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MULTIDIMENSIONAL DATA: SOME WORDS ON NOTATION 581<br />

Box 25.16<br />

Reliability coefficients for peer descriptions<br />

Liked peers<br />

Disliked peers<br />

Dimension Younger Older Younger Older<br />

subjects subjects subjects subjects<br />

Descriptiveness 0.83 0.91 0.80 0.84<br />

Personal 0.76 0.80 0.84 0.77<br />

involvement<br />

Depth 0.65 0.71 0.65 0.75<br />

Evaluative<br />

consistency<br />

0.69 0.92 0.76 0.69<br />

Box 25.19<br />

Expected frequencies in sex, voting preference and<br />

social class<br />

Middle class<br />

Working class<br />

Conservative Labour Conservative Labour<br />

Men 55.4 61.7 77.0 85.9<br />

Women 53.4 59.5 74.3 82.8<br />

Source:adaptedfromWhiteley1983<br />

Chapter 25<br />

Source:PeeversandSecord1973<br />

Box 25.17<br />

Sex, voting preference and social class: a three-way<br />

classification table<br />

Middle class<br />

Working class<br />

Conservative Labour Conservative Labour<br />

Men 80 30 40 130<br />

Women 100 20 40 110<br />

Source:adaptedfromWhiteley1983<br />

Box 25.18<br />

Sex, voting preference and social class: a three-way<br />

notational classification<br />

Middle class<br />

Working class<br />

Conservative Labour Conservative Labour<br />

Men n 111 n 121 n 112 n 122<br />

Women n 211 n 221 n 212 n 222<br />

working class we assume that these variables are<br />

statistically independent (that is to say, there is no<br />

relationship between them) and simply apply the<br />

multiplication rule of probability theory:<br />

p 222 =(p 2++ )(p +2+ )(p ++2 )<br />

=(0.49)(0.53)(0.58) = 0.15<br />

This can be expressed in terms of the expected<br />

frequency in cell n 222 as:<br />

N(p 2++ )(p +2+ )(p ++2 ) = 550 (0.49)(0.53)(0.58)<br />

= 82.8<br />

Similarly, the expected fre quency in cell n 112 is:<br />

and<br />

and<br />

N(p 1++ )(p +1+ )(p ++2 )where :<br />

p 1++ = n 1++<br />

n +++<br />

= 280<br />

550 = 0.51<br />

p +1+ = n +1+<br />

n +++<br />

= 260<br />

550 = 0.47<br />

p ++2 = n ++2<br />

n +++<br />

= 320<br />

550 = 0.58<br />

Thus N(p 1++ )(p +1+ )(p ++2 )<br />

= 550 (0.51)(0.47)(0.58) = 77.0<br />

Box 25.19 gives the expected frequencies for the<br />

data shown in Box 25.17.<br />

With the observed frequencies and the expected<br />

frequencies to hand, chi-square is calculated in the<br />

usual way:<br />

χ 2 = ∑ (O − E) 2<br />

= 159.41<br />

E<br />

Whiteley (1983) observes that degrees of freedom<br />

in a three-way contingency table is more complex<br />

than in a 2 × 2classification.Essentially,however,<br />

degrees of freedom refer to the freedom with which<br />

the researcher is able to assign values to the cells,<br />

given fixed marginal totals. This can be computed

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