Question

A clue to the amount of organic waste in a lake was the number of bacteria colonies in 100 milliliters of water. The number of colonies, in hundreds, for n = 30 samples of water from one portion of the lake yielded 93, 140, 8, 120, 3, 120, 33, 70, 91, 61, 7, 100, 19, 98, 110, 23, 14, 94, 57, 9, 66, 53, 28, 76, 58, 9, 73, 49, 37, 92 Construct an approximate 95% confidence interval for the mean number µ of colonies in 100 milliliters of water in this portion of the lake.

Answer 1

Solution:

Given the sample of size n = 30

93, 140, 8, 120, 3, 120, 33, 70, 91, 61, 7, 100, 19, 98, 110, 23, 14, 94, 57, 9, 66, 53, 28, 76, 58, 9, 73, 49, 37, 92

Using this, first we find sample mean($\bar x$) and sample standard deviation(s).

$\bar x$=   

sumofobservations

=   

93 +140 + .……. 492 30

$\bar x$ = 60.3667

Now ,

s=   

I-U z(x - x) 3

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

s= 39.6219

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, n = 30

$\therefore$ d.f= n-1 = 29

$\therefore$    
ta/2.0.f.
=   
ta/2,1-1
=
10.025,29
= 2.045

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

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

  

I-ta/2.n-1 * (s/ n) < < + ta/2.n-1 * (s/ n)

60.3667 - 2.045*(39.6219/ $\sqrt{}$ 30)   $< \mu <$ 60.3667 + 2.045*(39.6219/ $\sqrt{}$ 30)

60.3667 - 14.7934  < $\mu$ <  60.3667 + 14.7934

45.5733   < $\mu$ <  75.1601

is the required 95% confidence interval for mean....