Question

A sample of 18 observations taken from a normally distributed population produced the following data:

28.5	27.5	25.3	25.3	31.3	23.3	26.2	24.5	28.3
37.4	23.9	28.8	27.5	25.5	27.1	25.5	22.7	22.7

Round your answers to three decimal places.

a. What is the point estimate of μ?

x¯=  

b. Make a 90% confidence interval for μ.

c. What is the margin of error of estimate for μ in part b?

E=

Answer 1

Solution :

We are given a data of sample size n = 18

28.5,27.5,25.3,25.3,31.3,23.3,26.2,24.5,28.3,37.4,23.9,28.8,27.5,25.5,27.1,25.5,22.7,22.7

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

=   

= (28.5 + 27.5.......+ 22.7)/18

= 26.739

Now ,

s=   

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

s = 3.519

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, d.f = n - 1 = 17

    =    =  0.05,17 = 1.74

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

The margin of error is given by

E =  /2,d.f. * ( / n)

= 1.74* ( 3.519/ 18 )

E = 1.443

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

( - E ) <   <  ( + E)

( 26.739 - 1.443 )   <   <  ( 26.739 + 1.443)

25.296 <   < 28.182

Required 90% Confidence interval is ( 25.296 , 28.182 )

The point estimate of μ = 26.739

The margin of error E = 1.443