , 0.55 , 0.87   

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

=   

= (0.51 + 0.70 + ....+ 0.87)/7

= 0.72428571

Now ,

s=   

Using given data, find Xi- for each term.take squre for each.then we can easily find s.

s= 0.40107593

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, n = 7

d.f= n-1 = 6

     =    =     = 1.943

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

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

  

0.72428571 - 1.943*(0.40107593/ 7)    0.72428571 + 1.943*(0.40107593/ 7)

-0.295   < < 0.295

Required interval is (0.4297 , 1.0188)

A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm).

Value 1 fall in this interval .

Does it appear that there is too much mercury in tuna​ sushi? ANSWER is YES