Methods of Communication Research and Statistics

This exercise uses data gathered by students on the statistics course in 2006. Where necessary, use parts of syntax that you have generated for previous exercises.

Database: MCO_1.2006.sav

a. Check whether there is an association between gender and preferred newspaper and, if so, discuss the nature and strength of the association.

In this example, there are two nominal variables; therefore we have to generate a contingency table in order to test the association.

Test this using a chi-square test, which you can request as a statistic from the contingency table. You can infer the strength of the association from an appropriate measure of association. The level of measurement of both variables here is nominal and there is an asymmetric association (gender can determine newspaper preference, but not vice versa), so we select Goodman & Kruskal's tau (check box for Lambda in SPSS).

All the cells appear to have an expected value of 5 or more, so the chi-square test may be applied. However, there does not appear to be any significant association, X² (8) = 9.09, p = .335.

Chi-Square Tests

Value df Asymp. Sig. (2-sided)

Pearson Chi-Square 9,089(a) 8 ,335

Likelihood Ratio 9,172 8 ,328

Linear-by-Linear Association 1,386 1 ,239

N of Valid Cases 1541

a 0 cells (,0%) have expected count less than 5. The minimum expected count is 17,99.

You therefore no longer need to look at the measures of association. However, we will do so for educational purposes, in order to see that Goodman & Kruskal’s tau is not significant. This means that we have to assume that there is no association in the survey population. The value of the measures of association can be zero here.

Directional Measures

Value Asymp. Std. Error(a) Approx. T(b) Approx. Sig.

Nominal by Nominal Lambda Symmetric ,021 ,014 1,472 ,141

Which paper do you read? Dependent ,000 ,000 .(c) .(c)

Are you male of female? Dependent ,053 ,035 1,472 ,141

Goodman and Kruskal tau Which paper do you read? Dependent ,001 ,001
,370(d)

Are you male of female? Dependent ,006 ,004
,335(d)

a Not assuming the null hypothesis.
b Using the asymptotic standard error assuming the null hypothesis.
c Cannot be computed because the asymptotic standard error equals zero.
d Based on chi-square approximation

Differences between men and women that emerge from the contingency table, e.g. a greater percentage of men read the NRC newspaper than women, are probably characteristic of the sample, but not the population since the chi-square test is not significant.

v7 Which paper do you read? * v1 Are you male of female? Crosstabulation

v1 Are you male of female? Total

0 vrouw 1 man

v7 Which paper do you read? 1 nrchandelsblad Count 75 91 166

Expected Count 83,1 82,9 166,0

Std. Residual -,9 ,9

2 volkskrant Count 110 112 222

Expected Count 111,1 110,9 222,0

Std. Residual -,1 ,1

3 telegraaf Count 111 108 219

Expected Count 109,6 109,4 219,0

Std. Residual ,1 -,1

4 algemeendagblad Count 53 60 113

Expected Count 56,5 56,5 113,0

Std. Residual -,5 ,5

5 trouw Count 12 24 36

Expected Count 18,0 18,0 36,0

Std. Residual -1,4 1,4

6 parool Count 44 37 81

Expected Count 40,5 40,5 81,0

Std. Residual ,5 -,5

7 spits/metro Count 208 178 386

Expected Count 193,1 192,9 386,0

Std. Residual 1,1 -1,1

8 nrc next Count 40 38 78

Expected Count 39,0 39,0 78,0

Std. Residual ,2 -,2

9 anders Count 118 122 240

Expected Count 120,1 119,9 240,0

Std. Residual -,2 ,2

Total Count 771 770 1541

Expected Count 771,0 770,0 1541,0

*Syntax from exercise a.
*Note: first check the data and clean it, if necessary.
FREQUENCIES
  VARIABLES=v1 v7
  /ORDER= ANALYSIS.
*Define Variable Properties.
*v1.
VALUE LABELS v1
  0 'woman'
  1 'man'.
EXECUTE.
*The contingency table with chi-square and the correct measure of association (nominal, asymmetric).
CROSSTABS
  /TABLES=v7 BY v1
  /FORMAT= AVALUE TABLES
  /STATISTIC=CHISQ LAMBDA
  /CELLS= COUNT EXPECTED SRESID
  /COUNT ROUND CELL.

b. Check whether there is an association between preferred newspaper and frequency watching the news; if so, discuss the nature and strength of the association.

We have a nominal variable (v7 - newspaper) and an ordinal one (v8 - frequency of watching the news on TV); therefore we have to use a contingency table again and measures of association at a nominal level of measurement. Currently, it is difficult to see an asymmetric association; therefore we select Cramer's V as the measure of association.

