Question

A random sample of 20 purchases showed the amounts in the table​ (in $). The mean is ​$49.71 and the standard deviation is ​$20.12 ​a) Construct a 99​% confidence interval for the mean purchases of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be ​$21?

40.97 68.94 75.05 61.69 84.52 15.84 63.79 30.72 57.29 43.93 51.53 88.75 44.14 51.94 38.04 24.95 32.42 58.63 26.71 34.41

​a) What is the confidence​ interval?

​(___ ​,____​)

Answer 1

Solution

Given that,

= 49.71 ....... Sample mean

s = 20.12 ........Sample standard deviation

n = 20   ....... Sample size

Note that, Population standard deviation() is unknown..So we use t distribution. Our aim is to construct 99% confidence interval.

c = 0.99

= 1- c = 1- 0.99 = 0.01

  /2 = 0.01 2 = 0.005

Also, n = 20

d.f= n-1 = 19

     =     = 2.861

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

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

  

49.71 - 2.861 *( 20.12 / 20) 49.71 + 2.861 *( 20.12 / 20)

Required interval is  (36.839,62.581)

b)Margin of error =  

=  2.861 *( 20.12 / 20)

= 12.8715

c) Now , the population SD is known . So , we use z distribution

= 21

Our aim is to construct 99% confidence interval.

c = 0.99

= 1- c = 1- 0.99 = 0.01

  /2 = 0.01 2 = 0.005 and 1- /2 = 0.995

Search the probability 0.995 in the Z table and see corresponding z value

= 2.576   

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

  

  

49.71 - 2.576 *( 21 / 20) 49.71 + 2.576 *( 21 / 20)

Required interval is (37.615,61.805)