In: Math
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.
Construct a 95% confidence interval for the population mean worth of coupons. Use a critical value of 2.16 from the t distribution.
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 |
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 = 43.32
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 = 19.52
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 = t0.025,13 = 2.16
Margin of error = E = t/2,df * (s /n)
= 2.16* (19.52/ 14)
= 11.27
Margin of error = 11.2
The 95% confidence interval estimate of the population mean is,
- E < < + E
43.32 - 11.2 < < 43.32 + 11.2
32.05< < 54.52
(32.05, 54.52 )