Question

The following data represent the concentration of dissolved organic carbon​ (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts​ (a) through​ (c) on the right. 15.42 29.80 27.10 16.51 10.30 8.81 10.30 20.46 14.90 33.67 30.91 14.86 11.40 15.35 9.72 19.80 14.86 8.09 5.30 18.30 ​(a) Find the sample mean. The sample mean is nothing . ​(Round to two decimal places as​ needed.) ​(b) Find the sample standard deviation. The sample standard deviation is nothing . ​(Round to two decimal places as​ needed.) ​(c) Construct a 95 ​% confidence interval for the population mean mu . The 95 ​% confidence interval for the population mean mu is ​(nothing ​,nothing ​,). ​(Round to two decimal places as​ needed.)

Answer 1

Solution:

We are given a data of sample size n=20.

15.42 29.80 27.10 16.51 10.30 8.81 10.30 20.46 14.90 33.67 30.91 14.86 11.40 15.35 9.72 19.80 14.86 8.09 5.30 18.30

a)Find sample mean

=   

=

=

= 16.793= 16.80

The sample mean is 16.80.

b)Find sample SD

s=   

=

= 8.06

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

The sample standard deviation is s= 8.06

c)Confidence interval for population mean() using t distribution  

Given that,

= 16.80 ....... Sample mean

s = 8.06 ........Sample standard deviation

n = 20 ....... Sample size

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, n = 20

d.f= n-1 = 19

     =    = = 2.093

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

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

  

16.80 - 2.093*(8.06/ 20)    16.80 + 2.093*(8.06/ 20)

16.80-3.77 < < 16.80+3.77

13.03 < < 20.57

The 95 ​% confidence interval for the population mean mu is ​(13.03,20.57)