Question

A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.88 and the standard deviation is $20.86.

B) What is the margin of error ?A) Construct a 80% confidence interval for the mean purchases of all customers, assuming that assumptions and conditions for confidence interval have been met. (What is the confidence interval) ?

C) How would the confidence interval change if you had assumed that the standard deviation was known to be $21 ?

Answer 1

Solution:

Given,

= 49.88

= 20.86

n = 20

Note that, Population standard deviation() is known. So we use z distribution.

Our aim is to construct 80% confidence interval.

c = 0.80

= 1- c = 1- 0.80 = 0.20

  /2 = 0.20 2 = 0.10

Search the probability 0.950 in the Z table and see corresponding z value

  = 1.282

The margin of error is given by

E =  /2 * ( / n )

= 1.282 * ( 20.86/ 20 )

= 5.98

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

( - E ) <   <  ( + E)

( 49.88 - 5.98 )   <   <  ( 49.88 + 5.98 )

43.902 <   < 55.858

Required 80% confidence interval is ( 43.902 ,55.858 )

Then Solution:

Given,

= 49.88

= 21

n = 20

Note that, Population standard deviation() is known. So we use z distribution.

Our aim is to construct 80% confidence interval.

c = 0.80

= 1- c = 1- 0.80 = 0.20

  /2 = 0.20 2 = 0.10

Search the probability 0.950 in the Z table and see corresponding z value

  = 1.282

The margin of error is given by

E =  /2 * ( / n )

= 1.282 * ( 21/ 20 )

= 6.02

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

( - E ) <   <  ( + E)

( 49.88 - 6.02 )   <   <  ( 49.88 + 6.02 )

43.860 <   < 55.898

Required 80% confidence interval is ( 43.860 ,55.898 )