Question

The following data were drawn from a normal population. Find a 98.4% confidence interval for the mean.

7 21 20 8 14 12 18 14 9 23

Answer 1

Solution:

Given that,

x	dx = x -A, A = 15	dx2
7	-8	64
21	6	36
20	5	25
8	-6	36
14	-1	1
12	-3	9
18	3	9
14	-1	1
9	-6	36
23	8	64
x = 146	dx = - 4	dx2 = 294

a ) The sample mean is

Mean   =  ( x/ n)

= (7 + 21+ 20+ 8 +14 +12 +18 +14 +9 +23 / 10 )

= 146 /10

= 14.6

Mean = 14.6

The population standard deviation is

= ( dx² ) - (( dx )² / n ) / n )

= ( 294 ( ( - 4 )² / 10 ) / 10

   = ( 294 -1.6 / 10 )

= (292.4 / 10 )

= 29.24

= 5.4

The population standard deviation is = 5.4

= 14.6

= 5.4

n = 10

At 98.4% confidence level the z is ,

  = 1 - 98.4% = 1 - 0.984 = 0.016

/ 2 = 0.016 / 2 = 0.008

Z_/2 = Z_0.008 = 2.409

Margin of error = E = Z_/2* (/n)

= 2.409 * (5.4 / 10 )

= 4.1

At 98.4% confidence interval estimate of the population mean is,

- E < < + E

14.6- 4.1 < < 14.6 + 4.1

10.5 < < 18.7

(10.5,18.7 )