AI HL · 13.4
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Combining categories in a chi-squared test
13.4.8
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A library records what 149 visitors borrowed and their age group. A chi-squared test for independence is to be used to see whether the item borrowed is linked to age. (a) Find the expected frequencies. (b) Combine columns so that every expected frequency is greater than 5, and give the new table. (c) Test, at the 5% significance level, whether the item borrowed and the age group are independent.
Have a go first. Then open the working one line at a time.
From the calculator: Fiction: 3.76, 42.8, 23.5; Non-fiction: 2.42, 27.5, 15.1; Magazines: 1.83, 20.8, 11.4
Expected frequencies come from the calculator's stored matrix. Row total × column total ÷ grand total gives each one.
Under 12: 3.76, 2.42, 1.83. All are 5 or less.
The whole Under 12 column is below 5. A chi-squared test needs every expected frequency greater than 5.
Under 12 + 12 to 15 = Under 16. New table: Fiction: 56, 14; Non-fiction: 24, 21; Magazines: 19, 15
Combine neighbouring age groups. Under 12 is next to 12 to 15, so they form one group. Add the observed counts.
New expected frequencies: Fiction: 46.5, 23.5; Non-fiction: 29.9, 15.1; Magazines: 22.6, 11.4. The smallest is 11.4.
Every expected frequency is now greater than 5, so the test can go ahead.
H0: item borrowed and age group are independent. H1: they are not independent. df = (3 − 1)(2 − 1) = 2
The degrees of freedom use the new, smaller table.
From the calculator: χ² = 10.9, p = 0.00421
Enter the combined table as the observed matrix.
0.00421 < 0.05, so reject H0. There is sufficient evidence, at the 5% significance level, that the item borrowed is not independent of age group.
Compare p with 0.05 and answer in context.
A clinic records how long 241 patients waited and whether they came to the morning or the evening clinic. The results are shown. A chi-squared test for independence is used to test whether waiting time is linked to the clinic. (a) Find the expected frequencies and say which rows must be combined. (b) Find the degrees of freedom after combining. (c) Carry out the test at the 5% significance level.
Have a go first. Then open the working one line at a time.
From the calculator: Under 10 min: 53.4, 49.6; 10 to 20 min: 2.59, 2.41; 20 to 30 min: 4.67, 4.33; Over 30 min: 64.3, 59.7
Read the whole expected matrix before you decide what to combine.
10 to 20 min (2.59, 2.41) and 20 to 30 min (4.67, 4.33) are below 5. Combine them: 10 to 30 min.
Both small rows are next to each other, so they join to make one natural group. Justify the choice by naming the rows and their values.
New table: Under 10 min: 54, 49; 10 to 30 min: 6, 8; Over 30 min: 65, 59. Smallest expected frequency: 6.74
Add the observed counts, then recalculate the expected values. All are greater than 5.
3 rows, 2 columns: df = (3 − 1)(2 − 1) = 2
Before combining it would have been (4 − 1)(2 − 1) = 3. Use the smaller table.
H0: waiting time and clinic are independent. H1: they are not independent. From the calculator: χ² = 0.483, p = 0.785
State the hypotheses in context, then use the calculator on the combined table.
0.785 > 0.05, so do not reject H0. There is insufficient evidence, at the 5% significance level, that waiting time is not independent of clinic.
A large p-value means the data are consistent with independence. It does not prove independence.
234 students were asked how often they buy lunch at school. The results are shown. Test, at the 5% significance level, whether the answer is independent of the year group. Combine categories first if you need to, and say why.
Have a go first. Then open the working one line at a time.
From the calculator: Year 10: 5.21, 6.78, 95.4, 14.6; Year 11: 4.79, 6.22, 87.6, 13.4
One expected frequency is 5 or less, in the Never column.
Never: 5.21 and 4.79. Rarely: 6.78 and 6.22. Combine Never and Rarely: Never or rarely.
The Never column has an expected frequency of 5 or less. It is next to Rarely and means almost the same thing, so combine them. Most days and Every day are already greater than 5 and stay as they are.
New table: Year 10: 11, 95, 16; Year 11: 12, 88, 12. Smallest expected: 11.0
Every expected frequency is greater than 5.
H0: how often a student buys lunch and year group are independent. H1: they are not independent. df = (2 − 1)(3 − 1) = 2. From the calculator: χ² = 0.456, p = 0.796
The calculator gives df too. Check it matches the new table.
0.796 > 0.05, so do not reject H0. There is insufficient evidence, at the 5% significance level, that how often a student buys lunch is not independent of year group.
Do not reject H0. Say what that means in context.