An article in Human Factors (June 1989) presented data on visual accommodation (a function of eye...

An article in Human Factors (June 1989) presented data on visual accommodation (a function of eye movement) when recognizing a speckle pattern on a high-resolution CRT screen. The data are as follows: 36.45, 67.90, 38.77, 42.18, 26.72, 50.77, 39.1, and 51.48. Calculate the sample mean and sample standard deviation. Round your answers to 2 decimal places.

Sample mean =

Sample standard deviation =

Solutions

Expert Solution

Solution:

Given that,

x	dx	dx²
36.45	-6.55	42.9025
67.90	24.9	620.01
38.77	-4.23	17.8929
42.18	-0.82	0.6724
26.72	-16.28	265.0384
50.77	7.77	60.3729
39.1	-3.9	15.21
x = 301.89	dx = 0.89	dx² = 1022.0991

a ) The sample mean is

Mean = x / n

= 36.45+67.9+38.77+42.18+26.72+50.77+39.1 / 7

= 301.89 / 7

= 43.1271

Mean = 43.13

b ) The sample standard is S

S = ( dx² ) - (( dx )² / n ) / 1 -n )

= (1022.0991 ( (- 0.89 )² / 7 ) / 6

= ( 1022.0991 -0.1132 / 6 )

= ( 1021.9859 / 6 )

= 170.331

= 13.05

The sample standard is = 13.05

orchestra answered 1 year ago

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