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:


19, 43, 12, 18, 19, 33, 20, 14.

Find the upper bound of a 95% 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:

given :

n = 8

  $x_{i}$ = 19, 43, 12, 18, 19, 33, 20, 14

mean $x^-$ = (19 + 43 + 12 +18+ 19 + 33 + 20 + 14 )/8 = 22.25 // ($x^-$ = x bar)

x   

19 10.562

43 430.562

12 105.062

18 18.062

19 10.562

33 115.562

20 5.062

14 68.062

$\Sigma x$ 178 $\Sigma (x - x^-)^2$ = 763.496

standard deviation s =  $\sqrt{\Sigma (x - x^-)^2/ n-1}$

= $\sqrt{763.496/ 7}$   

= 10.443

: using t distribution since n < 30

from the t table we get the t value for 95% confidence interval and 7 degrees of freedom = 2.365

upper bound =  $x^- + t * (s/\sqrt n)$

= 22.25 + 2.365 * ( 10.443 / 2.828)

= 30.983