Question

A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal.

(e) Construct a 95% confidence interval for the population mean worth of coupons.  Use a critical value of 2.16 from the t distribution.

What is the lower bound?

(f)  Construct a 95% confidence interval for the population mean worth of coupons .

What is the upper bound? ( Round to 3 decimal places )

Answer 1

Solution:

We are given a data of sample size n = 14

20, 75, 50, 65, 30,55,40,40,30,55,1.50,40,65,40

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

=   

= 43.3214

Now ,

s=   

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

s= 19.5188

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, d.f = n - 1 = 13

    =    =  _0.025,13 = 2.16

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

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.16 * (19.5188 / 14 )

= 11.2679

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

( - E ) <   <  ( + E)

(43.3214 - 11.2679)   <   <  (43.3214 + 11.2679)

32.053 <   <  54.619

Answer :

e) What is the lower bound? =  32.053

f) What is the upper bound? =  54.619