We find a significant association, X₂ (40) = 77.67, p < .001, but it appears that there are many cells in the contingency table with too few expected observations: 24.0% have less than five expected observations. There are also some cells with an expected value of under 1. For this reason, we cannot use the chi-square test.

One alternative is to reduce the number of cells. The easiest thing to do would be to reduce the number of categories in the 'frequency of watching the news' variable, because adjacent values resemble each other here. Moreover, some of the categories occur very infrequently overall (e.g. 'never' occurs just 11 times); therefore, here we find the expected values that are too low.

The categories for watching TV news can be grouped together in different ways. In this case, we take the categories 'never', 'seldom', and 'sometimes' together.

We now see a significant association, X² (24) = 58.44, p < .001, and all of the cells have the minimum number of expected observations. We can now therefore rely on the chi-square test. However, the association is weak (Cramer's V = 0.11). Several notable differences include that relatively large percentage of Trouw readers (36.1%) watch the television news very often, whereas many Spits/Metro readers watch it very infrequently (31.1%).

* Syntax
FREQUENCIES
  VARIABLES=v8
  /ORDER= ANALYSIS.
* Contingency table with measures of association.
CROSSTABS
  /TABLES=v7 BY v8
  /FORMAT= AVALUE TABLES
  /STATISTIC=CHISQ PHI
  /CELLS= COUNT EXPECTED SRESID
  /COUNT ROUND CELL.
* Merge v8 categories.
RECODE
  v8
  (1 thru 3=3) (ELSE=Copy) INTO v8re.
VARIABLE LABELS v8re 'watching TV news (4 values)'.
EXECUTE.

* Define Variable Properties.
* v8re.
VALUE LABELS v8re
  3 'never - sometimes'
  4 'fairly often'
  5 'often'
  6 'very often'.
EXECUTE.
*New contingency table with measures of association.
CROSSTABS
  /TABLES=v7 BY v8re
  /FORMAT= AVALUE TABLES
  /STATISTIC=CHISQ PHI
  /CELLS= COUNT EXPECTED SRESID
  /COUNT ROUND CELL.

v7 Which newspaper do you read? * v8 How often do you watch the news on television? Cross tabulation

v8 How often do you watch the news on television? Total

1 never 2 seldom 3 sometimes 4 fairly often 5 often 6 very often

v7 Which news-paper do you read? 1 nrchandelsblad Count 0 11 18 38 57 41 165

Expected Count 1,2 7,3 28,6 38,0 51,3 38,7 165,0

Std. Residual -1,1 1,4 -2,0 ,0 ,8 ,4

2 volkskrant Count 1 11 30 42 73 65 222

Expected Count 1,6 9,8 38,5 51,1 69,0 52,0 222,0

Std. Residual -,5 ,4 -1,4 -1,3 ,5 1,8

3 telegraaf Count 0 6 35 64 58 55 218

Expected Count 1,6 9,6 37,8 50,2 67,8 51,1 218,0

Std. Residual -1,2 -1,2 -,5 1,9 -1,2 ,6

4 algemeendagblad Count 1 4 12 28 43 25 113

Expected Count ,8 5,0 19,6 26,0 35,1 26,5 113,0

Std. Residual ,2 -,4 -1,7 ,4 1,3 -,3

5 trouw Count 0 2 6 5 10 13 36

Expected Count ,3 1,6 6,2 8,3 11,2 8,4 36,0

Std. Residual -,5 ,3 -,1 -1,1 -,4 1,6

6 parool Count 1 0 7 21 33 21 83

Expected Count ,6 3,7 14,4 19,1 25,8 19,4 83,0

Std. Residual ,5 -1,9 -1,9 ,4 1,4 ,4

7 spits/metro Count 5 19 96 92 108 66 386

Expected Count 2,8 17,0 66,9 88,9 120,0 90,4 386,0

Std. Residual 1,4 ,5 3,6 ,3 -1,1 -2,6

8 nrc next Count 0 7 12 20 25 14 78

Expected Count ,6 3,4 13,5 18,0 24,2 18,3 78,0

Std. Residual -,7 1,9 -,4 ,5 ,2 -1,0

9 Other Count 3 8 51 45 72 61 240

Expected Count 1,7 10,6 41,6 55,3 74,6 56,2 240,0

Std. Residual 1,0 -,8 1,5 -1,4 -,3 ,6

Total Count 11 68 267 355 479 361 1541

Expected Count 11,0 68,0 267,0 355,0 479,0 361,0 1541,0

SPSS exercise 4.5