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As we have seen, chi-square can be used with nominal-type data (so long as frequencies are assigned to the di€erent categories speci®ed). We have also seen that it can be used to test the di€erence between an actual sample and some hypothetical distribution, and also to test di€erences between two or more actual examples. Very few alternatives to chi-square exist for testing data that is nominal.

The fundamental equation for x2_{is given as follows}

x2ˆXn

iˆ1

…Oi Ei†2

Ei

where Oi are the observed values, Ei are the expected values and n is the

number of values. The reader must note the following rule very closely and be sure to take account of it in carrying out calculations. Whenever x2 _is

calculated from 1  2, or 2  2 tables, an adjustment known as Yates correction for continuity must be used. This involves subtracting 0.5 from the absolute value (ignoring the + or ± sign) of the numerator before it is squared on each x2_{calculation. Since the formula above must be calculated}

for each cell and then all of these values be added up, it would make some di€erence to the ®nal x2 _{value if the Yates correction were carelessly}

forgotten.

Thus, for the cases involving 1  2, or 2  2 tabular displays we have x2ˆXn

iˆ1

…jOi Eij 12†2

Ei

Three problems involving x2_{will now be carried out in full.}

Problem 1

A sociologist is faced with the data shown below, and he wishes to ascertain if the ®gures for 1970 di€er signi®cantly from ®gures for the previous ten years. The people in social classes A and B are from the same street.

Social class A Social class B

Number of felony convictions for 1970 45 15

Non-parametric statistics 125

Solution

He has observed data on 60 cases from the two social classes concerned in a given neighbourhood. Of these 60 people, he could expect 40% (or 24 of them) to be to be from social class A and 36 of them to be from social class B.

x2ˆ…j45 24j 12†2 24 ‡ …j15 36j 1 2†2 36 ˆ 17:5 ‡ 11:7 ˆ 29:2

Notice that this was a 1  2 table, so the Yates correction for continuity was used. In using the x2 _{table to see if our computed value was signi®cant, we}

have to know the number of degrees of freedom to use ± for a one-sample case, the number of cells less one. Since there are only two cells in this case, the number of degrees of freedom is 1. We now look up the chi-square table in AppendixI and observe that with one degree of freedom a x2_{value equal to}

or greater than 3.84 is needed to reject the null hypothesis at the 0.05 level of signi®cance, while at the 0.01 level of signi®cance, a value equal to or greater than 6.64 is sucient to reject H0. Since our computed value is 29.2, we can

certainly reject H0, even at the 0.01 level of signi®cance.

That means that the sociologist is justi®ed in believing that some new factor has entered the picture, causing social class A people to be convicted more often than before and/or causing social class B people to be convicted less often than before. After carrying out this test and obtaining a result like this, he might begin an investigation into the reasons behind this signi®cant change in conviction patterns.

Problem 2

This problem illustrates the use of chi-square in handling di€erences between two samples. A test of attitude toward a certain political question was administered to 100 soldiers who had voluntered to serve and to 100 soldiers who had been conscripted. The data obtained were as shown below.

Agree Neutral Disagree Total

Volunteers 40 40 20 100

Conscripts 60 20 20 100

126 Basic concepts in statistics and epidemiology

Do a chi-square test to see if volunteers di€er signi®cantly from conscripted soldiers with respect to the question concerned.

Solution

There are an equal number of conscripts and volunteers, so of the 100 who agree we would expect 50 from each category, of the 60 who were neutral we would expect 30 in each category and of the 40 who disagree we would expect 20 in each category ± if H0is to be upheld. Our calculation of x2, then, should

look as follows* x2ˆ…40 50†₅₀ 2‡…40 30†₃₀ 2‡…20 20†₂₀ 2‡ …60 50†2 50 ‡ …20 30†2 30 ‡ …20 20†2 20 ˆ 2:0 ‡ 3:3 ‡ 0 ‡ 2:0 ‡ 3:3 ‡ 0 ˆ 10:6

For two or more samples, the degrees of freedom necessary to use the chi- square table is equal to the number of rows minus one times the number of columns minus one, which is …2 1†…3 1† ˆ 2.

With 2 degrees of freedom, the x2_{table tells us that at the 0.01 level, values}

equal to or in excess of 9.210 are signi®cant. Since our computed value is 10.6, our results are signi®cant and we reject H0.

