Results

Contingency Tables

Task 8: To conduct a chi-squared test, drag the variable employer to Rows, score to Columns, and frequency to Counts.

Task 9: Compute and interpret the odds ratio for Task 8.

Contingency Tables
		score
employer		Yes	No	Total
Sage Publications	Count	5.000	19.000	24.000
	Expected count	8.727	15.273	24.000
	% within row	20.833 %	79.167 %	100.000 %
	% within column	17.857 %	38.776 %	31.169 %
	% of total	6.494 %	24.675 %	31.169 %
	Pearson residuals	-1.262	0.954
University of Sussex	Count	23.000	30.000	53.000
	Expected count	19.273	33.727	53.000
	% within row	43.396 %	56.604 %	100.000 %
	% within column	82.143 %	61.224 %	68.831 %
	% of total	29.870 %	38.961 %	68.831 %
	Pearson residuals	0.849	-0.642
Total	Count	28.000	49.000	77.000
	Expected count	28.000	49.000	77.000
	% within row	36.364 %	63.636 %	100.000 %
	% within column	100.000 %	100.000 %	100.000 %
	% of total	36.364 %	63.636 %	100.000 %

The Contingency Table contain the number of cases that fall into each combination of categories. We can see that in total 28 people scored goals and of these 5 were from Sage Publications and 23 were from Sussex; 49 people didn’t score at all (63.6% of the total) and, of those, 19 worked for Sage (38.8% of the total that didn’t score) and 30 were from Sussex (61.2% of the total that didn’t score).

Before moving on to look at the test statistic itself we check that the assumption for chi-square has been met. The assumption is that in 2 × 2 tables (which is what we have here), all expected frequencies should be greater than 5. The smallest expected count is 8.7 (for Sage editors who scored). This value exceeds 5 and so the assumption has been met.

Chi-Squared Tests
	Value	df	p
Χ²	3.634	1	0.057
N	77

Odds Ratio
		95% Confidence Intervals
	Odds Ratio	Lower	Upper	p
Odds ratio	0.343	0.111	1.057
Fisher's exact test	0.348	0.088	1.158	0.075

Task 8: Pearson’s chi-square test examines whether there is an association between two categorical variables (in this case the job and whether the person scored or not). The value of the chi-square statistic is 3.63. This value has a two-tailed significance of p = .057, which is bigger than .05 (hence, non-significant). Because we made a specific prediction (that Sussex people would score more than Sage people), there is a case to be made that we can halve this p-value, which would give us a significant association (because p = .0285, which is less than .05). However, as explained in the book, I’m not a big fan of one-tailed tests. Also, bear i mind that this should definitely not be done after seeing the data, if you want to do honest science. In any case, we’d be well-advised to look for other information such as an effect size. Which brings us neatly onto the next task …

Task 9: The odds of someone scoring given that they were employed by SAGE are:

$ $

The odds of someone scoring given that they were employed by Sussex are:

$ $

Therefore, the odds ratio is:

$ $

The odds of scoring if you work for Sage are 0.34 times as high as if you work for Sussex; another way to express this is that if you work for Sussex, the odds of scoring are 1/0.34 = 2.95 times higher than if you work for Sage.

Write it up!

There was a non-significant association between the type of job and whether or not a person scored a goal, $ $ (1) = 3.63, p = .057, OR = 2.95. Despite the non-significant result, the odds of Sussex employees scoring were 2.95 times higher than that for Sage employees.