Question

Problem 2

The D-Mart department store has sampled 225 sales records on July 14. The average sale was Yen4200, with an observed sample standard deviation of Yen3000 per customer.

(a) (10 points)

Construct a 90% confidence interval for the mean sale value.

(b) (10 points)

How many sales records would need to be sampled for the 90% confidence interval to be within Yen100 of the sample mean?

Answer 1

Solution:

Solution:

Given that,

n = 225

= 4200   

s = 3000

a)  

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 = 225 - 1 = 224

    =    =  _0.05,224 = 1.652

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

The margin of error is given by

E =  /2,d.f. * ( / n)

= 1.652 * (3000 / 225)

= 330.4

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

( - E ) <   <  ( + E)

(4200 - 330.4)   <   <  (4200 + 330.4)

3869.6 <   <  4530.4

(3869.6 , 4530.4)

b)

E = 100

Take sample SD as a estimate of population SD

So ,

= 3000

c = 90% = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05

Using Z table ,

= 1.645

Now, sample size (n) is given by,

=  {(1.645 * 3000)/ 100}²

=  2435.4225

= 2436 ..(round to the next whole number)

n = 2436