Question

A survey of several 11 to 13 year olds recorded the following amounts spent on a trip to the mall:

$28.43,$25.23,$23.98,$24.79,$29.05

Construct the 95% confidence interval for the average amount spent by 11 to 13 year olds on a trip to the mall. Assume the population is approximately normal.

Step 3 of 4 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Solution:

Given that,

x	dx =x -A	dx2
28.43	2.43	5.9049
25.23	-0.77	0.5929
23.98	-2.02	4.04804
24.79	-1.21	1.4641
29.05	3.05	9.3025
x = 131.48	(dx) = 1.48	(dx)2 = 21.3448

The sample mean is

Mean = (  x / n )

= ( 28.43 + 25.23 + 23.98 + 24.79 + 29.05 / 5 )

= ( 131.48 / 5 )

= 26.296

Mean = 26.30

The sample standard is S

S = ( dx² ) - (( dx )² / n ) / 1 -n )

= ( 21.3448 ( ( 1.48 )² / 5 ) / 4

   = (21.3448 - 0.4381 / 4 )

= (20.9067/ 4 )

= 5.2267

= 2.2862

The sample standard is = 2.29

= 26.30

s =2.29

n = 5

Degrees of freedom = df = n - 1 = 5 - 1 = 4

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.005,24 =2.777

The critical value = 2.777

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

= 2.777 * (2.29 / 5)

= 2.843

Margin of error = 2.843

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

- E < < + E

26.30- 2.843 < < 26.30+ 2.843

23.457 < < 29.143

(23.457, 29.143 )