A word should be said here about the number of cells it is advisable to have in using chi-square. In general, the larger the number of cells or categories, the more sensitive will be the x2_{test ± so the more categories that we can have,}

the better. but if when working out the expected values, some cells have fewer than 5 in them as an expected frequency, that invalidates the test. That is, it does not matter if fewer than 5 are observed in a given category, as long as 5 or more are expected in each category. Now the question arises as to what to do if it happens that after an experimenter has set out his categories, his expected frequencies violate the rule in some categories.

In that case, we reduce the number of categories until the frequencies expected in each category exceed ®ve. For instance, in an attitude test like the one discussed in Problem 2, there might have been ®ve original choices for

* We do not need the Yates correction for continuity here because the tabular array is 3  2, not 1  2 or 2  2.

Non-parametric statistics 127

each question, namely: strongly agree, agree, neutral, disagree, strongly disagree. Then, when the expected values were calculated, it turned out that there were fewer than ®ve in some cells, the number of categories would be reduced by combing `strongly disagree' with `agree' and `disagree' with `strongly disagree'.

Problem 3

We shall now present and solve a problem involving more than two samples. In a plastics factory employing 1200 people there are three departments: moulding, tinting and assembling. The company psychologists have worked out about how many people from each department can be expected to have been absent in a year for the following periods: never, 1±3 days, 4±6 days, a week or more. In the data that follows the psychologists' ®gures are presented in parentheses, while the actual absence ®gures are given without parentheses. We wish to compare the two sets of ®gures.

Department Never 1±3 days 4±6 days A week or more Total

Moulding 400 (400) 100 (87.5) 50 (50) 50 (62.5) 600

Tinting 300 (266.6) 50 (58.3) 25 (33.3) 25 (41.7) 400

Assembling 100 (133.3) 25 (29.2) 25 (16.7) 50 (20.8) 200

Total 800 175 100 125 1200

Now the reader may wonder how the expected frequencies were calculated by the psychologists. This is important and we shall illustrate how it is done by considering the expected frequency of the upper right corner cell. You take the proportion between the ®rst row total and the total (which is 600

1200and

multiply that by the fourth column total (125)). This gives 125 ₁₂₀₀600 ˆ 62:5

Solution

128 Basic concepts in statistics and epidemiology x2ˆ…400 400†₄₀₀ 2‡…100 87:5†_87:5 2‡…50 50†₅₀ 2‡ …50 62:5†2 62:5 ‡ …300 266:6†2 266:6 ‡ …50 58:3†2 58:3 ‡ …25 33:3†2 33:3 ‡ …25 441:7†2 41:7 ‡ …100 133:3†2 133:3 ‡ …25 29:2†2 29:2 ‡ …25 16:7†2 16:7 ‡ …50 20:8†2 20:8 ˆ 0 ‡ 1:79 ‡ 0 ‡ 2:50 ‡ 4:16 ‡ 1:18 ‡ 2:07 ‡ 6:69 ‡ 8:32 ‡ 0:60 ‡ 4:13 ‡ 40:99 ˆ 72:43

The number of degrees of freedom involved is the number of rows less one multiplied by the number of columns less one. That gives …4 1†…3 1† ˆ 6 degrees of freedom. Even at the 0.001 level of signi®cance, the value we have obtained is signi®cant ± as discovered by using the chi-square table in the usual way.

Study exercises 6

1 The measurements of the heights (in inches) of sixadults of each of two di€erent nationalities are given below

Nationality A: 64, 67, 68, 65, 62, 61 Nationality B: 69, 70, 66, 63, 71, 72

Use Wilcoxon's two-sample test to determine whether there is a signi®cant di€erence in mean heights for these two nationalities (5% level).

2 In an experiment consisting of 25 pairs, the number of plus signs exceeds the number of minus signs. If a two-tailed test is applied, how many minus signs can there be, at most, to allow us to consider the results signi®cant on the basis of the 5% level of signi®cance?

3 In a paired feeding experiment, the gains in weight (in pounds) of subjects fed on two di€erent diets are as listed below. Can either diet be considered superior at the 5% level of signi®cance? (Use the Wilcoxon test.)

Non-parametric statistics 129

Pair Diet A Diet B

1 27 19 2 30 32 3 31 21 4 34 33 5 23 27 6 31 34 7 39 34 8 34 28 9 33 26 10 35 26

Chapter 10