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: 13.66, 41.35, 21.43, 10.49, 25.57, 37.04, 17.5, 27.07. 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
13.66	186.5956
41.35	1709.8225
21.43	459.2449
10.49	110.0401
25.57	653.8249
37.04	1371.9616
17.5	306.25
27.07	732.7849
x=194.11	x2=5530.5245

The sample mean is

Mean    = (x / n) )

= (13.66+ 41.3 + 21.43+ 10.49+ 25.57+ 37.04 +17.5+ 27.07 / 8 )

= 194.11 / 8

= 24.2637

Mean    = 24.26

The sample standard is S

  S = ( x² ) - (( x)² / n ) n -1

= (5530.5245 ( (187.21 )² / 8 ) 7

   = ( 5530.5245- 4709.8365 / 7)

= (820.688 / 7 )

= 117.2411

= 10.8278

The sample standard is 10.83

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 * (10.83/ 8)

= 11.479

Margin of error = 11.479

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

- E < < + E

24.26 - 11.479 < < 24.26 + 11.479

112.781< < 35.739

(12.781, 35.739 )