Page:Biometrika - Volume 6, Issue 1.djvu/15

This page has been proofread, but needs to be validated.

By Student

15

Standard deviation of standard deviations:—

	Calculated	.8556	±.015
	Observed	.9066
Difference=+		.0510

Comparison of Fit. Theoretical Equation: $y={\frac {16\times 750}{{\sqrt {2\pi }}\sigma ^{3}}}x^{2}e^{-{\frac {2x^{2}}{\sigma ^{2}}}}$ .

Scale in terms
of standard
deviation of
population

0 to .1

.1 to .2

.2 to .3

.3 to .4

.4 to .5

.5 to .6

.6. to .7

.7 to .8

.8 to .9

.9. to 1.0

1.0 to 1.1

1.1 to 1.2

1.2 to 1.3

1.3 to 1.4

1.4 to 1.5

1.5 to 1.6

1.6 to 1.7

Greater
than 1.7

Calculated
frequency

112

1012

27

4512

6412

7812

87

88

8112

71

58

45

33

23

15

912

512

7

Observed
frequency

3

1412

2412

3712

107

67

73

77

7712

64

5212

4912

35

28

1212

9

1112

7

Difference

+112

+4

−212

−8

+4212

−1112

−14

−11

−4

−7

−512

+412

+2

+5

−212

−12

+6

0

whence $\chi ^{2}=48.06$ , $P=.000,06$ (about).

In tabling the observed frequency, values between .0125 and .0875 were included in one group, while between .0875 and .0125 they were divided over the two groups. As an instance of the irregularity due to grouping I may mention that there were 31 cases of standard deviations 1.30 (in terms of the grouping) which is .5117 in terms of the standard deviation of the population, and they were therefore divided over the groups .4 to .5 and .5 to .6. Had they all been counted in groups .5 to .6 $\chi ^{2}$ would have fallen to 29.85 and $P$ would have risen to .03. The $\chi ^{2}$ test presupposes random sampling from a frequency following the given law, but this we have not got owing to the interference of the grouping.

When, however, we test the $z$ ’s where the grouping has not had so much effect we find a close correspondence between the theory and the actual result.

There were three cases of infinite values of $z$ which, for the reasons given above, were given the next largest values which occurred, namely $+6$ or $-6$ . The rest were divided into groups of .1; .04, .05 and .06, being divided between the two groups on either side.

The calculated value for the standard deviation of the frequency curve was $1(\pm .017)$ while the observed was 1.039. The value of the standard deviation is really infinite, as the fourth moment coefficient is infinite, but as we have arbitrarily limited the infinite cases we may take as an approximation ${\frac {1}{\sqrt {1500}}}$ from which the value of the probable error given above is obtained. The fit of the curve is as follows:—