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.

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.

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

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

2 = The value of chi squareO = The observed valueE = The expected value (O E)2 = all the values of (O E) squared then added together

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.

Construct a table with the information you have observed or obtained.Observed Frequencies (O)(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.)

Post CodesLE1LE2LE3LE4LE5 & LE6Row TotalOld Industry9131010850Food Industry43592142Column Total131615192992

Work out the expected frequency.Eg: expected frequency for old industry in LE1 = (50 x 13) / 92 = 7.07

Post CodesLE1LE2LE3LE4LE5 & LE6Row TotalOld Industry7.07Food IndustryColumn Total

Post CodesLE1LE2LE3LE4LE5 & LE6Row TotalOld Industry7.078.708.1510.3315.7650Food Industry5.937.306.858.6713.2442Column Total131615192992

For each of the cells calculate.Eg: Old industry in LE1 is (9 7.07)2 / 7.07 = 0.53

Post CodesLE1LE2LE3LE4LE5 & LE6Row TotalOld Industry0.53Food IndustryColumn Total

Add up all of the above numbers to obtain the value for chi square: 2 = 15.14.

Post CodesLE1LE2LE3LE4LE5 &L E6Old Industry0.532.130.420.013.82Food Industry0.632.540.500.014.55

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 15.14 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.

The distribution of old established industry and food processing industries in Leicester is significantly different.Now you have to look for geographical factors to explain your findings

The Vice president (sales) of a garment company wants to determine whether sales of the companys brand of jeans is independent of age group. He has appointed a marketing researcher for this purpose. The marketing researcher has taken a random sample of 703 customers who have purchased jeans. The researcher conducted survey for three brands of the jeans, namely Brand 1, Brand 2, and Brand 3. The researcher has also divided the age groups in to four categories: 15 to 26, 26 to 35,36 to 45 and 46 to 55. The observation of the researcher are provided

Determine whether brand preference is independent of age group. Use =0.05

Brand

Age Brand 1Brand 2Brand 3Row total15 to 2565757221226 to 3560406416436 to 4545525014746 to 55556560180Column Total225232246703

A firm is interested in knowing whether preference for its three brands : Brand 1, Brand 2, and Brand 3 is independent of type of occupation: government job, private job and own business. Data collected from the consumers are given in the following table. Use =0.05 to test whether brand preference is independent of type of occupation.

Brand

Type of occupationBrand 1Brand 2Brand 3Government Job 788790Private Job110120125Own Business111123127