Question

Using the info below, answer the next following questions:

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.

(a) Determine the sample mean in cents (Round to 3 decimal places)

(b) Determine the standard deviation from the sample . (Round to 3 decimal places)

(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? ( Round to 3 decimal places )

(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:

x	x2
20	400
75	5625
50	2500
65	4225
30	900
55	3025
40	1600
40	1600
30	900
55	3025
1.5	2.25
40	1600
65	4225
40	1600
x=606.5	x2=31227.25

a )The sample mean is

Mean     = (x / n) )

= (20+75+50+65+30+55+40+40+30+55+1.50+40+65+40/ 14 )

= 606.5/ 14

= 43.3214

Mean     = 43.321

b )The sample standard is S

  S = ( x² ) - (( x)² / n ) n -1

= (31227.25( (39.27)² / 14 ) 13

   = ( 31227.25- 26274.4464 / 13)

= (4952.8036/ 13)

= 380.6849

= 19.5188

The sample standard is = 19.519

Degrees of freedom = df = n - 1 = 14- 1 = 13

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,13 = 2.16

Margin of error = E = t_/2,df * (s /n)

= 2.16* (19.519/ 14)

= 11.268

Margin of error = 11.268

The 95% confidence interval estimate of the population mean is,

- E < < + E

43.321 - 11.268 < < 43.321 + 11.268

32.051< < 54.589

e )The lower bound = 32.051

f )The upper bound = 54.589