Question

The following sample data are from a normal population: 10, 9, 12, 14, 13, 11, 7, 4.

a. What is the point estimate of the population mean?

b. What is the point estimate of the population standard deviation (to 2 decimals)?

c. With 95% confidence, what is the margin of error for the estimation of the population mean (to 1 decimal)?

d. What is the 95% confidence interval for the population mean (to 1 decimal)?
( , )

Answer 1

Solution:

We are given a data of sample size n=10

10, 9, 12, 14, 13, 11, 7, 4

a) we know, the sample mean is the point estimate of the population mean.

=   

= (10+9+12+14+13+11+7+4)/8

= 80/8

= 10.0 is the point estimate of the population mean.

b)The sample standard deviation(s) is the point estimate of the population SD.

s=   

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

s= 3.30 is the point estimate of the population SD.

c)MARGIN OF ERROR FOR 95% CONFIDENCE

Given that,

= 10.0 ....... Sample mean

s = 3.30 ........Sample standard deviation

n = 8 ....... 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 = 8

d.f= n-1 = 7

     =    = = 2.365

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

  

=2.365*(3.30/8)

= 2.8

d)95% confidence interval for population mean

confidence interval for mean() is given by:

10.0-2.8 < < 10.0+2.8

7.2 < < 12.8

Answer is ( 7.2 , 12.8) is the required interval