What goes in a results section? Raw data Measure of central tendency Measure of dispersion Graph Statistical test

Our raw data for today: A student is interested to see whether finding reality TV programmes entertaining is related to being either male or female. Her results are shown in the table below This is called a contingency table What is the level of measurement? Men Women Total Finds reality TV entertaining 19 35 54 Does not find reality TV entertaining 41 25 66 60 120

Data analysis Show the results of the contingency as a bar chart What percentage of men and women find reality TV entertaining?

Would be used: Test of difference/association Nominal data Chi-Square Would be used: Test of difference/association Nominal data

Chi-Square and the Null Hypothesis The chi-squared test tests the null hypothesis. In this example, the null is that there’s no association between finding reality TV entertaining and being male or female – this is shown by the ‘expected frequency’. If the null hypothesis is true, the expected frequency would show that equal numbers of men and women find reality TV entertaining, and equal amounts do not.

Step 1: Expected frequency The results are shown in the table below (expected frequency) The expected frequencies are worked out using the following formula E = row total x column total overall total Men Women Total Finds reality TV entertaining 19 (27) 35 54 Does not find reality TV entertaining 41 (33) 25 66 60 120

Step 2 The Chi-square is calculated using this formula: O = observed frequency (the score in the table) E = expected frequency (what you would expect based on the null hypothesis and your sample size)

Step 2 The Chi-square is calculated using this formula: So for each pair of observed and expected frequencies you subtract the expected score away from the observed score, square it, and then divide it by the expected score

Step 2 Men Women Total Finds reality TV entertaining 19 (27) 35 54 Does not find reality TV entertaining 41 (33) 25 66 60 120 Finds reality TV entertaining Men Women Does not find reality TV entertaining

Step 2 Men Women Total Finds reality TV entertaining 19 (27) 35 54 Does not find reality TV entertaining 41 (33) 25 66 60 120 Finds reality TV entertaining Men (19-27)2 27 = 2.37 Women (35-27)2 Does not find reality TV entertaining (41-33)2 33 = 1.94 (25-33)2

Step 3 Then add up all of your answers to give your observed value 2.37 + 2.37 + 1.94 + 1.94 8.62 Finds reality TV entertaining Men (19-27)2 27 = 2.37 Women (35-27)2 Does not find reality TV entertaining (41-33)2 33 = 1.94 (25-33)2

Step 4 Before you can compare your observed value to your critical value you need one more thing… An extra thing for Chi-squared, rather than knowing N (number of ppts) you have to know df (degrees of freedom)

Step 4 Degrees of freedom (df) = (No. of rows - 1) x (No. of columns – 1) In this example (and in the exam) : df = (2-1) x (2-1) = 1 x 1 = 1 Df=1 Men Women Total Finds reality TV entertaining 19 35 54 Does not find reality TV entertaining 41 25 66 60 120

Step 5 Compare your observed value to the critical value to find out whether your results are statistically significant (and if you can reject the null hypothesis!) Level of significance for two-tailed test df 0.20 0.10 0.05 0.02 1 1.64 2.71 3.84 5.41 2 3.22 4.60 5.99 7.82 3 4.64 6.25 9.84 4 7.78 9.49 11.67 5 7.29 9.24 11.07 13.39 6 8.56 10.64 12.59 15.03

Step 5 Compare your observed value to the critical value to find out whether your results are statistically significant (and if you can reject the null hypothesis!) Level of significance for two-tailed test df 0.20 0.10 0.05 0.02 1 1.64 2.71 3.84 5.41 2 3.22 4.60 5.99 7.82 3 4.64 6.25 9.84 4 7.78 9.49 11.67 5 7.29 9.24 11.07 13.39 6 8.56 10.64 12.59 15.03

Step 5 Observed value = 8.62 Critical value = 3.84 CHI-SQUARE= Observed > Critical to be significant (if there’s an R in the name, the observed needs to be GREATER than the critical)

Step 6 How to write this up: The results are significant suggesting there is an association between ?? The observed value (8.62) is greater than the critical value (3.84) at p<0.05, two-tailed, df=1. Therefore, we can reject the null hypothesis.

In the exam You will NEVER have to calculate the test But you do need to be able to use the observed value to decide whether the result is significant, and write a concluding statement. You also need to be able to draw a contingency table. Let’s try 

Exam Practice June 2011 Psychological research suggests an association between birth order and certain abilities. For example, first-born children are often logical in their thinking whereas later-born children tend to be more creative. A psychologist wonders whether this might mean that birth order is associated with different career choices. She decides to investigate and asks 50 artists and 65 lawyers whether they were the first-born child in the family or not. (a) Write a non-directional hypothesis for this study (2) (b) Identify an appropriate sampling method for this study and explain how the psychologist might have obtained such a sample. (3)

Exam Practice The psychologist found the following results: •    20 of the 50 artists were first-born children •    35 of the 65 lawyers were first-born children. She analysed her data using a statistical test and calculated a value of x2 = 2.27. She then looked at the relevant table to see whether this value was statistically significant. An extract from the table is provided below. (c) Imagine that you are writing the results section of the report on this investigation. Using information from the description of the study above and the relevant information from the statistical table, provide contents suitable for the results section. (12) You must provide all of the following: an appropriately labelled contingency table a sketch of an appropriately labelled bar chart identification of the appropriate statistical test with justification for its use identification of an appropriate significance level a statement of the results of the statistical test in relation to the hypothesis. (Total 17 marks)

Quick Quiz Name two reasons why you would use a Chi-square test Does the observed value have to greater, or less than, the critical value to be significant?

Using a spread sheet to do the hard work for you