Question

In the late summer of 2008, a brief war broke out between the two countries—Russia and Georgia. Suppose you are a researcher interested in nationalistic attitudes in these two countries. You decided to use data from the World Values Survey, which is available at the following URL: http://www.worldvaluessurvey.org/. The data of interest are presented below on Russian respondents.

	Age
Proud of Nationality?	15–29	30–49	50+
Very proud	738	132	279
Quite proud	940	227	388
Not very proud	523	133	238
Not proud at all	194	47	83

Exactly 940 persons between the ages of 15 and 29 indicated that they were “quite proud” of their nationality. Suppose for the moment that only 840 persons between the ages of 15 and 29 were “quite proud” of their nationality. Keeping the other cells as they were originally, recalculate the value of the chi-square statistic and determine whether it is statistically significant.

Answer 1

Replace 940 by 840

The formula of Chi square test statistics is,

Where O - observed frequencies which are given in the table

E - Expected frequencies

The formula to find the expected frequency is,

To find the expected values, find the row, column and grand(overall total)

Expected value for 738, the row total is 1149, column total 2295 and the overall total is 3822

Similarly the expected values for the remaining

Expected value for 132

Expected value for 279

Expected value for 840

Expected value for 227,

Expected value for 388

Expected value for 523,

Expected value for 133,

Expected value for 238,

Expected value for 194,

Expected value for 47,

Expected value for 83,

The table for calculation of chi-square test statistics is,

O	E
738	689.9411	2309.655	3.347612
132	162.0385	902.3092	5.568488
279	297.0202	324.7351	1.093309
840	873.6852	1134.696	1.298746
227	205.1923	475.5754	2.317706
388	376.1224	141.0762	0.375081
523	536.821	191.021	0.355837
133	126.0769	47.92899	0.380157
238	231.102	47.58184	0.205891
194	194.5526	0.305356	0.00157
47	45.69231	1.710059	0.037426
83	83.7551	0.570179	0.006808
Total			14.98863

Chi-square test statistics = 14.989

Chi - square critical value

Degrees of freedom = (number of row categories - 1) * (number of column categories - 1) = (4-1) * (3-1) = 3 * 2 = 6

Alpha(level of significance) = 0.05

The chi-square critical value for DF 6 with alpha 0.05 is 12.592

Chi-square critical value = 12.592

Decision: Critical value is < test statistics value, so reject the null hypothesis.

Reject the null hypothesis, that is there is a statistical significant result.