Question

Determine the 95% confidence interval for the following scores: 83, 92, 81, 87, 79, 93, 88, 87, 83. Hint: You will need to determine the sample mean and standard deviation and then construct your confidence interval around these.

Answer 1

Solution:

Given that,

x	dx=x-A = x = 86	dx2
83	-3	9
92	6	36
81	-5	25
87	1	1
79	-7	49
93	7	49
88	2	4
87	1	1
83	-3	9
x =773	dx = -1	dx2 = 183

Mean =  x /n

=(83+92+81+ 87+ 79+ 93+ 88+ 87+ 83 / 8 )

= 733 / 9

= 85.8889

Mean = 86

Sample standard deviation is S

=  ( dx² - (( dx )² / n) n -1 )

= ( 183 ) - (( -1 )² / 9/8)

= (183- 0.1111 ) / 8

= 182.8889 / 8

= 22.8611

= 4.7813

Sample standard deviation is S = 5

= 86

s = 5

n = 8

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

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,7 = 2.365

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

= 2.365 * (5 8)

= 4.18

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

- E < < + E

86- 4.18< < 86+ 4.18

81.82 < < 90.18

(81.82 , 90.18 )