# Inferential Statistics 3: The Chi Square Test

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Inferential Statistics 3: The Chi Square Test
Advanced Higher Geography Statistics Inferential Statistics 3: The Chi Square Test

Introduction (1) We often have occasions to make comparisons between two characteristics of something to see if they are linked or related to each other. One way to do this is to work out what we would expect to find if there was no relationship between them (the usual null hypothesis) and what we actually observe.

Introduction (2) The test we use to measure the differences between what is observed and what is expected according to an assumed hypothesis is called the chi-square test.

For Example Some null hypotheses may be:
‘there is no relationship between the height of the land and the vegetation cover’. ‘there is no difference in the location of superstores and small grocers shops’ ‘there is no connection between the size of farm and the type of farm’

Important The chi square test can only be used on data that has the following characteristics: The frequency data mush have a precise numerical value and must be organised into categories or groups. The data must be in the form of frequencies The expected frequency in any one cell of the table must be greater than 5. The total number of observations must be greater than 20.

Formula χ 2 = ∑ (O – E)2 E χ2 = The value of chi square
O = The observed value E = The expected value ∑ (O – E)2 = all the values of (O – E) squared then added together

Worked Example Write down the NULL HYPOTHESIS and ALTERNATIVE HYPOTHESIS and set the LEVEL OF SIGNIFICANCE. NH ‘ there is no difference in the distribution of old established industries and food processing industries in the postal district of Leicester’ AH ‘There is a difference in the distribution of old established industries and food processing industries in the postal district of Leicester’ We will set the level of significance at 0.05.

Table Time! Observed Frequencies (O)
Construct a table with the information you have observed or obtained. Observed Frequencies (O) Post Codes LE1 LE2 LE3 LE4 LE5&LE6 Row Total Old Industry 9 13 10 8 50 Food Industry 4 3 5 21 42 Column Total 16 15 19 29 92 (Note: that although there are 3 cells in the table that are not greater than 5, these are observed frequencies. It is only the expected frequencies that have to be greater than 5.)

Now Work out the expected frequency.
Expected frequency = row total x column total Grand total Eg: expected frequency for old industry in LE1 = (50 x 13) / 92 = 7.07 Post Codes LE1 LE2 LE3 LE4 LE5&LE6 Row Total Old Industry 7.07 Food Industry Column Total

Check your answers Post Codes LE1 LE2 LE3 LE4 LE5&LE6 Row Total
Old Industry 7.07 8.70 8.15 10.33 15.76 50 Food Industry 5.93 7.30 6.85 8.67 13.24 42 Column Total 13 16 15 19 29 92

Now (O – E)2 E For each of the cells calculate.
Eg: Old industry in LE1 is (9 – 7.07)2 / 7.07 = 0.53 Post Codes LE1 LE2 LE3 LE4 LE5&LE6 Row Total Old Industry 0.53 Food Industry Column Total

Old Industry 0.53 2.13 0.42 0.01 3.82 Food Industry 0.63 2.54 0.50 4.55 Then Add up all of the above numbers to obtain the value for chi square: χ2 =

Finally Look up the significance tables. These will tell you whether to accept the null hypothesis or reject it. The number of degrees of freedom to use is: the number of rows in the table minus 1, multiplied by the number of columns minus 1. This is (2-1) x (5-1) = 1 x 4 = 4 degrees of freedom. We find that our answer of is greater than the critical value of 9.49 (for 4 degrees of freedom and a significance level of 0.05) and so we reject the null hypothesis.

Now you have to look for geographical factors to explain your findings
In other words ‘The distribution of old established industry and food processing industries in Leicester is significantly different.’ The hard bit! Now you have to look for geographical factors to explain your findings