Question

Dr. Mack Lemore, an expert in consumer behavior, wants to estimate the average amount of money that people spend in thrift shops. He takes a small sample of 8 individuals and asks them to report how much money they had in their pockets the last time they went shopping at a thrift store. Here are the data:


14.73,   28.89, 27.73, 16.35,   21.56,   22.34,   28.73,   26.88.

Find the lower bound of a 98% confidence interval for the true mean amount of money individuals carry with them to thrift stores, to two decimal places. Take all calculations toward the final answer to three decimal places.

Answer 1

Solution:

x	x2
14.73	216.9729
28.89	834.6321
27.73	768.9529
16.35	267.3225
21.56	464.8336
22.34	499.0756
28.73	825.4129
26.88	722.5344
x=187.21	x2=4599.7369

The sample mean is

Mean   = (x / n) )

= (14.73 + 28.89 +27.73 +16.35 + 21.56 + 22.34 + 28.73 + 26.88 / 8 )

= 187.21 / 8

= 23.4012

Mean   = 23.401

The sample standard is S

  S = ( x² ) - (( x)² / n ) n -1

= (4599.7369 ( (187.21 )² / 8 ) 7

   = ( 4599.7369 - 4380.948 / 7)

= (218.7889 / 7 )

= 31.2556

= 5.5907

The sample standard is 5.591

Degrees of freedom = df = n - 1 = 8 - 1 = 7

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,7 =2.998

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

= 2.998 * (5.591/ 8)

= 5.926

Margin of error = 5.926

The 98% confidence interval estimate of the population mean is,

- E < < + E

23.401 - 5.926 < < 23.401 + 5.926

17.475 < < 29.327

(17.475, 29.327 )